W11 Homework B

Due date: Tuesday 3/31, 11:59pm

Conditional distribution

01

02

From conditional to joint, and back again

Suppose we have the following data about random variables $X$ and $Y$:

$$\begin{array}{c}{f}_X(x)=\{\begin{array}{cc}3x^2& 0\le x\le 1\\ 0& \text{ otherwise }\end{array}\\ \\ f_{Y|X}(y|x)=\{\begin{array}{cc}2y/x^2& 0\le y\le x\\ 0& \text{ otherwise }\end{array}\end{array}$$

(a) Find the joint distribution $f_{X,Y}(x,y)$.

(b) Find $f_{X|Y}(x|y)$.

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Solution

Solutions - 5250-02

(a)

Obtain the joint from the conditional and marginal:

$$\begin{array}{c}{f}_{X,Y}(x,y)=f_{Y|X}(y|x)\cdot f_X(x)\\ \\ \gg \gg 3x^2\cdot \frac{2y}{x^2}\gg \gg 6y\end{array}$$ $$f_{X,Y}(x,y)=\{\begin{array}{cc}6y& 0\le y\le x\le 1\\ 0& \text{otherwise}\end{array}$$

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(b)

(1) Compute the marginal $f_Y(y)$:

$$\begin{array}{c}{f}_Y(y)={\int }_y^{1}6ydx\\ \\ \gg \gg {6yx|}_y^1\gg \gg 6y(1-y)\end{array}$$

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(2) Apply the conditional density formula:

$$\begin{array}{c}{f}_{X|Y}(x|y)=\frac{f_{X,Y}(x,y)}{f_Y(y)}\\ \\ \gg \gg \frac{6y}{6y(1-y)}\gg \gg \frac{1}{1-y}\end{array}$$ $$f_{X|Y}(x|y)=\{\begin{array}{cc}{\displaystyle \frac{1}{1-y}}& 0\le y\le x\le 1\\ 0& \text{otherwise}\end{array}$$Link to original

02

04

Cashews in a can

A nut company markets cans of mixed nuts containing almonds, cashews, and peanuts. Suppose the net weight of each can is exactly $1\mathrm{lb}$, but the weight contribution of each type of nut is random. Let $X$ be the weight of almonds in a selected can and $Y$ the weight of cashews. The joint PDF of $X$ and $Y$ is given below:

$$f_{X,Y}(x,y)=\{\begin{array}{cc}24xy& x,y\in [\mathrm{0,1}],x+y\le 1\\ 0& \text{otherwise}\end{array}$$

Suppose that the weight of cashews in a particular can is $0.5\mathrm{lbs}$. Calculate the probability that the weight of almonds in this can is more than $0.3\mathrm{lbs}$.

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Solution

Solutions - 5250-04

(1) Compute the marginal $f_Y(y)$:

$$\begin{array}{c}{f}_Y(y)={\int }_0^{1-y}24xydx\\ \\ \gg \gg {12x^{2}y|}_0^{1-y}\gg \gg 12{(1-y)}^{2}y\end{array}$$

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(2) Compute $f_{X|Y}(x|0.5)$:

$$\begin{array}{c}{f}_{X|Y}(x|0.5)=\frac{f_{X,Y}(x,0.5)}{f_Y(0.5)}\\ \\ \gg \gg \frac{24x\cdot \frac{1}{2}}{12{\left(\frac{1}{2}\right)}^2\cdot \frac{1}{2}}\gg \gg \frac{12x}{3/2}\gg \gg 8x\end{array}$$

The support is $x+0.5\le 1$, so $0\le x\le 0.5$.

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(3) Compute $P[X>0.3|Y=0.5]$:

$$\begin{array}{c}P[X>0.3|Y=0.5]={\int }_{0.3}^{0.5}8xdx\\ \\ \gg \gg {4x^2|}_{0.3}^{0.5}\\ \\ \gg \gg 4{(0.5)}^2-4{(0.3)}^2\gg \gg 1-0.36\gg \gg 0.64\end{array}$$Link to original

03

05

New and returning customers

A sales representative will randomly select and call 2 customers. The representative’s goal is to get each customer to complete a satisfaction survey. Each of these customers is categorized as “new” or “returning.” 70% of customers are new and 30% are returning. Let $X$ be the number of new customers that are called.

(a) Construct the PMF of $X$, $P_X(x)$.

The probability of any new customer completing the survey is 0.15, and the probability of any returning customer completing the survey is 0.20. (Customers operate independently.) Let $Y$ be the number of new customers that complete the survey.

(b) Construct $P_{Y|X}(y|0)$, $P_{Y|X}(y|1)$, and $P_{Y|X}(y|2)$. (You should construct 3 separate PMFs.)

(c) Construct the joint PMF of $X$ and $Y$.

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Solution

Solutions - 5250-05

(a)

$$\begin{array}{cccc}x& 0& 1& 2\\ P_X(x)& {0.3}^2& 2(0.7)(0.3)& {0.7}^2\\ & =0.09& =0.42& =0.49\end{array}$$

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(b)

$$\begin{array}{cc}y& 0\\ P_{Y|X}(y|0)& 1\end{array}\begin{array}{ccc}y& 0& 1\\ P_{Y|X}(y|1)& 0.85& 0.15\end{array}\begin{array}{cccc}y& 0& 1& 2\\ P_{Y|X}(y|2)& {(0.85)}^2& 2(0.85)(0.15)& {0.15}^2\end{array}$$

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(c)

Multiply $P_{Y|X}(y|x)\cdot P_X(x)$ for each cell:

$$\begin{array}{cccc}Y\downarrow X\to & 0& 1& 2\\ 0& 0.09& \begin{array}{c}0.42(0.85)\\ =0.357\end{array}& \begin{array}{c}0.49({0.85}^2)\\ =0.354\end{array}\\ 1& 0& \begin{array}{c}0.42(0.15)\\ =0.063\end{array}& \begin{array}{c}0.49\cdot 2(0.85)(0.15)\\ =0.125\end{array}\\ 2& 0& 0& \begin{array}{c}0.49{(0.15)}^2=0.011\end{array}\end{array}$$Link to original

Conditional expectation

04

02

Conditional distribution and expectation from joint PDF

Suppose that $X$ and $Y$ have the following joint PDF:

$$f_{X,Y}(x,y)=\{\begin{array}{cc}cxy& 0<y<1,0<x<y\\ 0& \text{ otherwise }\end{array}$$

Notice that the range of possibilities for $x$ depends on the value of $y$.

First, show that $c=8$. Then compute:

(a) $f_X$ $$ (b) $f_{Y|X}$ $$ (c) $E[Y|X=0.5]$ $$ (d) $E[Y|X]$

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Solution

Solutions - 5260-02

Integrate the joint PDF and solve for $c$:

$$\begin{array}{c}{\int }_0^1{\int }_0^{y}cxydxdy=1\\ \\ \gg \gg {\int }_0^1\frac{c{y}^3}{2}dy\gg \gg \frac{c}{8}=1\gg \gg c=8\end{array}$$

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(a)

Integrate the joint PDF with respect to $y$ to obtain $f_X$:

$$\begin{array}{c}{f}_X(x)={\int }_x^{1}8xydy\\ \\ \gg \gg 8x{\left[\frac{y^2}{2}\right]|}_x^1\gg \gg 4x(1-x^2)\end{array}$$

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(b)

Apply the conditional density formula:

$$\begin{array}{c}{f}_{Y|X}(y|x)=\frac{f_{X,Y}(x,y)}{f_X(x)}\\ \\ \gg \gg \frac{8xy}{4x(1-x^2)}\gg \gg \frac{2y}{1-x^2},x<y\end{array}$$

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(c)

(1) Plug in $x=0.5$ into the conditional density from part (b):

$$\begin{array}{c}{f}_{Y|X}(y|0.5)=\frac{2y}{1-{0.5}^2}\gg \gg \frac{8y}{3}\end{array}$$

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(2) Compute the conditional expectation:

$$\begin{array}{c}E[Y|X=0.5]={\int }_{0.5}^1\frac{8y^2}{3}dy\\ \\ \gg \gg \frac{8}{3}{\left[\frac{y^3}{3}\right]|}_{0.5}^1\gg \gg {\displaystyle \frac{7}{9}}\end{array}$$

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(d)

Integrate the conditional density from part (b) to find $E[Y|X=x]$:

$$\begin{array}{c}E[Y|X=x]={\int }_x^1\frac{2y^2}{1-x^2}dy\\ \\ \gg \gg \frac{2}{1-x^2}{\left[\frac{y^3}{3}\right]|}_x^1\gg \gg {\displaystyle \frac{2(1-x^3)}{3(1-x^2)}}\end{array}$$Link to original