Question

Let Y1 and Y2 be independent exponentially distributed random variables, each with mean 7. Find P(Y1 > Y2 | Y1 < 2Y2). (Enter your probability as a fraction.)

Answer 1

Y₁ and Y₂ are independent, so their joint density is

$f_{Y_1,Y_2}(y_1,y_2)=f_{Y_1}(y_1)f_{Y_2}(y_2)=\begin{cases}\frac1{49}e^{-\frac{y_1+y_2}7}&\text{for }y_1\ge0,y_2\ge0\\0&\text{otherwise}\end{cases}$

By definition of conditional probability,

P(Y₁ > Y₂ | Y₁ < 2 Y₂) = P((Y₁ > Y₂) and (Y₁ < 2 Y₂)) / P(Y₁ < 2 Y₂)

Use the joint density to compute the component probabilities:

• numerator:

$P((Y_1>Y_2)\text{ and }(Y_1$

$=\displaystyle\frac1{49}\int_0^\infty\int_{\frac{y_1}2}^{y_1}e^{-\frac{y_1+y_2}7}\,\mathrm dy_2\,\mathrm dy_1$

$=\displaystyle-\frac17\int_0^\infty\int_{-\frac{3y_1}{14}}^{-\frac{2y_1}7}e^u\,\mathrm du\,\mathrm dy_1$

$=\displaystyle-\frac17\int_0^\infty\left(e^{-\frac{2y_1}7} - e^{-\frac{3y_1}{14}}\right)\,\mathrm dy_1$

$=\displaystyle-\frac17\left(-\frac72e^{-\frac{2y_1}7} + \frac{14}3 e^{-\frac{3y_1}{14}}\right)\bigg|_0^\infty$

$=\displaystyle-\frac17\left(\frac72 - \frac{14}3\right)=\frac16$

• denominator:

$P(Y_1$

(I leave the details of the second integral to you)

Then you should end up with

P(Y₁ > Y₂ | Y₁ < 2 Y₂) = (1/6) / (2/3) = 1/4