91Ó°ÊÓ

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

We are given joint pmf of \({\rm{X}}\) and \({\rm{Y}}\) in exercise \({\rm{1}}{\rm{.}}\)

The conditional probability density function of \({\rm{Y}}\) given that \({\rm{X = x}}\) is

1. \({{\rm{f}}_{{\rm{Y}}\mid {\rm{X}}}}{\rm{(y}}\mid {\rm{x) = }}\frac{{{\rm{f(x,y)}}}}{{{{\rm{f}}_{\rm{X}}}{\rm{(x)}}}}{\rm{,}}\;\;\;{\rm{ - ¥< y < ¥}}\)when \({\rm{X}}\) and \({\rm{Y}}\) are continuous rv's,

2. \({{\rm{p}}_{{\rm{Y}}\mid {\rm{X}}}}{\rm{(y}}\mid {\rm{x) = }}\frac{{{\rm{p(x,y)}}}}{{{{\rm{p}}_{\rm{X}}}{\rm{(x)}}}}{\rm{,}}\;\;\;{\rm{ - ¥< y < ¥}}\)when \({\rm{X}}\) and \({\rm{Y}}\) are discrete rv's.

In order to determine conditional probability density function of \({\rm{Y}}\) given \({\rm{X = 1}}\) we need to determine marginal pmf of \({\rm{X}}\) and find values\({{\rm{p}}_{\rm{X}}}{\rm{(1)}}\).

The marginal probability mass function of discrete random variable \({\rm{X}}\) is

\({{\rm{p}}_{\rm{X}}}{\rm{(x) = }}\sum\limits_{\rm{y}} {\rm{p}} {\rm{(x,y), for every x,}}\)

By the definition, for\({\rm{x\^I \{ 0,1,2\} }}\), the following is true

\({{\rm{p}}_{\rm{X}}}{\rm{(x) = }}\sum\limits_{{\rm{yI \{ 0,1,2\} }}} {\rm{p}} {\rm{(x,y) = p(x,0) + p(x,1) + p(x,2)}}{\rm{.}}\)

For every \({\rm{xI \{ 0,1,2\} }}\) the marginal probability is the sum of a particular row. Therefore, we have

\(\begin{aligned}{l}{{\rm{p}}_{\rm{X}}}{\rm{(0) = 0}}{\rm{.1 + 0}}{\rm{.04 + 0}}{\rm{.02 = 0}}{\rm{.16}}\\{{\rm{p}}_{\rm{X}}}{\rm{(1) = 0}}{\rm{.08 + 0}}{\rm{.2 + 0}}{\rm{.06 = 0}}{\rm{.34}}\\{{\rm{p}}_{\rm{X}}}{\rm{(2) = 0}}{\rm{.06 + 0}}{\rm{.14 + 0}}{\rm{.3 = 0}}{\rm{.5}}\end{aligned}\)

which determines the marginal pmf of\({\rm{X}}\).

The marginal pmf of \({\rm{X}}\) is

\(\begin{aligned}{l}{{\rm{p}}_{\rm{X}}}{\rm{(0) = 0}}{\rm{.16}}\\{{\rm{p}}_{\rm{X}}}{\rm{(1) = 0}}{\rm{.34}}\\{{\rm{p}}_{\rm{X}}}{\rm{(2) = 0}}{\rm{.5}}\\{{\rm{p}}_{\rm{X}}}{\rm{(x) = 0}}{\rm{.}}\;\;\;{\rm{xI \{ 0,1,2\} }}\end{aligned}\)

For \({\rm{(a)}}\) we need only\({{\rm{p}}_{\rm{X}}}{\rm{(1) = 0}}{\rm{.34}}\).

The conditional pmf of \({\rm{Y}}\) given \({\rm{X = 1}}\) is

\(\begin{aligned}{{\rm{p}}_{{\rm{Y}}\mid {\rm{X}}}}{\rm{(0}}\mid {\rm{1) = }}\frac{{{\rm{p(1,0)}}}}{{{{\rm{p}}_{\rm{X}}}{\rm{(1)}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.08}}}}{{{\rm{0}}{\rm{.34}}}}\\{\rm{ = 0}}{\rm{.2353}}\\{{\rm{p}}_{{\rm{Y}}\mid {\rm{X}}}}{\rm{(1}}\mid {\rm{1) = }}\frac{{{\rm{p(1,1)}}}}{{{{\rm{p}}_{\rm{X}}}{\rm{(1)}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.2}}}}{{{\rm{0}}{\rm{.34}}}}\\{\rm{ = 0}}{\rm{.5882}}\\{{\rm{p}}_{{\rm{Y}}\mid {\rm{X}}}}{\rm{(2}}\mid {\rm{1) = }}\frac{{{\rm{p(1,2)}}}}{{{{\rm{p}}_{\rm{X}}}{\rm{(1)}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.06}}}}{{{\rm{0}}{\rm{.34}}}}\\{\rm{ = 0}}{\rm{.1765}}\\{{\rm{p}}_{{\rm{Y}}\mid {\rm{X}}}}{\rm{(y}}\mid {\rm{1) = }}\frac{{{\rm{p(1,y)}}}}{{{{\rm{p}}_{\rm{X}}}{\rm{(1)}}}}\\{\rm{ = }}\frac{{\rm{0}}}{{{\rm{0}}{\rm{.34}}}}{\rm{ = 0,}}\;\;\;{\rm{yI \{ 0,1,2\} }}\end{aligned}\)

To summarize,

the conditional pmf of \({\rm{Y}}\) given \({\rm{X = 1}}\)

is

\({{\rm{p}}_{{\rm{Y}}\mid {\rm{X}}}}{\rm{(y}}\mid {\rm{1) = }}\left\{ {\begin{aligned}{*{20}{l}}{{\rm{0}}{\rm{.2353}}}&{{\rm{,y = 0}}}\\{{\rm{0}}{\rm{.5882}}}&{{\rm{,y = 1}}}\\{{\rm{0}}{\rm{.1765}}}&{{\rm{,y = 2}}}\\{\rm{0}}&{{\rm{,y = 0}}}\end{aligned}} \right.\)

(b) :

For \({\rm{(b)}}\) we need\({{\rm{p}}_{\rm{X}}}{\rm{(2) = 0}}{\rm{.5}}\). This is because we need to determine conditional pmf of \({\rm{Y}}\) given\({\rm{X = 2}}\). Similarly, as in\({\rm{(a)}}\), the following is true

\(\begin{aligned}{{\rm{p}}_{{\rm{Y}}\mid {\rm{X}}}}{\rm{(0}}\mid {\rm{2) = }}\frac{{{\rm{p(1,0)}}}}{{{{\rm{p}}_{\rm{X}}}{\rm{(2)}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.06}}}}{{{\rm{0}}{\rm{.5}}}}\\{\rm{ = 0}}{\rm{.12}}\\{{\rm{p}}_{{\rm{Y}}\mid {\rm{X}}}}{\rm{(1}}\mid {\rm{2) = }}\frac{{{\rm{p(1,1)}}}}{{{{\rm{p}}_{\rm{X}}}{\rm{(2)}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.14}}}}{{{\rm{0}}{\rm{.5}}}}\\{\rm{ = 0}}{\rm{.28}}\\{{\rm{p}}_{{\rm{Y}}\mid {\rm{X}}}}{\rm{(2}}\mid {\rm{2) = }}\frac{{{\rm{p(1,2)}}}}{{{{\rm{p}}_{\rm{X}}}{\rm{(2)}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.3}}}}{{{\rm{0}}{\rm{.5}}}}\\{\rm{ = 0}}{\rm{.6}}\\{{\rm{p}}_{{\rm{Y}}\mid {\rm{X}}}}{\rm{(y}}\mid {\rm{2) = }}\frac{{{\rm{p(1,y)}}}}{{{{\rm{p}}_{\rm{X}}}{\rm{(2)}}}}\\{\rm{ = }}\frac{{\rm{0}}}{{{\rm{0}}{\rm{.5}}}}{\rm{ = 0,}}\;\;\;{\rm{yI \{ 0,1,2\} }}\end{aligned}\)

To summarize,

the conditional pmf of \({\rm{Y}}\) given \({\rm{X = 2}}\)

is

\({{\rm{p}}_{{\rm{Y}}\mid {\rm{X}}}}{\rm{(y}}\mid {\rm{2) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{0}}{\rm{.12}}}&{{\rm{,y = 0}}}\\{{\rm{0}}{\rm{.28}}}&{{\rm{,y = 1}}}\\{{\rm{0}}{\rm{.6}}}&{{\rm{,y = 2}}}\\{\rm{0}}&{{\rm{,y = 0}}}\end{array}} \right.\)

(c):

The following holds

\(\begin{aligned}{\rm{P(Y£1}}\mid {\rm{X = 2)}}\\{\rm{ = P(Y = 0}}\mid {\rm{X = 2) + P(Y = 1}}\mid {\rm{X = 2)}}\\{\rm{ = }}{{\rm{p}}_{{\rm{Y}}\mid {\rm{X}}}}{\rm{(0}}\mid {\rm{2) + }}{{\rm{p}}_{{\rm{Y}}\mid {\rm{X}}}}{\rm{(1}}\mid {\rm{2)}}\\{\rm{ = 0}}{\rm{.12 + 0}}{\rm{.28}}\\{\rm{ = 0}}{\rm{.4}}\end{aligned}\)

(d)

Now we need to compute conditional pmf of \({\rm{X}}\) given\({\rm{Y = 2}}\). In order to do that, we need only\({{\rm{p}}_{\rm{Y}}}{\rm{(2)}}\). The following holds

\({{\rm{p}}_{\rm{Y}}}{\rm{(2) = 0}}{\rm{.02 + 0}}{\rm{.06 + 0}}{\rm{.3 = 0}}{\rm{.38}}\)

The conditional pmf of \({\rm{X}}\) given \({\rm{Y = 2}}\) is

\(\begin{aligned}{{\rm{p}}_{{\rm{X}}\mid {\rm{Y}}}}{\rm{(0}}\mid {\rm{2) = }}\frac{{{\rm{p(9,2)}}}}{{{{\rm{p}}_{\rm{Y}}}{\rm{(2)}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.02}}}}{{{\rm{0}}{\rm{.38}}}}\\{\rm{ = 0}}{\rm{.0526}}\\{{\rm{p}}_{{\rm{X}}\mid {\rm{Y}}}}{\rm{(1}}\mid {\rm{2) = }}\frac{{{\rm{p(1,2)}}}}{{{{\rm{p}}_{\rm{Y}}}{\rm{(2)}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.06}}}}{{{\rm{0}}{\rm{.38}}}}\\{\rm{ = 0}}{\rm{.1579}}\\{{\rm{p}}_{{\rm{X}}\mid {\rm{Y}}}}{\rm{(2}}\mid {\rm{2) = }}\frac{{{\rm{p(2,2)}}}}{{{{\rm{p}}_{\rm{Y}}}{\rm{(2)}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.3}}}}{{{\rm{0}}{\rm{.38}}}}\\{\rm{ = 0}}{\rm{.7895}}\\{{\rm{p}}_{{\rm{X}}\mid {\rm{Y}}}}{\rm{(x}}\mid {\rm{2) = }}\frac{{{\rm{p(x,2)}}}}{{{{\rm{p}}_{\rm{Y}}}{\rm{(2)}}}}\\{\rm{ = }}\frac{{\rm{0}}}{{{\rm{0}}{\rm{.38}}}}{\rm{ = 0,}}\;\;\;{\rm{xI \{ 0,1,2\} }}\end{aligned}\)

To summarize,

the conditional pmf of \({\rm{Y}}\) given \({\rm{X = 2}}\)

is

\({{\rm{p}}_{{\rm{X}}\mid {\rm{Y}}}}{\rm{(x}}\mid {\rm{2) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{0}}{\rm{.0526}}}&{{\rm{,x = 0}}}\\{{\rm{0}}{\rm{.1579}}}&{{\rm{,x = 1}}}\\{{\rm{0}}{\rm{.7895}}}&{{\rm{,x = 2}}}\\{\rm{0}}&{{\rm{,x = 0}}}\end{array}} \right.\)