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‎Continuous Probability Distributions

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2013-09-20T14:20:20Z

‎Outlook

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2013-08-13T11:32:50Z

Rb: /* Probability Density Functions (pdf) */

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‎Probability Density Functions (pdf)

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Rb: /* Properties of the pdf */

2013-08-13T10:26:49Z

‎Properties of the pdf

Rb: Created page with " = Continuous Probability Distributions = In the Discrete random variables Section, we introduced the notion of a ''random variable'' and, in part..."

2013-08-13T10:22:48Z

Created page with " = Continuous Probability Distributions = In the Discrete random variables Section, we introduced the notion of a ''random variable'' and, in part..."

New page

= Continuous Probability Distributions =

In the [[Probability_DiscreteRV|Discrete random variables]] Section, we introduced the notion of a ''random variable'' and, in particular a ''discrete'' random variable. It was then discussed how to use mathematical functions in order to assign probabilities to the various possible numerical values of such a random variable. A probability distribution is a method by which such probabilities can be assigned and in the discrete case this can be achieved via a probability mass function (''pmf''). The following Figure illustrates the ''pmf'' for a particular ''binomial random variable''.

[[File:ProbDistCont_binom.jpg|frameless|400px]]

Notice how ''masses'' of probability are dropped onto the possible discrete (isolated) outcomes. You should also recall that, if you added up the probabilities for all possible outcomes you would obtain 1. We now develop mathematical functions which can used to describe probability distributions associated with a ''continuous random variable''.

'''HEALTH WARNING: Before reading this section you MUST revise your undertanding of integration.'''

== Additional Resources ==

Khan Academy:

* If you need an introduction to integration you could do worse than watching this clip. [https://www.khanacademy.org/math/calculus/integral-calculus/indefinite_integrals/v/antiderivatives-and-indefinite-integrals]. But a note of caution in terms of language. Salman Khan often uses the term "anti-derivative" for what we often call "integrating". If you need more details on tegration you may want to continue the trail of clips following the above intrductory video.

= Continuous random variables =

Recall that a random variable is a function applied on a sample space, by which we mean that physical attributes of a sample space are mapped (by this function) into a number. When a ''continuous'' random variable is applied on a sample space, a range of possible numbers is implied (not just isolated numbers as with a discrete random variable).

As an example, let <math>X=</math> ‘''the contents of a reservoir''’, where the appropriate sample space under consideration allows for the reservoir being just about ''empty'', just about ''full'' or ''somewhere in between''. Here we might usefully define the range of possible values for <math>X</math> as <math>0<x<1,</math> with <math>0</math> signifying ‘empty’ and <math>1</math> signifying ‘full’. As noted before when talking about the characteristics of continuous variables, theoretically, we can’t even begin to list possible numerical outcomes for <math>X</math>; any value in the interval is possible.

How do we thus distribute probability in this case? Well, presumably the probability must be distributed only over the range of possible values for <math>X</math>; which, in the reservoir example, is over the unit interval <math>\left(0,1\right)</math>. However, unlike the discrete case where a specific ''mass'' of probability is dropped on each of the discrete outcomes, for continuous random variables probability is distributed more smoothly, rather like brushing paint on a wall, over the ''whole interval'' of defined possible outcomes. Therefore in some areas the distribution of probability is quite thick and others it can be relatively thin. This is depicted in this Figure.

[[File:ProbDistCont_density1.jpg|frameless|400px]]

Such a function will later be called a ''probability density function'' (''pdf''). Some thought should convince you that for a ''continuous'' random variable, <math>X</math>, it must be the case that <math>\Pr (X=c)=0</math> for all real numbers <math>c</math> contained in the range of possible outcomes of <math>X</math>. If this were not the case, then the axioms of probability would be violated. However, there should be a positive probability of <math>X</math> being close, or in the ''neighbourhood'', of <math>c</math>. (A neighbourhood of <math>c</math> might be <math>c\pm 0.01</math>, say.) For example, although the probability that the reservoir is ''exactly'' <math>90\%</math> full must be ''zero ''(who can measure <math>90\%</math> exactly?), the axioms of probability require that there is non-zero probability of the reservoir being between, say, <math>85\%</math> and <math>95\%</math> full. (Indeed, being able to make judgements like this is an eminently reasonable requirement if we wish to investigate, say, the likelihood of water shortages.) This emphasizes that we can not think of probability being assigned at specific points (numbers), rather probability is ''distributed'' over intervals of numbers.

We therefore must confine our attention to assigning probabilities of the form <math>\Pr (a<X\leq b)</math>, for some real numbers <math>a<b</math>; i.e., what is the probability that <math>X</math> takes on values between <math>a</math> and <math>b</math>. For an appropriately defined mathematical function describing how probability is distributed, as depicted in the above Figure, this would be the area under that function between the values <math>a</math> and <math>b</math>. Such functions can be constructed and are called ''probability density functions''. Let us emphasize the role of ''area'' again: it is the ''area'' under the probability density function which provides probability, ''not'' the probability density function itself. Thus, by the axioms of probability, if <math>a</math> and <math>b</math> are in the range of possible values for the continuous random variable, <math>X</math>, then <math>\Pr \left( a<X\leq b\right) </math> must always return a positive number, lying between <math>0</math> and <math>1</math>, ''no matter how close'' <math>b</math> is to <math>a</math> (provided only that <math>b>a</math>).

What sorts of mathematical functions can usefully serve as probability density functions? To develop the answer to this question, we begin by considering another question: ''what mathematical functions would be appropriate as cumulative distribution functions''?

= Cumulative distribution function (''cdf'') =

For a continuous random variable, <math>X</math>, the ''cdf'' is a ''smooth'' continuous function defined as <math>F(x)=\Pr (X\leq x)</math>, for ''all'' real numbers <math>x</math>; e.g., <math>F(0.75)=\Pr (X\leq 0.75)</math>. Therefore, the type of probabilities we are looking at now are a special case of the interval probabilities we discussed previously, as <math>F(0.75)=\Pr (X\leq 0.75)=Pr(-\infty<X\leq x)</math>. But looking at this special case will make our job somewhat easier for starters. The following should be observed:

* such a function is defined for all real numbers <math>x</math>, not just those which are possible realisations of the random variable <math>X</math>;
* we use <math>F(.)</math>, rather than <math>P(.)</math>, to distinguish the cases of ''continuous'' and ''discrete'' random variables, respectively.

Let us now establish the ''mathematical properties'' of such a function. We can do this quite simply by making <math>F(.)</math> adhere to the axioms of probability. Firstly, since <math>F(x)</math> is to be used to return probabilities, it must be that

<math>0\leq F(x)\leq 1,\quad \text{for all }x.</math>

Secondly, it must be a smooth, increasing function of <math>x</math> (over intervals where possible values of <math>X</math> can occur). To see this, consider again the reservoir example and any arbitrary numbers <math>a</math> and <math>b</math>, satisfying <math>0<a<b<1. </math> Notice that <math>a<b;</math> <math>b</math> can be as a close as you like to <math>a</math>, but it must always be strictly greater than <math>a</math>. Therefore, the axioms of probability imply that <math>\Pr \left( a<X\leq b\right) >0</math>, since the event ‘<math>a<X\leq b</math>’ is possible. Now divide the real line interval <math>(0,b]</math> into two mutually exclusive intervals, <math>(0,a]</math> and <math>(a,b].</math> Then we can write the event ‘<math>X\leq b</math>’ as

<math>\left( X\leq b\right) =\left( X\leq a\right) \cup \left( a<X\leq b\right) .</math>

Assigning probabilities on the left and right, and using the axiom of probability concerning the allocation of probability to mutually exclusive events, yields

<math>\Pr \left( X\leq b\right) =\Pr \left( X\leq a\right) +\Pr \left( a<X\leq b\right)</math>

or

<math>\Pr \left( X\leq b\right) -\Pr \left( X\leq a\right) =\Pr \left( a<X\leq b\right) .</math>

Now, since <math>F(b)=\Pr \left( X\leq b\right) </math> and <math>F\left( a\right) =\Pr \left( X\leq a\right) </math>, we can write

<math>F(b)-F(a)=\Pr \left( a<X\leq b\right) >0.</math>

Thus <math>F(b)-F(a)>0</math>, for all real numbers <math>a</math> and <math>b</math> such that <math>b>a</math>, no matter how close. You should note that we have now reconstructed the interval probability which we discussed in the previous Section, <math>\Pr \left( a<X\leq b\right)</math>, as a function of what we now call a cumulative density function. We also previously discussed that <math>\Pr \left( a<X\leq b\right) > 0</math> on the range on which <math>X</math> is defined (i.e. between 0 and 1 for the reservoir example) and therefore we know that <math>F(x)</math> must be an increasing function. A little more delicate mathematics shows that it must be a ''smoothly'' increasing function<ref>This is due to the fact that <math>\Pr (X=b)=0</math> which implies that we are not getting any discrete changes from <math>F(b-\epsilon)</math> to <math>F(b)</math>.
</ref>. All in all then, <math>F(x)</math> appears to be smoothly increasing from <math>0</math> to <math>1</math> over the range of possible values for <math>X</math>. In the reservoir example, the very simple function <math>F(x)=x</math> would appear to satisfy these requirements, provided <math>0<x<1</math>.

More generally, we now formally state the properties of a ''cdf''. For complete generality, <math>F(x)</math> must be defined over the whole real line even though in any given application the random variable under consideration may only be defined on an interval of that real line.

== Properties of a cdf ==

A cumulative distribution function is a mathematical function, <math>F(x)</math>, satisfying the following properties:

# <math>0\leq F(x)\leq 1</math>.
# If <math>b>a</math> then <math>F(b)\geq F(a)</math>; i.e., <math>F</math> is increasing. In addition, over all intervals of possible outcomes for a continuous random variable, <math>F(x)</math> is smoothly increasing; i.e., it has no ''sudden jumps''.
# <math>F(x)\rightarrow 0</math> as <math>x\rightarrow -\infty </math>; <math>F(x)\rightarrow 1</math> as <math>x\rightarrow \infty </math>; i.e., <math>F\left( x\right) </math> decreases to <math>0</math> as <math>x</math> falls, and increases to <math>1</math> as <math>x</math> rises.

Any function satisfying the above may be considered suitable for modelling cumulative probabilities, <math>\Pr \left( X\leq x\right) </math>, for a continuous random variable. Careful consideration of these properties reveals that <math>F(x)</math> can be ''flat'' (i.e., non-increasing) over some regions. This is perfectly acceptable since the regions over which <math>F(x)</math> is flat correspond to those where values of <math>X</math> can not occur and. therefore, zero probability is distributed over such regions. In the reservoir example, <math>F(x)=0</math>, for all <math>x\leq 0</math>, and <math>F(x)=1</math>, for all <math>x\geq 1;</math> it is therefore flat over these two regions of the real line. This particular examples also demonstrates that the last of the three properties can be viewed as completely general; for example, the fact that <math>F(x)=0</math>, in this case, for all <math>x\leq 0</math> can be thought of as simply a special case of the requirement that <math>F(x)\rightarrow 0</math> as <math>x\rightarrow -\infty </math>.

Some possible examples of ''cdf''s, in other situations, are depicted in the following Figure.

[[File:ProbDistCont_density2.jpg|frameless|400px]]

The first of these is strictly increasing over the whole real line, indicating possible values of <math>X</math> can fall anywhere. The second is increasing, but only strictly over the interval <math>x>0</math>; this indicates that the range of possible values for the random variable is <math>x>0</math> with the implication that <math>\Pr \left( X\leq 0\right) =0</math>. The third is only strictly increasing over the interval <math>\left( 0<x<2\right) </math>, which gives the range of possible values for <math>X</math> in this case; here <math>\Pr \left( X\leq 0\right) =0</math>, whilst <math>\Pr \left( X\geq 2\right) =0</math>.

In practice we will often be interested in interval probabilities of the type <math>\Pr (a<X\leq b)</math>. Therefore, let us end this discussion by re-iterating how probabilities and the ''cdf'' are related to each other:

* <math>\Pr (a<X\leq b)=F(b)-F(a)</math>, for any real numbers <math>a</math> and <math>b</math>;
* <math>\Pr (X > a)=1-F(a)</math>, since <math>F(a) = Pr(X \leq a)</math> and <math>\Pr \left( X\leq a\right) +\Pr \left( X>a\right) =1</math>, for any real number <math>a</math>;
* <math>\Pr \left( X<a\right) =\Pr \left( X\leq a\right) </math>, since <math>\Pr \left(
X=a\right) =0</math>.

Our discussion of the ''cdf'' was introduced as a means of developing the idea of a ''probability density function'' (''pdf''). Visually the ''pdf'' illustrates how all of the probability is distributed over possible values of the continuous random variable; we used the analogy of paint being brushed over the surface of a wall. The ''pdf'' is also a mathematical function satisfying certain requirements in order that the axioms of probability are not violated. We also pointed out that it would be the ''area'' under that function which yielded probability. Note that this is in contrast to the ''cdf'', <math>F(x)</math>, where the function itself gives probability. We shall now investigate how a ''pdf'' should be defined.

= Probability Density Functions (''pdf'') =

For a continuous random variable, it is well worth reminding ourselves of the following:

* There is no function which gives <math>Pr(X=x)</math> for some number <math>x</math>, since all such probabilities are ''identically zero''.

However, and as discussed in the previous section, there is a smooth, increasing, function <math>F(x)</math>, the ''cdf'', which provides <math>\Pr \left(X\leq x\right)</math>. In particular

<math>\Pr \left( a<X\leq b\right) =F(b)-F(a),</math>

for real numbers <math>b>a</math>. Also, since <math>F(x)</math> is smoothly continuous and differentiable over the range of possible values for <math>X</math> (see the above Figure with different ''cdf''s), then there must exist a function <math>f\left( x\right) =dF(x)/dx</math>, the derivative of <math>F(x)</math>. Note that <math>f(x)</math> must be positive over ranges where <math>F(x)</math> is increasing; i.e., over ranges of possible values for <math>X</math>. On the other hand, <math>f(x)=0</math>, over ranges where <math>F(x)</math> is flat; i.e., over ranges where values of <math>X</math> can ''not'' occur.

Moreover, the ''Fundamental Theorem of Calculus'' (which simply states that ''differentiation'' is the opposite of ''integration'') implies that if <math>f(x)=dF(x)/dx</math>, then

<math>F(b)-F(a)=\int_{a}^{b}f(x)dx.</math>

We therefore have constructed a function <math>f(x)=dF(x)/dx</math>, such that the area under it yields probability (recall that the integral of a function between two specified limits gives the area under that function). Such a function, <math>f\left( x\right) </math>, is the ''probability density function''.

In general, <math>\lim_{a\rightarrow -\infty }F(a)=0</math>, so by letting <math>a\rightarrow -\infty </math> in the above we can define the fundamental relationship between the ''cdf'' and ''pdf'' as:

<math>\begin{aligned}
F(x) &=&\Pr \left( X\leq x\right) =\int_{-\infty }^{x}f(t)dt, \\
f(x) &=&dF(x)/dx;\end{aligned}</math>

i.e., <math>F(x)</math> is the area under the curve <math>f(t)</math> up to the point <math>t=x</math>. Now, letting <math>x\rightarrow \infty </math>, and remembering that <math>\lim_{x\rightarrow \infty }F(x)=1</math>, we find that

<math>\int_{-\infty }^{\infty }f(x)dx=1;</math>

i.e., ''total area under'' <math>f(x)</math> ''must equal'' <math>1</math> (rather like the total area of the sample space as depicted by a Venn Diagram).

These definitions are all quite general so as to accommodate a variety of situations; however, as noted before, implicit in all of the above is that <math>f(x)=0</math> over intervals where no probability is distributed; i.e., where <math>F(x) </math> is flat. Thus, in the reservoir example we could be more explicit with the limits and write

<math>F\left( x\right) =\int_{0}^{x}f(t)dt,\quad 0<x<1,</math>

for some suitable function <math>f(.)</math>, since <math>f(t)=0</math> for all <math>t\leq 0</math>, and all <math>t\geq 1</math>. For example, suppose the contents of the reservoir can be modelled by the continuous random variable which has probability density function

<math>f(x)=\left\{
\begin{array}{cc}
3(1-x)^{2}, & 0\leq x\leq 1 \\
0, & \text{otherwise.}\end{array}
\right.</math>

We can then calculate the probability that the reservoir will be over <math>75\%</math> full as:

<math>\begin{aligned}
\Pr \left( X>0.75\right) &=&\int_{0.75}^{1}3(1-x)^{2}dx \\
&=&-\left[ (1-x)^{3}\right] _{0.75}^{1} \\
&=&\left( \dfrac{1}{4}\right) ^{3} \\
&=&\dfrac{1}{64}.\end{aligned}</math>

The following Figure also gives a simple example of how the ''cdf'' and ''pdf'' are related to each other. Here probability is distributed ''uniformly'' over a finite interval (in this case, it is the ''unit'' interval <math>\left[ 0,1\right] </math>). Such a distribution is therefore said to be uniform.

[[File:ProbDistCont_pdfcdf.jpg|frameless|400px]]

For example,

<math>\Pr \left( 0.25<X\leq 0.5\right) =F\left( 0.5\right) -F\left( 0.25\right) =\frac{1}{2}-\frac{1}{4}=\frac{1}{4}.</math>

Alternatively, using the ''pdf'',

<math>\Pr \left( 0.25<X\leq 0.5\right) =\int_{0.25}^{0.5}f(x)dx=0.25.</math>

Also note that the total area under <math>f(x)</math>, over the unit interval, is clearly equal to <math>1</math>.

To recap, then, let us list the properties of the ''pdf.''

== Properties of the ''pdf'' ==

A ''pdf'' for a continuous random variable is a mathematical function which must satisfy,

# <math>f(x)\geq 0</math>
# <math>\int_{-\infty }^{\infty }f(x)dx=1</math>

Probabilities can be calculated as:

* <math>\Pr (a<X\leq b)=\int_{a}^{b}f(x)dx</math>
** i.e., it is the ''area'' under the ''pdf ''which gives probability

and the relationship with ''cdf'' is given by:

* ** <math>f(x)=dF(x)/dx</math>
** <math>F(x)=\int_{-\infty }^{x}f(t)dt.\bigskip </math>

== Outlook ==

With the theory out of the way, let’s think of examples for continuous distributions. In fact we are spoiled for choice. Check out this [http://en.wikipedia.org/wiki/Category:Continuous_distributions|list] of different continuous distributions from Wikipedia. The ''uniform'' distribution example we just discussed features on this list as well [http://en.wikipedia.org/wiki/Uniform_distribution_(continuous)]. It is the simplest of all examples. It should be noted though that there is not only one ''uniform'' distribution, there is an infinite number of uniform distributions. We chose the ''uniform'' distribution defined on the interval <math>\left[ 0,1\right] </math>. But we could have chosen any interval <math>\left[ a,b\right] </math>. The main properties wouldn’t have changed, but the actual probabilities, e.g. <math>\Pr \left( 0.25<X\leq 0.5\right)</math> would. In the context of distributions we call the <math>a</math> and <math>b</math> ''parameters''. Almost all distributions do have parameters changing them will have different effects.

We shall pick two more examples. Here the ''exponential'' distribution and then the ''normal'' distribution. The latter is so important that we shall dedicated its own [[ProbabilityNorm|page]] to it. We need this large range of different distributions as any random variable will have different properties and we will, in all cases, have to try and find that distribution that has properties that best represent that of the random variable we are interested in. Just briefly consider again the case of the level in our reservoir. Do you think this may be uniformly distributed? Most likely not. We would expect the cases where it is almost empty (<math>X</math> close to 0) or full (<math>X</math> close to 1) much less likely than some values around its ''usual'' level (say somewhere between 0.4 and 0.6). So, a ''uniform'' distribution appears inadequate.

= Exponential Distribution =

Let the continuous random variable, denoted <math>X</math>, monitor the elapsed time, measured in minutes, between successive cars passing a junction on a particular road. Traffic along this road in general flows freely, so that vehicles can travel independently of one another, not restricted by the car in front. Occasionally, there are quite long intervals between successive vehicles, while more often there are smaller intervals. To accommodate this, the following ''pdf'' for <math>X</math> is defined:

<math>f(x)=\left\{
\begin{array}{ll}
\frac{1}{\theta} \exp (-\frac{x}{\theta}), & x>0, \\
0, & x\leq 0.\end{array}
\right.</math>

In this case we have one ''parameter'', <math>\theta</math>. For this initial discussion we shall set that parameter to <math>\lambda = 1</math>, but you should keep in mind that it could be any other positive value.

The graph of <math>f(x)</math>, with <math>\lambda = 1</math>, looks as follows:

[[File:ProbDistCont_exponential.jpg|frameless|400px]]

Now, you might care to verify that <math>\int_{0}^{\infty }\exp (-x)dx=1</math>, and from the diagram it is clear that

<math>\Pr \left( a<X\leq a+1\right) >\Pr \left( a+1<X\leq a+2\right) ,</math>

for any number <math>a>0</math>. By setting <math>a=1</math>, say, this implies that an elapsed time of between <math>1</math> and <math>2</math> minutes has greater probability than an elapsed time of between <math>2</math> and <math>3</math> minutes; and, in turn, this has a greater probability than an elapsed time of between <math>3</math> and <math>4</math> minutes, etc; i.e., smaller intervals between successive vehicles will occur more frequently than longer ones. <ref>What other sort of problem may this distribution be relevant for? Imagine you uploaded a video on YouTube and you were interested in how long any viewer stick with your video clip. This is a random variable and the ''exponential'' distribution is likely to be a good representation of this random variable. Of course, you would have to choose an appropriate value for the parameter .
</ref>

Suppose we are interested in the probability that <math>1<X\leq 3?</math> i.e., the probability that the elapsed time between successive cars passing is somewhere between <math>1</math> and <math>3</math> minutes? To find this, we need

<math>\begin{aligned}
\Pr \left( 1<X\leq 3\right) &=&\int_{1}^{3}\exp (-x)dx \\
&=&\left[ -\exp (-x)\right] _{1}^{3} \\
&=&e^{-1}-e^{-3} \\
&=&0.318.\end{aligned}</math>

One might interpret this as meaning that about <math>32\%</math> of the time, successive vehicles will be between <math>1</math> and <math>3</math> minutes apart.

The above distribution is known as the ''unit'' exponential distribution (as we chose <math>\theta = 1</math>). In general, we say that the continuous random variable <math>X</math> has an ''exponential distribution'' if it has probablity density function given by

<math>f(x)=\frac{1}{\theta }\exp \left( -\frac{x}{\theta }\right) ,\quad x>0;\quad
\theta >0</math>

where <math>\theta </math> is called the parameter of the distribution (sometimes the ''mean'', which is discussed in this [[Probability_MomentsExp|Section]]. Note that when <math>\theta =1</math>, we get back to the special case of the unit exponential distribution. Notice that the random variable, here, can only assume positive values.

Note that

<math>\begin{aligned}
\int_{0}^{\infty }f(x)dx &=&\int_{0}^{\infty }\frac{1}{\theta }\exp \left( -\frac{x}{\theta }\right) dx \\
&=&\left[ -\exp \left( -\frac{x}{\theta }\right) \right] _{0}^{\infty } \\
&=&1\end{aligned}</math>

so that it is a proper probability density function (clearly, also, <math>f\left( x\right) >0</math> for all <math>x>0</math>).

== Revision: Properties of the Exponential Function ==

When working with the exponential distribution it is essential to you remember rules of calculation with the exponential function

# <math>y=e^{x}</math> is a strictly positive and strictly increasing function of <math>x</math>
# <math>\frac{dy}{dx}=e^{x}</math> and <math>\frac{d^{2}y}{dx^{2}}=e^{x}</math>
# <math>\ln y=x</math>
# When <math>x=0</math>, <math>y=1</math>, and <math>y>1</math> when <math>x>0</math>, <math>y<1</math> when <math>x<0</math>
# By the chain rule of differentiation, if <math>y=e^{-x}</math> then <math>\frac{dy}{dx}=-e^{-x}</math>.

== Additional Resources ==

* Wikipedia: [http://en.wikipedia.org/wiki/Exponential_distribution|Exponential Distribution], but note that they use slightly different notation in terms of the parameter of the exponential distribution. They use <math>\lambda</math> which is related to our <math>\theta</math> in the following way: <math>\lambda = 1 / \theta</math>.
* Wolfram MathWorld: [http://mathworld.wolfram.com/ExponentialDistribution.html|Exponential Distribution], but note that they use slightly different notation in terms of the parameter of the exponential distribution. They use <math>\lambda</math> which is related to our <math>\theta</math> in the following way: <math>\lambda = 1 / \theta</math>.
* How to use EXCEL to calculate probabilities from an exponential distribution []

= Exercise 3 =

<ol>
<li>In an experiment, if a mouse is administered dosage level <math>A</math> of a certain (harmless) hormone then there is a <math>0.2</math> probability that the mouse will show signs of aggression within one minute. For dosage levels <math>B</math> and <math>C</math>, the probabilities are <math>0.5</math> and <math>0.8</math>, respectively. Ten mice are given exactly the same dosage level of the hormone and, of these, exactly <math>6</math> shows signs of aggression within one minute of receiving the dose.
<ol>
<li>Calculate the probability of this happening for each of the three dosage levels, <math>A,B</math> and <math>C</math>. (This is essentially a Binomial random variable problem, so you can check your answers using EXCEL.)</li>
<li>Assuming that each of the three dosage levels was equally likely to have been administered in the first place (with a probability of <math>1/3</math>), use Bayes’ Theorem to evaluate the likelihood of each of the dosage levels ''given ''that <math>6</math> out of the <math>10</math> mice were observed to react in this way.</li></ol>
</li>
<li>Let <math>X</math> be the random variable indicating the number of incoming planes every <math>k</math> minutes at a large international airport, with probability mass function given by <math>p(x)=\Pr (X=x)=\frac{(0.9k)^{x}}{x!}\exp (-0.9k),\quad x=0,1,2,3,4,..</math>. . Find the probabilities that there will be
<ol>
<li>exactly <math>9</math> incoming planes during a period of <math>5</math> minutes (i.e., find <math>\Pr (X=9)</math> when <math>k=5)</math>;</li>
<li>fewer than <math>5</math> incoming planes during a period of <math>4</math> minutes (i.e., find <math>\Pr (X<5)</math> when <math>k=4)</math>;</li>
<li>at least <math>4</math> incoming planes during an <math>2</math> minute period (i.e., find <math>\Pr (X\geq 4)</math> when <math>k=2)</math>.</li></ol>

Check all your answers using EXCEL.</li>
<li>The random variable <math>Y</math> is said to be ''Geometric'' if it has probability mass function given by<math>p(y)=\Pr (Y=y)=(1-\theta )\theta ^{y-1},\quad y=1,2,3,...;\quad 0<\theta
<1;\medskip </math>where <math>\theta </math> is an unknown ‘parameter’.Show that the cumulative distribution function can be expressed as
<math>P(y)=\Pr (Y\leq y)=1-\theta ^{y},\quad y=1,2,3,...</math>
with <math>P(y)=0</math> for <math>y\leq 0</math> and <math>P(y)\rightarrow 1</math> as <math>y\rightarrow \infty
. </math>
(Note that <math>P(y)=p(1)+p(2)+...+p(y)=\sum_{t=1}^{y}p(t)</math> can be written in longhand as
<math>P(y)=\left( 1-\theta \right) \left( 1+\theta +\theta ^{2}+\theta ^{3}+\ldots
+\theta ^{y-1}\right) .</math>
The term in the second bracket on the right-hand side is the sum of a ''Geometric Progression.'')</li>
<li>The weekly consumption of fuel for a certain machine is modelled by means of a continuous random variable, <math>X</math>, with probability density function
<math>g(x)=\left\{
\begin{array}{c}
3(1-x)^{2},\quad 0\leq x\leq 1, \\
0,\quad \text{otherwise}.\end{array}
\right.</math>
Consumption, <math>X</math>, is measured in hundreds of gallons per week.
<ol>
<li>Verify that <math>\int_{0}^{1}g(x)dx=1</math> and calculate <math>\Pr (X\leq 0.5)</math>.</li>
<li>How much fuel should be supplied each week if the machine is to run out fuel <math>10\%</math> of the time at most? (Note that if <math>s</math> denotes the supply of fuel, then the machine will run out if <math>X>s</math>.)</li></ol>
</li>
<li>The lifetime of a electrical component is measured in <math>100</math>s of hours by a random variable <math>T</math> having the following probability density function
<math>f(t)=\left\{
\begin{array}{c}
\exp (-t),\quad t>0, \\
0,\quad \text{otherwise}.\end{array}
\right.</math>
<ol>
<li>Show that the cumulative distribution function, <math>F(t)=\Pr (T\leq t)</math> is given by
<math>F(t)=\left\{
\begin{array}{ll}
1-\exp (-t), & t>0 \\
0 & t\leq 0.\end{array}
\right.</math></li>
<li>Show the probability that a component will operate for at least <math>200</math> hours without failure is <math>\Pr (T\geq 2)\cong 0.135</math>.?</li>
<li>Three of these electrical components operate independently of one another in a piece of equipment and the equipment fails if ANY ONE of the individual components fail. What is the probability that the equipment will operate for at least <math>200</math> hours without failure? (Use the result in (b) in a binomial context).</li></ol>
</li></ol>

=== Additional resources ===

Khan Academy:

* This is a set of two clips to explain a poisson random variables. Here is the link to the first clip: [https://www.khanacademy.org/math/probability/random-variables-topic/poisson_process/v/poisson-process-1].
* How to use EXCEL to calculate discrete probabilities (Binomial and Poisson) [http://youtu.be/4atJiXLqzwA].

= Footnotes =

=Footnotes=

<references />

@@ Line 190: / Line 190: @@
 == Outlook ==
-With the theory out of the way, let’s think of examples for continuous distributions. In fact we are spoiled for choice. Check out this [http://en.wikipedia.org/wiki/Category:Continuous_distributions|list] of different continuous distributions from Wikipedia. The ''uniform'' distribution example we just discussed features on this list as well [http://en.wikipedia.org/wiki/Uniform_distribution_(continuous)]. It is the simplest of all examples. It should be noted though that there is not only one ''uniform'' distribution, there is an infinite number of uniform distributions. We chose the ''uniform'' distribution defined on the interval <math>\left[ 0,1\right] </math>. But we could have chosen any interval <math>\left[ a,b\right] </math>. The main properties wouldn’t have changed, but the actual probabilities, e.g. <math>\Pr \left( 0.25<X\leq 0.5\right)</math> would. In the context of distributions we call the <math>a</math> and <math>b</math> ''parameters''. Almost all distributions do have parameters changing them will have different effects.
+With the theory out of the way, let’s think of examples for continuous distributions. In fact we are spoiled for choice. Check out this [http://en.wikipedia.org/wiki/Category:Continuous_distributions list] of different continuous distributions from Wikipedia. The ''uniform'' distribution example we just discussed features on this list as well [http://en.wikipedia.org/wiki/Uniform_distribution_(continuous)]. It is the simplest of all examples. It should be noted though that there is not only one ''uniform'' distribution, there is an infinite number of uniform distributions. We chose the ''uniform'' distribution defined on the interval <math>\left[ 0,1\right] </math>. But we could have chosen any interval <math>\left[ a,b\right] </math>. The main properties wouldn’t have changed, but the actual probabilities, e.g. <math>\Pr \left( 0.25<X\leq 0.5\right)</math> would. In the context of distributions we call the <math>a</math> and <math>b</math> ''parameters''. Almost all distributions do have parameters changing them will have different effects.
 We shall pick two more examples. Here the ''exponential'' distribution and then the ''normal'' distribution. The latter is so important that we shall dedicated its own [[Probability<sub>N</sub>orm|page]] to it. We need this large range of different distributions as any random variable will have different properties and we will, in all cases, have to try and find that distribution that has properties that best represent that of the random variable we are interested in. Just briefly consider again the case of the level in our reservoir. Do you think this may be uniformly distributed? Most likely not. We would expect the cases where it is almost empty (<math>X</math> close to 0) or full (<math>X</math> close to 1) much less likely than some values around its ''usual'' level (say somewhere between 0.4 and 0.6). So, a ''uniform'' distribution appears inadequate.

@@ Line 140: / Line 140: @@
 &=&\dfrac{1}{64}.\end{aligned}</math>
-The following Figure also gives a simple example of how the ''cdf'' and ''pdf'' are related to each other. Here probability is distributed ''uniformly'' over a finite interval (in this case, it is the ''unit'' interval <math>\left[ 0,1\right] </math>). Such a distribution is therefore said to be uniform.
+The following Figure also gives a simple example of how the ''cdf'' and ''pdf'' are related to each other. Here probability is distributed ''uniformly'' over a finite interval (in this case, it is the ''unit'' interval <math>\left[ 0,1\right] </math>). Such a distribution is therefore said to be uniform and
 [[File:ProbDistCont_pdfcdf.jpg|frameless|600px]]

← Older revision		Revision as of 10:33, 13 August 2013
Line 245:		Line 245:
	= Footnotes =		= Footnotes =

−	~~=Footnotes=~~

	<references />		<references />

@@ Line 9: / Line 9: @@
 Notice how ''masses'' of probability are dropped onto the possible discrete (isolated) outcomes. You should also recall that, if you added up the probabilities for all possible outcomes you would obtain 1. We now develop mathematical functions which can used to describe probability distributions associated with a ''continuous random variable''.
-'''HEALTH WARNING: Before reading this section you MUST revise your undertanding of integration.'''
+'''HEALTH WARNING: Before reading this section you MUST revise your understanding of integration.'''
 == Additional Resources ==

@@ Line 142: / Line 142: @@
 The following Figure also gives a simple example of how the ''cdf'' and ''pdf'' are related to each other. Here probability is distributed ''uniformly'' over a finite interval (in this case, it is the ''unit'' interval <math>\left[ 0,1\right] </math>). Such a distribution is therefore said to be uniform.
-[[File:ProbDistCont_pdfcdf.jpg|frameless|400px]]
+[[File:ProbDistCont_pdfcdf.jpg|frameless|600px]]
 For example,

@@ Line 165: / Line 165: @@
 Probabilities can be calculated as:
-* <math>\Pr (a<X\leq b)=\int_{a}^{b}f(x)dx</math>
+* <math>\Pr (a<X\leq b)=\int_{a}^{b}f(x)dx</math> i.e., it is the ''area'' under the ''pdf ''which gives probability
 and the relationship with ''cdf'' is given by:
-* ** <math>f(x)=dF(x)/dx</math>
+* <math>f(x)=dF(x)/dx</math>
-** <math>F(x)=\int_{-\infty }^{x}f(t)dt. </math>
+* <math>F(x)=\int_{-\infty }^{x}f(t)dt.</math>
 == Outlook ==
@@ Line 240: / Line 239: @@
 == Additional Resources ==
-* Wikipedia: [http://en.wikipedia.org/wiki/Exponential_distribution|Exponential Distribution], but note that they use slightly different notation in terms of the parameter of the exponential distribution. They use <math>\lambda</math> which is related to our <math>\theta</math> in the following way: <math>\lambda = 1 / \theta</math>.
+* Wikipedia: [http://en.wikipedia.org/wiki/Exponential_distribution|Exponential Distribution], but note that they use slightly different notation in terms of the parameter of the exponential distribution. They use <math>\lambda</math> which is related to our <math>\theta</math> in the following way, <math>\lambda = 1 / \theta</math>.
-* Wolfram MathWorld: [http://mathworld.wolfram.com/ExponentialDistribution.html|Exponential Distribution], but note that they use slightly different notation in terms of the parameter of the exponential distribution. They use <math>\lambda</math> which is related to our <math>\theta</math> in the following way: <math>\lambda = 1 / \theta</math>.
+* Wolfram MathWorld: [http://mathworld.wolfram.com/ExponentialDistribution.html|Exponential Distribution], but note that they use slightly different notation in terms of the parameter of the exponential distribution. They use <math>\lambda</math> which is related to our <math>\theta</math> in the following way, <math>\lambda = 1 / \theta</math>.
-* How to use EXCEL to calculate probabilities from an exponential distribution []
+* How to use EXCEL to calculate probabilities from an exponential distribution [http://youtu.be/7SixLei1L_8]
 = Footnotes =