Question

Suppose thatthe joint density of X and Y is given by
(x,y)=e-(x/y).e(-y) ,0<x<∞,0<y<∞ and is o otherwise. Find P(X>1 / Y=y).

Answer 1

Answer on Question #44951 – Math – Statistics and Probability

Suppose that the joint density of $X$ and $Y$ is given by

$$f(x, y) = e^{-\left(\frac{x}{y}\right)} \cdot e^{-y}, \quad 0 < x < \infty, \quad 0 < y < \infty$$

and is otherwise. Find $P(X > 1 \mid Y = y)$.

Solution

$$P(X > 1 \mid Y = y) = \int_{1}^{\infty} f_{X \mid Y}(x \mid y) \, dx.$$ $$f_{X \mid Y}(x \mid y) = \frac{f(x, y)}{f_{Y}(y)} = \frac{e^{-\left(\frac{x}{y}\right)} \cdot e^{-y}}{\int_{0}^{\infty} e^{-\left(\frac{x}{y}\right)} \cdot e^{-y} \, dx} = \frac{e^{-\left(\frac{x}{y}\right)} \cdot e^{-y}}{e^{-y} \int_{0}^{\infty} e^{-\left(\frac{x}{y}\right)} \, dx} = \frac{e^{-\left(\frac{x}{y}\right)}}{y}.$$ $$P(X > 1 \mid Y = y) = \int_{1}^{\infty} f_{X \mid Y}(x \mid y) \, dx = \int_{1}^{\infty} \frac{e^{-\left(\frac{x}{y}\right)}}{y} \, dx = \frac{1}{y} \int_{1}^{\infty} e^{-\left(\frac{x}{y}\right)} \, dx = \frac{1}{y} \left(y e^{-\frac{1}{y}}\right) = e^{-\frac{1}{y}}.$$

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