W11 Homework A

Due date: Friday 3/27, 11:59pm

Conditional distribution

01

01

Conditional density from joint density

Suppose that $X$ and $Y$ have joint probability density given by:

$$f_{X,Y}(x,y)=\{\begin{array}{cc}{\displaystyle \frac{12}{5}x(2-x-y)}& x,y\in [\mathrm{0,1}]\\ & \\ 0& \text{ otherwise }\end{array}$$

(a) Compute $f_{X|Y}(x|y)$, for $y\in [\mathrm{0,1}]$.

(b) Compute $P[X>1/2|Y=y]$.

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Solution

Solutions - 5250-01

(a)

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

$$\begin{array}{c}{f}_Y(y)=\frac{12}{5}{\int }_0^1(2x-x^2-xy)dx\\ \\ \gg \gg \frac{12}{5}{[x^2-\frac{x^3}{3}-\frac{x^{2}y}{2}]|}_0^1\\ \\ \gg \gg \frac{12}{5}(\frac{2}{3}-\frac{y}{2})\end{array}$$

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(2) Apply the conditional density formula:

$$\begin{array}{c}{f}_{X|Y}(x|y)=\frac{f_{X,Y}(x,y)}{f_Y(y)}\\ \\ \gg \gg \frac{\frac{12}{5}x(2-x-y)}{\frac{12}{5}(\frac{2}{3}-\frac{y}{2})}\gg \gg \frac{6x(2-x-y)}{4-3y}\end{array}$$ $$f_{X|Y}(x|y)=\{\begin{array}{cc}{\displaystyle \frac{6x(2-x-y)}{4-3y}}& x,y\in [0,1]\\ 0& \text{otherwise}\end{array}$$

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(b)

Integrate the conditional density over $(\frac{1}{2},1)$:

$$\begin{array}{c}P[X>\frac{1}{2}|Y=y]={\int }_{1/2}^1\frac{6x(2-x-y)}{4-3y}dx\\ \\ \gg \gg \frac{6}{4-3y}{\int }_{1/2}^1(2x-x^2-xy)dx\\ \\ \gg \gg \frac{6}{4-3y}{[x^2-\frac{x^3}{3}-\frac{x^{2}y}{2}]|}_{1/2}^1\\ \\ \gg \gg {\displaystyle \frac{11-9y}{4(4-3y)}}\end{array}$$ $$P[X>\frac{1}{2}|Y=y]=\{\begin{array}{cc}{\displaystyle \frac{11-9y}{4(4-3y)}}& y\in [0,1]\\ 0& \text{otherwise}\end{array}$$Link to original

02

03

Time till recharge

Let $X$ denote the amount of time (in hours) that a battery on a solar calculator will operate properly before needing to be recharged by exposure to light. The function below is the PDF of $X$.

$$f(x)=\{\begin{array}{cc}{\displaystyle \frac{50}{6}}{x}^{-3}& 2<x<10\\ & \\ 0& \text{ otherwise }\end{array}$$

Suppose that a calculator has already been in use for 5 hours. Find the probability it will operate properly for at least another 2 hours.

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Solution

Solutions - 5250-03

(1) Compute $P[B]=P[X>5]$:

$$\begin{array}{c}P[X>5]={\int }_5^{10}\frac{50}{6}{x}^{-3}dx\\ \\ \gg \gg {\frac{-25}{6}{x}^{-2}|}_5^{10}\\ \\ \gg \gg \frac{-25}{6}({10}^{-2}-5^{-2})\gg \gg 0.125\end{array}$$

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(2) Compute the conditional density $f_{X|B}$:

$$f_{X|B}(x)=\frac{\frac{50}{6}{x}^{-3}}{0.125},5<x<10$$

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(3) Compute $P[X>7|X>5]$:

$$\begin{array}{c}P[X>7|X>5]={\int }_7^{10}\frac{\frac{50}{6}{x}^{-3}}{0.125}dx\\ \\ \gg \gg \frac{1}{0.125}\cdot \frac{50}{6}{\left[\frac{-1}{2}{x}^{-2}\right]}_7^{10}\\ \\ \gg \gg \frac{-100}{3}({10}^{-2}-7^{-2})\gg \gg \approx 0.3469\end{array}$$Link to original

Conditional expectation

03

01

Conditional distribution and expectation from joint PMF

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

$$P_{X,Y}(k,\ell )=\{\begin{array}{cc}c& k=\mathrm{1,2,3,4};\ell =1,\dots ,k\\ 0& \text{ otherwise }\end{array}$$

Notice that the possibilities for $\ell$ depend on the choice of $k$.

First, show that $c=1/10$. Then compute:

(a) $P_X$ $$ (b) $P_{Y|X}$ $$ (c) $E[Y|X=4]$ $$ (d) $E[Y|X]$

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Solution

Solutions - 5260-01

Find the possible combinations of $k$ and $\ell$. For a given value of $k$, there are $k$ possible options for $\ell$. In total there are $1+2+3+4=10$ combinations, so $c=\frac{1}{10}$.

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(a)

$$P_X(k)=\{\begin{array}{cc}{\displaystyle \frac{k}{10}}& k=1,2,3,4\\ 0& \text{otherwise}\end{array}$$

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(b)

(1) Apply the conditional PMF formula:

$$\begin{array}{c}{P}_{Y|X}(\ell |k)=\frac{P_{X,Y}(k,\ell )}{P_X(k)}\\ \\ \gg \gg \frac{c}{kc}\gg \gg \frac{1}{k}\end{array}$$

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(2) State the conditional PMF:

$$P_{Y|X}(\ell |k)=\{\begin{array}{cc}{\displaystyle \frac{1}{k}}& \ell =1,\dots ,k\\ 0& \text{otherwise}\end{array}$$

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(c)

Compute the conditional expectation:

$$\begin{array}{c}E[Y|X=4]=\sum _{\ell =1}^4\frac{\ell }{4}\gg \gg {\displaystyle \frac{5}{2}}\end{array}$$

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(d)

Find the general formula for conditional expectation:

$$\begin{array}{c}E[Y|X]=\sum _{\ell =1}^x\frac{\ell }{x}\\ \\ \gg \gg \frac{1}{x}\sum _{\ell =1}^x\ell \gg \gg \frac{1}{x}\cdot \frac{x(x+1)}{2}\gg \gg {\displaystyle \frac{x+1}{2}}\end{array}$$Link to original