Solutions - 5200-05

(a)

Since $X$ and $Y$ are independent, the joint PDF is the product of the marginals:

$$f_{X,Y}(x,y)=\{\begin{array}{cc}(0.1e^{-0.1x})(0.1e^{-0.1y})=0.01e^{-0.1(x+y)}& x,y>0\\ 0& \text{otherwise}\end{array}$$

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(b)

(1) Expand the conditional probability:

$$\begin{array}{c}P[X-5>Y-15|X>5\text{AND}Y>15]=\frac{P[X>Y-10\text{AND}X>5\text{AND}Y>15]}{P[X>5\text{AND}Y>15]}\end{array}$$

The three conditions in the numerator are partially redundant because $Y>15$ implies $Y-10>5$ and therefore $X>5$ follows automatically. So we may drop the middle term:

$$\gg \gg \frac{P[X>Y-10\text{AND}Y>15]}{P[X>5\text{AND}Y>15]}$$

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(2) Evaluate the denominator using independence:

$$\begin{array}{c}P[X>5\text{AND}Y>15]=P[X>5]\cdot P[Y>15]\\ \\ \gg \gg e^{-0.1(5)}\cdot e^{-0.1(15)}\gg \gg e^{-0.5}\cdot e^{-1.5}\gg \gg e^{-2}\end{array}$$

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(3) Evaluate the numerator:

The region is $x>y-10$ and $y>15$, i.e. above $y=15$ and below $y=x+10$.

For the outer variable $x$, the lower bound is $x=5$ (where the line $y=x+10$ meets $y=15$). For a given $x$ value, $y$ ranges from $15$ to $x+10$.

Inner integral over $y$:

$$\begin{array}{c}{\int }_5^{\infty }{\int }_{15}^{x+10}0.01e^{-0.1(x+y)}dydx\\ \\ \gg \gg {\int }_5^{\infty }0.01e^{-0.1x}{\left[\frac{e^{-0.1y}}{-0.1}\right]}_{y=15}^{x+10}dx\\ \\ \gg \gg {\int }_5^{\infty }0.1e^{-0.1x}(e^{-1.5}-e^{-0.1x-1})dx\end{array}$$

Outer integral over $x$:

$$\begin{array}{c}0.1e^{-1.5}{\int }_5^{\infty }{e}^{-0.1x}dx-0.1e^{-1}{\int }_5^{\infty }{e}^{-0.2x}dx\\ \\ \gg \gg 0.1e^{-1.5}\cdot \frac{e^{-0.5}}{0.1}-0.1e^{-1}\cdot \frac{e^{-1}}{0.2}\\ \\ \gg \gg e^{-2}-0.5e^{-2}\gg \gg 0.5e^{-2}\end{array}$$

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(4) Plug into original ratio:

$$\begin{array}{c}\frac{0.5e^{-2}}{e^{-2}}\gg \gg {\displaystyle \frac{1}{2}}\end{array}$$

This result is what we would expect to obtain by applying the memoryless property of exponential distributions.