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 pdf of \({\rm{X}}\) and\({\rm{Y}}\).

We will first calculate value of \({\rm{K}}\) as follows (you can skip this part is you are not interested in this). From relation

\(\int_{{\rm{ - ¥}}}^{\rm{¥}} {\int_{{\rm{ - ¥}}}^{\rm{¥}} {\rm{f}} } {\rm{(x,y)dxdy = 1}}\)

we can find value of \({\rm{K}}\).

The area \({\rm{20ï¿¡xï¿¡30,20ï¿¡yï¿¡30}}\) is a simple square, therefore the following holds

\(\begin{aligned}\int_{ - ¥}^¥{\int_{ - ¥}^¥f } (x,y)dxdy &= \int_{20}^{30} {\int_{20}^{30} K } \left( {{x^2} + {y^2}} \right)dxdy \\ &= K\int_{20}^{30} {\int_{20}^{30} {{x^2}} } dydx + K\int_{20}^{30} {\int_{20}^{30} {{y^2}} } dxdy \\ &= K\int_{20}^{30} {{x^2}} \left( {\left. y \right|_{20}^{30}} \right)dx + K\int_{20}^{30} {{y^2}} \left( {\left. x \right|_{20}^{30}} \right)dy \\ &= \left. {10K \times \frac{{{x^3}}}{3}} \right|_{20}^{30} + \left. {10K \times \frac{{{y^3}}}{3}} \right|_{20}^{30} \\ &= 2 \times 10K\left( {\frac{{30}}{3} - \frac{{{{20}^3}}}{3}} \right) \\ &= 20K \times \frac{{19,000}}{3} \\ &= \frac{{380,000}}{3} \times K \\ &= 120 \\\end{aligned} \)

Hence, we have

\(\frac{{{\rm{380,000}}}}{{\rm{3}}}{\rm{ \times K = 1}}\)

or equally

\({\rm{K = }}\frac{{\rm{3}}}{{{\rm{380,000}}}}\)

(a):

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.

We have the joint pdf; we need the marginal pdf of \({\rm{X}}\).

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

\({{\rm{f}}_{\rm{X}}}{\rm{(x) = }}\int_{{\rm{ - ¥}}}^{\rm{¥}} {\rm{f}} {\rm{(x,y)dy,}}\;\;\;{\rm{ for - ¥< x < ¥}}\)

The marginal probability mass function of continuous random variable \({\rm{Y}}\) is

\({{\rm{f}}_{\rm{Y}}}{\rm{(x) = }}\int_{{\rm{ - ¥}}}^{\rm{¥}} {\rm{f}} {\rm{(x,y)dx,}}\;\;\;{\rm{ for - ¥< y < ¥}}\)

The following holds

\({{\rm{f}}_{\rm{X}}}{\rm{(x) = }}\int_{{\rm{20}}}^{{\rm{30}}} {\rm{K}} \left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right){\rm{dy = 10K}}{{\rm{x}}^{\rm{2}}}{\rm{ + }}\left. {{\rm{K}}\frac{{{{\rm{y}}^{\rm{3}}}}}{{\rm{3}}}} \right|_{{\rm{20}}}^{{\rm{30}}}\)

\({\rm{ = 10 \times K}}{{\rm{x}}^{\rm{2}}}{\rm{ + 0}}{\rm{.05}}\),

°À(µ÷°À°ù³¾µ÷20ï¿¡xï¿¡30°¨°¨°À),

\({{\rm{f}}_{\rm{X}}}{\rm{(x) = 0}}\),

\({\rm{xI (20,30)}}{\rm{.}}\)

We would get the same marginal distribution for \({\rm{Y}}\) if we substitute \({\rm{x}}\) with\({\rm{y}}\).

Therefore, the conditional pdf of \({\rm{Y}}\) given that \({\rm{X = x}}\) is

\({{\rm{f}}_{{\rm{Y}}\mid {\rm{X}}}}{\rm{(y}}\mid {\rm{x) = }}\frac{{{\rm{f(x,y)}}}}{{{{\rm{f}}_{\rm{X}}}{\rm{(x)}}}}{\rm{ = }}\frac{{{\rm{K}}\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right)}}{{{\rm{10K}}{{\rm{x}}^{\rm{2}}}{\rm{ + 0}}{\rm{.05}}}}{\rm{,}}\;\;\;{\rm{20ï¿¡yï¿¡30}}{\rm{.}}\)

Similarly, the conditional pdf of \({\rm{X}}\) given that \({\rm{Y = y}}\) is

\({{\rm{f}}_{{\rm{X}}\mid {\rm{Y}}}}{\rm{(x}}\mid {\rm{y) = }}\frac{{{\rm{f(x,y)}}}}{{{{\rm{f}}_{\rm{Y}}}{\rm{(y)}}}}{\rm{ = }}\frac{{{\rm{K}}\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right)}}{{{\rm{10K}}{{\rm{y}}^{\rm{2}}}{\rm{ + 0}}{\rm{.05}}}}{\rm{,}}\;\;\;{\rm{20ï¿¡xï¿¡30}}{\rm{.}}\)

(b):

The probability of the first event, \(\text{ }\!\!\{\!\!\text{ Y }\!\!{}^\text{3}\!\!\text{ 25}\mid \text{X=22 }\!\!\}\!\!\text{ }\), is

\(\begin{aligned}P(Y25\mid X = 22) &= \int_{25}^{30} {{f_{Y\mid X}}} (y\mid 22)dy \\ &= \int_{25}^{30} {\frac{{K\left( {{{22}^2} + {y^2}} \right)}}{{10K \times {{22}^2} + 0.05}}} dy \\ &= \int_{25}^{30} {\frac{{K \times {{22}^2}}}{{10K \times {{22}^2} + 0.05}}} dy + \int_{25}^{30} {\frac{{K{y^2}}}{{10K \times {{22}^2} + 0.05}}} dy \\ &= \frac{{K \times {{22}^2}}}{{10K \times {{22}^2} + 0.05}}(30 - 25) + \left. {\frac{K}{{10K \times {{22}^2} + 0.05}}\frac{{{y^3}}}{3}} \right|_{25}^{30} \\ &= 0.56. \\ \end{aligned} \)

The probability of event \(\text{ }\!\!\{\!\!\text{ Y }\!\!{}^\text{3}\!\!\text{ 25 }\!\!\}\!\!\text{ }\) is

\(\begin{aligned}P(Y25) &= \int_{25}^{30} {{f_Y}} (y)dy = \int_{25}^{30} {\left( {10K{y^2} + 0.05} \right)} dy \\ &= \int_{25}^{30} 1 0K{y^2}dy + \int_{25}^{30} 0 .05dy \\ &= \left. {10K\frac{{{y^3}}}{3}} \right|_{25}^{30} + 0.05 \times (30 - 25) \\ &= 0.75. \\\end{aligned} \)

Probability that the left tire pressure is at least \(\text{25}\) psi is bigger than probability that the left tire pressure is at least \(\text{25}\) psi given that the right tire pressure is \(\text{22}\) psi.

(c):

The Expected Value (mean value) of a continuous random variable \({\rm{X}}\) with pdf \({\rm{f(x)}}\) is

\({\rm{E(X) = }}{{\rm{\mu }}_{\rm{X}}}{\rm{ = }}\int_{{\rm{ - ¥}}}^{\rm{¥}} {\rm{x}} {\rm{ \times f(x)dx}}{\rm{.}}\)

Therefore, we have

\(\begin{aligned}E(Y\mid X = 22) &= \int_{ - ¥}^¥ y \times {f_{Y\mid X}}(y\mid 22)dy \\ &= \int_{20}^{30} y \times \frac{{K\left( {{{22}^2} + {y^2}} \right)}}{{10K \times {{22}^2} + 0.05}}dy \\ &= \int_{20}^{30} y \times \frac{{K \times {{22}^2}}}{{10K \times {{22}^2} + 0.05}}dy + \int_{20}^{30} {\frac{{K{y^3}}}{{10K \times {{22}^2} + 0.05}}} dy \\ &= \frac{{K \times {{22}^2}}}{{10K \times {{22}^2} + 0.05}}(30 - 20) + \left. {\frac{K}{{10K \times {{22}^2} + 0.05}}\frac{{{y^3}}}{3}} \right|_{20}^{30} \\ &= 25.37. \\\end{aligned} \)

The Variance of \({\rm{X}}\), denoted by \({\rm{V(X)}}\left( {{\rm{\sigma }}_{\rm{X}}^{\rm{2}}} \right.\) or \(\left. {{{\rm{\sigma }}^{\rm{2}}}} \right)\) is

\({\rm{V(X) = }}{{\rm{\sigma }}_{\rm{X}}}{\rm{ = E}}\left( {{{{\rm{(X - E(X))}}}^{\rm{2}}}} \right){\rm{ = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ - (E(X)}}{{\rm{)}}^{\rm{2}}}\)

c)

The Standard Deviation of \({\rm{X}}\) is

\({{\rm{\sigma }}_{\rm{X}}}{\rm{ = }}\sqrt {{\rm{\sigma }}_{\rm{X}}^{\rm{2}}} \)

In order to compute standard deviation, we need the following expectation

\(\begin{aligned}E\left( {{Y^2}\mid X = 22} \right) &= \int_{ - ¥}^¥{{y^2}} \times {f_{Y\mid X}}(y\mid 22)dy \\ &= \int_{20}^{30} {{y^2}} \times \frac{{K\left( {{{22}^2} + {y^2}} \right)}}{{10K \times {{22}^2} + 0.05}}dy \\ &= \int_{20}^{30} {{y^2}} \times \frac{{K \times {{22}^2}}}{{10K \times {{22}^2} + 0.05}}dy + \int_{20}^{30} {\frac{{K{y^4}}}{{10K \times {{22}^2} + 0.05}}} dy \\ &= \left. {\frac{{K \times {{22}^2}}}{{10K \times {{22}^2} + 0.05}}\frac{{{y^3}}}{3}} \right|_{20}^{30} + \left. {\frac{K}{{10K \times {{22}^2} + 0.05}}\frac{{{y^5}}}{5}} \right|_{20}^{30} \\ &= 652.03. \\\end{aligned} \)

Therefore, the variance is

\(\begin{aligned}{\rm{V(Y}}\mid {\rm{X = 22) = }}{{\rm{\sigma }}_{\rm{Y}}}\mid {\rm{X = 22 = E}}\left( {{{\rm{Y}}^{\rm{2}}}\mid {\rm{X = 22}}} \right){\rm{ - (E(Y}}\mid {\rm{X = 22)}}{{\rm{)}}^{\rm{2}}}\\{\rm{ = 652}}{\rm{.03 - 25}}{\rm{.3}}{{\rm{7}}^{\rm{2}}}\\{\rm{ = 8}}{\rm{.3931}}\end{aligned}\)

Finally, the standard deviation of pressure in the tire is

\({{\rm{\sigma }}_{{\rm{Y}}\mid {\rm{X = 22}}}}{\rm{ = }}\sqrt {{\rm{8}}{\rm{.3931}}} {\rm{ = 2}}{\rm{.8971}}{\rm{.}}\)