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‎The Binomial random variable

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‎The Binomial random variable

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‎Probability mass function

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‎Additional resources

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= Discrete Probability Distributions =

The axioms of probability tell us how we should combine and use probabilities in order to make sensible statements concerning uncertain events. To a large extent this has assumed an initial allocation of probabilities to events of interest, from which probabilities concerning related events (unions, intersections and complements) can be computed. The question we shall begin to address in the next two sections is how we might construct ''models'' which assign probabilities in the first instance.

The ultimate goal is the development of tools which enable statistical analysis of ''data''. Any data under consideration (after, perhaps, some coding) are simply a set of ''numbers ''which describe the appropriate members of a sample in meaningful way. Therefore, in wishing to assign probabilities to outcomes generated by sampling, we can equivalently think of how to assign probabilities to the numbers that are explicitly generated by the same sampling process. When dealing with numbers, a natural line of enquiry would be to characterise possible ''mathematical functions'' which, when applied to appropriate numbers, yield probabilities satisfying the three basic axioms. A mathematical functionwhich acts in this fashion is termed a mathematical or ''statistical model'':

* ''mathematical/statistical models'': mathematical functions which may be useful in assigning probabilities in a gainful way.

If such models are to have wide applicability, we need a ‘general’ approach. As noted above, and previously, events (on a sample space of interest) are often described in terms of physical phenomena; see Sections 3 and 4. Mathematical functions require ''numbers''. We therefore need some sort of ''mapping'' from the physical attributes of a sample space to real numbers, before we can begin developing such models. The situation is depicted in following Figure, in which events of interest defined on a physical sample space, <math>S,</math> are mapped into numbers, <math>x,</math> on the real line. Note that there is only one number for each physical event, but that two different events could be assigned the same number. Thus, this mapping can be described by a ''function''; it is this function, mapping from the sample space to the real line, which defines a ''random variable''. A further function, <math>f(.),</math> is then applied on the real line in order to generate probabilities.

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

The initial task, then, is the mapping from <math>S</math> to the real line and this is supplied by introducing the notion of a ''random variable'':

= Random Variable =

For our purposes, we can think of a ''random variable'' as having '''two''' components:

* a label/description which defines the variable of interest
* the definition of a procedure which assigns numerical values to events on the appropriate sample space.

Note that:

* often, but not always, how the numerical values are assigned will be implicitly defined by the chosen label
* A random variable is '''neither''' RANDOM or VARIABLE! Rather, it is device which describes how to assign numbers to physical events of interest: “''a random variable is a real valued function defined on a sample space''”.
* A random variable is indicated by an ''upper case'' letter (<math>X</math>,<math>\,\,Y</math>,<math>\,\,Z</math>, <math>T\,\,</math>etc). The strict mathematical implication is that since <math>X</math> is a function, when it is applied on a sample space (of physical attributes) it yields a number

Just in case you didn’t realise yet, the above is somewhat abstract, so let us now consider some examples of ''random variables'':

== Examples of random variables ==

<ul>
<li>Let <math>X=</math> ‘''the number of HEADs obtained when a fair coin is flipped'' 3 times’. This definition of <math>X</math> implies a function on the physical sample space which generates particular numerical values. Thus <math>X</math> is a random variable and the values it can assume are:
<math>X(</math>H,H,H<math>)=3;\,\,X(</math>T,H,H<math>)=2;\,\,X(</math>H,T,H<math>)=2;\,\,X(</math>H,H,T<math>)=2;\,\,</math> <math>X(</math>H,T,T<math>)=1;\,\,X(</math>T,H,T<math>)=1;\,\,X(</math>T,T,H<math>)=1;\,\,X(</math>T,T,T<math>)=0.</math>
This is an example (as indicated in the above Figure) where two different outcomes in the sample space (e.g. <math>{T,H,H}</math> and <math>{H,T,H}</math>) are mapped into the same number, here 2.</li>
<li>Let the random variable <math>Y</math> indicate whether or not a household has suffered some sort of property crime in the last <math>12</math> months, with <math>Y(</math>''yes''<math>)=1</math> and <math>Y(</math>''no''<math>)=0.</math> Note that we could have chosen the numerical values of <math>1</math> or <math>2</math> for ''yes'' and ''no'' respectively. However, the mathematical treatment is simplified if we adopt the ''binary ''responses of 1 and 0.</li>
<li>Let the random variable <math>T</math> describe the length of time, measured in weeks, that an unemployed job-seeker waits before securing permanent employment. So here, for example,
<math>T(15</math> ''weeks unemployed''<math>)=15,</math><math>T(31</math> ''weeks unemployed''<math> )=31,</math> etc.</li></ul>

Once an experiment is carried out, and the random variable (<math>X</math>) is applied to the outcome, a number is ''observed'', or ''realised''; i.e., the value of the function at that point in the sample space. This is called a ''realisation'', or possible outcome, of <math>X</math> and is denoted by a ''lower case'' letter, <math>x</math>.

In the above examples, the possible realisations of the random variable <math>X</math> (i.e., possible values of the function defined by <math>X</math>) are <math>x=0,1,2</math> or <math>3</math>. For <math>Y,</math> the possible realisations are <math>y=0,1;</math> and for <math>T</math> they are <math>t=1,2,3,...</math>.

The examples of <math>X,</math> <math>Y</math> and <math>T</math> given here all are applications of ''discrete'' random variables (the outcomes, or values of the function, are all integers). Technically speaking, the functions <math>X,</math> <math>Y</math> and <math>T</math> are not continuous.

= Discrete random variables =

In general, a ''discrete random variable'' can only assume discrete realisations which are easily listed prior to experimentation. Having defined a discrete random variable, probabilities are assigned by means of a ''probability distribution.'' A probability distribution is essentially a function which maps from <math>x</math> (the real line) to the interval <math>\left[ 0,1\right]</math>; thereby generating probabilities.

In the case of discrete random variable, we shall use what is called a ''probability mass function'':

== Probability mass function ==

The probability mass function ''(pmf)'' is defined for a '''DISCRETE''' random variable, <math>X,</math> only and is the ''function'':

<math>p(x)=\Pr (X=x),\quad \text{for all }x.</math>

Note that:

<ul>
<li>We use <math>p(x)</math> here to emphasize that probabilities are being generated for the outcome <math>x;</math> e.g., <math>p(1)=\Pr (X=1),\,\,</math>etc.</li>
<li>Note that <math>p(r)=0,</math> if the number <math>r</math> is NOT a possible realisation of <math>X.</math> Thus, for the property crime random variable <math>Y</math>, with <math>p(y)=\Pr (Y=y)</math>, it must be that <math>p(0.5)=0</math> since a realisation of <math>0.5</math> is impossible for the random variable.</li>
<li>If <math>p(x)</math> is to be useful, then it follows from the ''axioms of probability'' that,
<math>p(x)\geq 0\quad and\quad \sum_{x}p(x)=1</math>
where the sum is taken over all possible values that <math>X</math> can assume.</li></ul>

For example, when <math>X=</math> ‘''the number of HEADs obtained when a fair coin is flipped'' <math>3</math> times’, we can write that <math>\sum_{j=0}^{3}p(j)=p(0)+p(1)+p(2)+p(3)=1</math> since the number of heads possible is either <math>0,1,2,</math> or <math>3</math>. Be clear about the notation being used here: <math>p(j)</math> is being used to give the probability that <math>j</math> heads are obtained in <math>3</math> flips; i.e., <math>p(j)=\Pr \left( X=j\right) ,</math> for values of <math>j</math> equal to <math>0,1,2,3</math>.

The ''pmf'' tells us how probabilities are distributed across all possible outcomes of a discrete random variable <math>X</math>; it therefore generates a ''probability distribution''. A ''probability distribution'' tells you how the probabilities ''distribute'' across all possible outcomes. We will encounter many examples of this soon.

== Cumulative distribution function ==

In the '''DISCRETE''' case the cumulative distribution function (''cdf)'' is a function which ''cumulates'' (adds up) values of <math>p(x)</math>, the ''pmf''. In general, it is defined as the function:

<math>P(x)=\Pr (X\leq x);</math>

e.g., <math>P(1)=\Pr (X\leq 1)</math>. Note the use of an upper case letter, <math>P\left(.\right) ,</math> for the ''cdf'', as opposed to the lower case letter, <math>p\left( .\right) ,</math> for the ''pmf''. In the case of a discrete random variable it is constructed as follows:

Suppose the discrete random variable, <math>X,</math> can take on possible values <math>x=a_{1},a_{2},a_{3},...,</math> etc, where the <math>a_{j}</math> are an increasing sequence of numbers <math>\left( a_{1}<a_{2}<\ldots \right) </math>. Then, for example, we can construct the following (cumulative) probability:

<math>\Pr \left( X\leq a_{4}\right)
=P(a_{4})=p(a_{1})+p(a_{2})+p(a_{3})+p(a_{4})=\sum_{j=1}^{4}p(a_{j}),</math>

i.e., we take all the probabilities assigned to possible values of <math>X,</math> up to the value under consideration (in this case <math>a_{4}</math>), and then add them up. It follows from the axioms of probability that <math>\sum_{j}p(a_{j})=1,</math> all the probabilities assigned must sum to unity, as noted before. Therefore,

<math>\begin{aligned}
\Pr \left( X\geq a_{4}\right) &=&p\left( a_{4}\right) +p\left( a_{5}\right)+p\left( a_{6}\right) +\ldots \\
&=&\left\{ \sum_{j}p(a_{j})\right\} -\left\{ p\left( a_{1}\right) +p\left(a_{2}\right) +p\left( a_{3}\right) \right\} \\
&=&1-\Pr \left( X\leq a_{3}\right) ,\end{aligned}</math>

and, similarly,

<math>\Pr \left( X>a_{4}\right) =1-\Pr \left( X\leq a_{4}\right)</math>

which is always useful to remember.

= Example Distributions =

In order to fill these theoretical constructs with some life we shall use a few examples. All these examples represent discrete random variables. There are many cases in which the random variable of interest is of a continuous nature. The same principles introduced here apply to that type of random variables and are discussed here [[ProbabilityContRV|Continuous Random Variables]].

== A Bernoulli random variable ==

A ''Bernoulli'' random variable is a particularly simple (but very useful) discrete random variable. The ‘property crime’ random variable, <math>Y</math>, introduced above is a particular example. A Bernoulli random variable can only assume one of two possible values: <math>y=0</math> or <math>y=1</math>; with probabilities <math>\left( 1-\pi \right) </math> and <math>\pi ,</math> respectively. Often, the value <math>1</math> might be referred to as a success and the value <math>0</math> a failure. Here, <math>\pi </math> is any number satisfying <math>0<\pi <1,</math> since it is a probability, and it is called a ''parameter'' for this this random variable. Note that, here, <math>\pi </math> is the Greek letter ''pi'' (lower case), with English equivalent ''p'', and is ''not'' used here to denote the number ''Pi'' <math>=3.14159...</math> . Clearly, different choices for <math>\pi </math> generate different probabilities for the outcomes of interest; it is an example of a very simple statistical model and the ''(pmf)'' can be written compactly as

<math>p(y)=\pi ^{y}\left( 1-\pi \right) ^{1-y},\quad 0<\pi <1,\quad y=0,1.</math>

For a variable with two discrete outcome only, as this one, the ''cdf'' is not very interesting. But we shall in any case state it

<math>\begin{aligned}
P(y \leq 0)= (1-\pi)\\
P(y \leq 1)= 1\\\end{aligned}</math>

The actual values for both, the ''pmf'' and the ''cdf'', will depend on the value of the ''parameter'' <math>\pi</math>.

== The Binomial random variable ==

Let’s use a different, but related, random variable that allow us to demonstrate the difference between ''pmf'' and ''cdf'' in a more insightful way. Consider the case where <math>X=</math> ''the number of HEADs obtained from '' 3 ''flips of a fair coin''. In this case, there are just four possible values of <math>X</math> with <math>a_{1}=0</math>, <math>a_{2}=1</math>, <math>a_{3}=2</math> and <math>a_{4}=3</math> in the notation used above. Furthermore, due to independence, we can write that

* <math>p(0)=\Pr (X=0)=\Pr (</math>T,T,T<math>)=\Pr \left( \text{T}\right) \times \Pr\left( \text{T}\right) \times \Pr \left( \text{T}\right) =(1/2)^{3}=1/8</math>
* <math>p(1)=\Pr (X=1)=\Pr (</math>H,T,T<math>)+\Pr (</math>T,H,T<math>)+\Pr (</math>T,T,H<math>)</math><math>\,=(1/2)^{3}+(1/2)^{3}+(1/2)^{3}=1/8+1/8+1/8=3/8</math>
* <math>p(2)=\Pr (X=2)=\Pr (</math>H,H,T<math>)+\Pr (</math>H,T,H<math>)+\Pr (</math>T,H,H<math>)</math><math>\,=(1/2)^{3}+(1/2)^{3}+(1/2)^{3}=1/8+1/8+1/8=3/8</math>
* <math>p(3)=\Pr (X=3)=\Pr (</math>H,H,H<math>)=(1/2)^{3}=1/8</math>

whilst

* <math>P(2)=\Pr (X\leq 2)=p(0)+p(1)+p(2)=1/8+3/8+3/8=7/8</math>
* <math>\Pr (X>1)=1-\Pr \left( X\leq 1\right) =1-\left( 1/8+3/8\right) =1/2</math>.
* Note also that <math>P(2.5)=P(X\leq 2.5)</math> must be identical to <math>P(2)=\Pr (X\leq 2)</math> - ''think about it''!

In fact, this is a very simple example of a '''Binomial distribution'''. A binomial random variable is based on several repetitions of independent but identical Bernoulli experiments, an experiment which can result in only one of two possible outcomes (''success'' with a probability denoted by <math>\pi </math> (<math>0<\pi <1</math>) and ''failure'' with probability denoted by <math>1-\pi </math>). Therefore a binomial random variable is based on several independent Bernoulli random variables with parameter <math>\pi</math>.

A good example is opinion polls, when they as simple "yes" or "no" questions, say "should the UK stay in the European Union?" Individuals answer ''yes ''<math>\left( 1\right) </math> or ''no ''<math>\left( 0\right)</math>. A random variable, <math>X</math>, is then said to have a '''BINOMIAL''' distribution (and is called a Binomial random variable) if it is defined to be the total number of successes in <math>n</math> ''independent and identical Bernoulli'' experiments; e.g. of the 1010 people randomly selected, how many were in favour of staying in the European Union.

Note that the possible realisations of <math>X</math> are <math>x=0,1,2,...,n</math>, where <math>n</math> is the number of Bernoulli experiments (<math>n=</math> 1010 in the above opinion poll example), The Binomial ''probability mass function'' is:

<math>p(x)=P(X=x)=\binom{n}{x}\pi ^{x}(1-\pi )^{n-x},\quad x=0,1,2,...,n;\quad
0<\pi <1.</math>

where

<math>\binom{n}{x}=\frac{n!}{x!(n-x)!},</math>

and <math>n!</math> denotes <math>n</math> ''factorial'': <math>n!=n(n-1)(n-2)...2\times 1;</math> e.g., <math>3!=6.</math> In the above formula, we '''define''' <math>0!=1.</math> Note that <math>\binom{n}{x}</math> is called a ''combinatorial coefficient ''and always gives an integer; it simply counts the total number of different ways that we can arrange exactly <math>x</math> “ones” and <math>\left( n-x\right) </math> “zeros” together. For example, consider in how may different ways we can arrange (or combine) 2 “ones” and 1 “zero”. The possibilities are

<math>\left( 1,1,0\right) ;\left( 1,0,1\right) ;\left( 0,1,1\right)</math>

and that’s it. There are only three ways we can do it. Using <math>\binom{n}{x}</math> we need to substitute <math>x=2</math> and <math>n=3,</math> since the total number of “ones” and “zeros” is <math>3.</math> This gives <math>\binom{3}{2}=\frac{3!}{2!1!}=3;</math> as we discovered above.

Consider, then, a Binomial random variable, with parameters <math>n=5</math> and <math>\pi=0.3</math>. The possible outcomes for the random variable <math>X</math> are 0, 1, ..., 5 and we could calculate the probability for each of these outcomes according to the above formula. Let’s do that for one outcome, let’s calculate <math>p(2) = \Pr (X=2)</math>:

<math>\begin{aligned}
p(2) &=&\Pr (X=2)=\binom{5}{2}\left( 0.3\right) ^{2}\left( 0.7\right) ^{3} \\
&=&\frac{5!}{2!3!}\left( 0.09\right) \left( 0.343\right) \\
&=&\frac{5\times 4}{2}\left( 0.09\right) \left( 0.343\right) \\
&=&0.3087.\end{aligned}</math>

We can now also think of the cumulative distribution function, i.e. probabilities of the type <math>P(2) = \Pr (X \leq 2)</math>:

<math>\begin{aligned}
P\left( 2\right) &=&\Pr \left( X\leq 2\right) =\Pr \left( X=0\right) +\Pr
\left( X=1\right) +\Pr \left( X=2\right) \\
&=&p\left( 0\right) +p\left( 1\right) +p\left( 2\right) \\
&=&\binom{5}{0}\left( 0.3\right) ^{0}\left( 0.7\right) ^{5}+\binom{5}{1}\left( 0.3\right) ^{1}\left( 0.7\right) ^{4}+\binom{5}{2}\left( 0.3\right)
^{2}\left( 0.7\right) ^{3} \\
&=&0.16807+0.36015+0.3087 \\
&=&0.83692.\end{aligned}</math>

It is apparent that the calculation of probabilities of this type, in particular the cumulative probabilities can be rather burdensome. It is therefore often useful to approximate discrete random variables with a continuous random variable. In this context the transformed random variable, <math>Z=X/n</math>, which defines the proportion of successes, will be useful. It should be obvious that for very large <math>n</math> this can indeed be thought of a continuous random variable. The advantage, which you will be able to appreciate after mastering Sections [[ProbabilityContRV|Continuous Random Variables]] and [[ProbabilityNorm|Normal Random Variables]], is that calculating these probabilities can become more straightforward.

== Geometric Random Variable ==

As in the Binomial case, consider repeating independent and identical Bernoulli experiments (each of which results in a success, with probability <math> \pi ,</math> or a failure, with probability <math>1-\pi </math>). Define the random variable <math>X</math> to be the number of Bernoulli experiments performed in order to achieve the first success. This is a ''Geometric'' random variable.

The probability mass function is

<math>p\left( x\right) =\Pr \left( X=x\right) =\pi \left( 1-\pi \right)
^{x-1},\quad x=1,2,3,...;\quad 0<\pi <1,</math>

and if a discrete random variable has such a probability mass function then we say that is a Geometric distribution with ''parameter'' <math>\pi</math>.

Suppose the probability of success is <math>\pi =0.3.</math> What is the probability that the first success is achieved on the second experiment? We require

<math>p\left( 2\right) =\Pr \left( X=2\right) =\left( 0.7\right) \left( 0.3\right) =0.21.</math>

The probability that the first success is achieved on or before the third experiment is

<math>\begin{aligned}
P\left( 3\right) &=&\Pr \left( X\leq 3\right) \\
&=&p\left( 1\right) +p\left( 2\right) +p\left( 3\right) \\
&=&\left( 0.7\right) ^{0} \left( 0.3\right) +\left(0.7\right) ^{1} \left( 0.3\right) +\left( 0.7\right) ^{2} \left( 0.3\right) \\
&=& \left( 1+0.7+0.7^{2}\right)\left( 0.3\right) \\
&=&0.657.\end{aligned}</math>

== Poisson Random Variable ==

The Poisson random variable is widely-used to count the number of events occurring in a given interval of time. Examples include (i) the number of cars passing an observation point, located on a long stretch of straight road, over a 10 minute interval; (ii) the number of calls received at a telephone exchange over a 10 second interval.

The probability mass function is

<math>p\left( x\right) =\Pr \left( X=x\right) =\frac{\lambda ^{x}}{x!}\exp \left(
-\lambda \right) ,\quad x=0,1,2,...;\quad \lambda >0</math>

in which, <math>\exp (.)</math> is the exponential function (<math>\exp \left( a\right)=e^{a}</math>) and <math>\lambda </math> is a positive real number (a ''parameter''). We say that <math>X</math> has a Poisson distribution with parameter <math>\lambda </math> (note that <math>\lambda</math> will often be referred to as the ''mean'' which will be discussed in a later Section).

Suppose that the number of calls, <math>X,</math> arriving at a telephone exchange in any 10 second interval follows a Poisson distribution with <math>\lambda =5</math>. What is <math>\Pr \left( X=2\right)</math>?

<math>\begin{aligned}
p\left( 2\right) &=&\Pr \left( X=2\right) =\frac{5^{2}}{2!}\exp \left(
-5\right) \\
&=&0.0842.\end{aligned}</math>

The probability of more than <math>2</math> calls is

<math>\begin{aligned}
\Pr \left( X>2\right) &=&1-\Pr \left( X\leq 2\right) \\
&=&1-\left\{ \Pr \left( X=0\right) +\Pr \left( X=1\right) +\Pr \left(
X=2\right) \right\} \\
&=&1-\left\{ e^{-5}+5\times e^{-5}+12.5\times e^{-5}\right\} \\
&=&1-0.1247 \\
&=&0.8753.\end{aligned}</math>

= Footnotes =

@@ Line 178: / Line 178: @@
 &=&0.83692.\end{aligned}</math>
-It is apparent that the calculation of probabilities of this type, in particular the cumulative probabilities can be rather burdensome. It is therefore often useful to approximate discrete random variables with a continuous random variable. In this context the transformed random variable, <math>Z=X/n</math>, which defines the proportion of successes, will be useful. It should be obvious that for very large <math>n</math> this can indeed be thought of a continuous random variable. The advantage, which you will be able to appreciate after mastering Sections [[Probability<sub>C</sub>ontRV|Continuous Random Variables]] and [[Probability<sub>N</sub>orm|Normal Random Variables]], is that calculating these probabilities can become more straightforward.
+It is apparent that the calculation of probabilities of this type, in particular the cumulative probabilities can be rather burdensome. It is therefore often useful to approximate discrete random variables with a continuous random variable. In this context the transformed random variable, <math>Z=X/n</math>, which defines the proportion of successes, will be useful. It should be obvious that for very large <math>n</math> this can indeed be thought of a continuous random variable. The advantage, which you will be able to appreciate after mastering Sections [[Probability_ContRV|Continuous Random Variables]] and [[Probability_Norm|Normal Random Variables]], is that calculating these probabilities can become more straightforward.
 === Additional resources ===

@@ Line 77: / Line 77: @@
 <p>where the sum is taken over all possible values that <math>X</math> can assume.</p></li></ul>
-For example, when <math>X=</math> ‘''the number of HEADs obtained when a fair coin is flipped'' <math>3</math> times’, we can write that <math>\sum_{j=0}^{3}p(j)=p(0)+p(1)+p(2)+p(3)=1</math> since the number of heads possible is either <math>0,1,2,</math> or <math>3</math>. Be clear about the notation being used here: <math>p(j)</math> is being used to give the probability that <math>j</math> heads are obtained in <math>3</math> flips; i.e., <math>p(j)=\Pr \left( X=j\right) ,</math> for values of <math>j</math> equal to <math>0,1,2,3</math>.
+For example, when <math>X=</math> ‘''the number of HEADs obtained when a fair coin is flipped'' <math>3</math> times’, we can write that <math>\sum_{j=0}^{3}p(j)=p(0)+p(1)+p(2)+p(3)=1</math> since the number of heads possible is either <math>0,1,2,</math> or <math>3</math>. Be clear about the notation being used here. <math>p(j)</math> is the probability that <math>j</math> heads are obtained in <math>3</math> flips; i.e., <math>p(j)=\Pr \left( X=j\right) ,</math> for values of <math>j</math> equal to <math>0,1,2,3</math>.
 The ''pmf'' tells us how probabilities are distributed across all possible outcomes of a discrete random variable <math>X</math>; it therefore generates a ''probability distribution''. A ''probability distribution'' tells you how the probabilities ''distribute'' across all possible outcomes. We will encounter many examples of this soon.

← Older revision		Revision as of 16:12, 9 August 2013
Line 243:		Line 243:

	* 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].		* 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 =

@@ Line 125: / Line 125: @@
 The actual values for both, the ''pmf'' and the ''cdf'', will depend on the value of the ''parameter'' <math>\pi</math>.
-== The Binomial random variable ==
+== <div id="Binomial"></div> The Binomial random variable ==
 Let’s use a different, but related, random variable that allow us to demonstrate the difference between ''pmf'' and ''cdf'' in a more insightful way. Consider the case where <math>X=</math> ''the number of HEADs obtained from '' 3 ''flips of a fair coin''. In this case, there are just four possible values of <math>X</math> with <math>a_{1}=0</math>, <math>a_{2}=1</math>, <math>a_{3}=2</math> and <math>a_{4}=3</math> in the notation used above. Furthermore, due to independence, we can write that

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 The examples of <math>X,</math> <math>Y</math> and <math>T</math> given here all are applications of ''discrete'' random variables (the outcomes, or values of the function, are all integers). Technically speaking, the functions <math>X,</math> <math>Y</math> and <math>T</math> are not continuous.
 = Discrete random variables =
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 In general, a ''discrete random variable'' can only assume discrete realisations which are easily listed prior to experimentation. Having defined a discrete random variable, probabilities are assigned by means of a ''probability distribution.'' A probability distribution is essentially a function which maps from <math>x</math> (the real line) to the interval <math>\left[ 0,1\right]</math>; thereby generating probabilities.
-In the case of discrete random variable, we shall use what is called a ''probability mass function'':
+== Additional Resources ==
 == Probability mass function ==
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 It is apparent that the calculation of probabilities of this type, in particular the cumulative probabilities can be rather burdensome. It is therefore often useful to approximate discrete random variables with a continuous random variable. In this context the transformed random variable, <math>Z=X/n</math>, which defines the proportion of successes, will be useful. It should be obvious that for very large <math>n</math> this can indeed be thought of a continuous random variable. The advantage, which you will be able to appreciate after mastering Sections [[Probability<sub>C</sub>ontRV|Continuous Random Variables]] and [[Probability<sub>N</sub>orm|Normal Random Variables]], is that calculating these probabilities can become more straightforward.
 == Geometric Random Variable ==
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 &=&1-0.1247 \\
 &=&0.8753.\end{aligned}</math>
 = Footnotes =