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Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

Let \({\rm{X}}\) and \({\rm{Y}}\) be discrete random variables. The joint probability mass function \({\rm{p(x,y)}}\) is

\({\rm{p(x,y) = P(X = x,Y = y)}}\)

for every pair\({\rm{(x,y)}}\), where and\(\sum\limits_{\rm{x}} {\sum\limits_{\rm{y}} {\rm{p}} } {\rm{(x,y) = 1}}\).

We first need to determine the joint probability mass function, \({\rm{p(x,y), of }}{{\rm{X}}_{\rm{1}}}\), and\({{\rm{X}}_{\rm{2}}}\). Because of the independence and the multiplication property we can calculate the pmf easy.

Multiplication Property: Two events \({\rm{A}}\) and \({\rm{B}}\) are independent if and only if

\({\rm{P(A{C}B) = P(A) \times P(B)}}\)

Both random variables can take values \({{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{I \{ 25,40,65\} }}\) with probabilities, respectively of values, \({\rm{0}}{\rm{.2,0}}{\rm{.5}}\), and\({\rm{0}}{\rm{.3}}\). One example is given below, and the rest of the values is given in the following table. For\({{\rm{x}}_{\rm{1}}}{\rm{ = 40}}\), and\({{\rm{x}}_{\rm{2}}}{\rm{ = 65}}\), the following holds

\(\begin{aligned}{\rm{p(40,65) = P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 40,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 65}}} \right)\\{\rm{ = P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 40}}} \right){\rm{P}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ = 65}}} \right)\\{\rm{ = 0}}{\rm{.5 \times 0}}{\rm{.3}}\\{\rm{ = 0}}{\rm{.15}}\end{aligned}\)

The pmf table is

(a):

Possible values for random variable \({\rm{\bar X}}\) are

\(\begin{aligned}{l}\frac{{{\rm{25 + 25}}}}{{\rm{2}}}{\rm{ = 25;}}\;\;\\\frac{{{\rm{25 + 40}}}}{{\rm{2}}}{\rm{ = 32}}{\rm{.5;}}\\\frac{{{\rm{25 + 65}}}}{{\rm{2}}}{\rm{ = 45;}}\\\frac{{{\rm{40 + 40}}}}{{\rm{2}}}{\rm{ = 40;}}\\\frac{{{\rm{40 + 65}}}}{{\rm{2}}}{\rm{ = 52}}{\rm{.5;}}\;\\\frac{{{\rm{65 + 65}}}}{{\rm{2}}}{\rm{ = 65}}{\rm{.}}\end{aligned}\)

The following is true

\(\begin{aligned}{\rm{P(\bar X = 25) = P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 25,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 25}}} \right)\\{\rm{ = p(25,25)}}\\{\rm{ = 0}}{\rm{.04}}\\{\rm{P(\bar X = 32}}{\rm{.5) = P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 25,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 40}}} \right){\rm{ + P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 40,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 25}}} \right)\\{\rm{ = 0}}{\rm{.1 + 0}}{\rm{.1}}\\{\rm{ = 0}}{\rm{.2}}\end{aligned}\)

Similarly, we calculate other probabilities given in the following table

The Expected Value (mean value) of a discrete random variable \({\rm{X}}\) with set of possible values \({\rm{S}}\) and pmf \({\rm{p(x)}}\) is

\({\rm{E(X) = }}{{\rm{\mu }}_{\rm{X}}}{\rm{ = }}\sum\limits_{{\rm{x\^I S}}} {\rm{x}} {\rm{ \times p(x)}}\)

Therefore, the expected value of random variable \({\rm{\bar X}}\) is

\(\begin{aligned}{\rm{E(\bar X) = }}\sum\limits_{{\rm{\bar x\^I S}}} {{\rm{\bar x}}} {\rm{ \times p(\bar x)}}\\{\rm{ = 25 \times 0}}{\rm{.04 + 32}}{\rm{.5 \times 0}}{\rm{.2 + \ldots + 65 \times 0}}{\rm{.09}}\\{\rm{ = 44}}{\rm{.5}}{\rm{.}}\end{aligned}\)

Let's compute \({\rm{\mu }}\) from the pmf given before\({\rm{(a)}}\). The following holds

\({\rm{\mu = 25 \times 0}}{\rm{.2 + 40 \times 0}}{\rm{.5 + 65 \times 0}}{\rm{.3 = 44}}{\rm{.5}}{\rm{.}}\)

Hence, the expected value of random variable \({\rm{\bar X}}\) is identical to \({\rm{\mu }}\) or

\({\rm{E(\bar X) = \mu }}\)

(b):

Similarly, as in\({\rm{(a)}}\), we first need to compute all possible values that random variable \({{\rm{S}}^{\rm{2}}}\) can take. Therefore, the following is true

\(\begin{aligned}{l}\frac{{\rm{1}}}{{{\rm{2 - 1}}}}\left( {{{{\rm{(25 - 25)}}}^{\rm{2}}}{\rm{ + (25 - 25}}{{\rm{)}}^{\rm{2}}}} \right){\rm{ = 0}}\\\frac{{\rm{1}}}{{{\rm{2 - 1}}}}\left( {{{{\rm{(25 - 32}}{\rm{.5)}}}^{\rm{2}}}{\rm{ + (40 - 32}}{\rm{.5}}{{\rm{)}}^{\rm{2}}}} \right){\rm{ = 112}}{\rm{.5}}\\\frac{{\rm{1}}}{{{\rm{2 - 1}}}}\left( {{{{\rm{(25 - 45)}}}^{\rm{2}}}{\rm{ + (65 - 45}}{{\rm{)}}^{\rm{2}}}} \right){\rm{ = 800}}\\\frac{{\rm{1}}}{{{\rm{2 - 1}}}}\left( {{{{\rm{(40 - 52}}{\rm{.5)}}}^{\rm{2}}}{\rm{ + (65 - 52}}{\rm{.5}}{{\rm{)}}^{\rm{2}}}} \right){\rm{ = 312}}{\rm{.5}}\end{aligned}\)

true

By calculating other values, you would always get one of those four. Note: here \({\rm{\bar x}}\) corresponds to the given pair, for example, if \({{\rm{x}}_{\rm{1}}}{\rm{ = 25}}\) and \({{\rm{x}}_{\rm{2}}}{\rm{ = 40}}\) the corresponding \({\rm{\bar x}}\) is equal to\({\rm{32}}{\rm{.5}}\).

The probabilities can be calculated as follows

\(\begin{aligned}{\rm{P}}\left( {{{\rm{S}}^{\rm{2}}}{\rm{ = 0}}} \right){\rm{ = p(25,25) + p(40,40) + p(65,65)}}\\{\rm{ = 0}}{\rm{.04 + 0}}{\rm{.25 + 0}}{\rm{.09}}\\{\rm{ = 0}}{\rm{.38}}\end{aligned}\)

and this is because we get \({\rm{0}}\) when computing value of statistic \({{\rm{S}}^{\rm{2}}}\) for given values. Similarly for the rest

\(\begin{aligned}{l}{\rm{P}}\left( {{{\rm{S}}^{\rm{2}}}{\rm{ = 112}}{\rm{.5}}} \right){\rm{ = p(25,40) + p(40,25)}}\\{\rm{ = 0}}{\rm{.1 + 0}}{\rm{.1}}\\{\rm{ = 0}}{\rm{.2P}}\left( {{{\rm{S}}^{\rm{2}}}{\rm{ = 312}}{\rm{.5}}} \right){\rm{ = p(40,65) + p(65,40)}}\\{\rm{ = 0}}{\rm{.15 + 0}}{\rm{.15}}\\{\rm{ = 0}}{\rm{.3P}}\left( {{{\rm{S}}^{\rm{2}}}{\rm{ = 800}}{\rm{.0}}} \right){\rm{ = p(25,65) + p(65,25)}}\\{\rm{ = 0}}{\rm{.06 + 0}}{\rm{.06}}\\{\rm{ = 0}}{\rm{.12}}\end{aligned}\)

The pmf of \({{\rm{S}}^{\rm{2}}}\) is given in the following table

We cannot calculate the expectation as follows

\(\begin{aligned}{\rm{E}}\left( {{{\rm{S}}^{\rm{2}}}} \right){\rm{ = 0 \times 0}}{\rm{.38 + 112}}{\rm{.5 \times 0}}{\rm{.2 + 312}}{\rm{.5 \times 0}}{\rm{.3 + 800 \times 0}}{\rm{.12}}\\{\rm{ = 212}}{\rm{.25}}{\rm{.}}\end{aligned}\)

lirom the given pmf before\({\rm{(a)}}\), we calculate \({{\rm{\sigma }}^{\rm{2}}}\)

\(\begin{aligned}{{\rm{\sigma }}^{\rm{2}}}{\rm{ = (25 - 44}}{\rm{.5}}{{\rm{)}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.2 + (40 - 44}}{\rm{.5}}{{\rm{)}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.5 + (65 - 44}}{\rm{.5}}{{\rm{)}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.3}}\\{\rm{ = 212}}{\rm{.25}}{\rm{.}}\end{aligned}\)

We can conclude that

\({\rm{E}}\left( {{{\rm{S}}^{\rm{2}}}} \right){\rm{ = }}{{\rm{\sigma }}^{\rm{2}}}\)