Question

given the following Y=X∙H, where X and H are two random variables with pdf
fX (x)=1/x e^(x^2 ) ,x≥0 ,and fH (h)=1/h e^(〖-3h^2〗^ )
What is the pdf fy|X (y|x)?
What is the pdf fY (y)?

Answer 1

Answer on Question #48725-Math-Statistics and Probability

Given the following $Y = X \cdot H$, where $X$ and $H$ are two random variables with pdf

$$f_X(x) = \frac{1}{x} e^{-x^2}, \quad x \geq 0, \quad \text{and} \quad f_H(h) = \frac{1}{h} e^{-3h^2}$$

What is the pdf $f_{Y|X}(y|x)$?

What is the pdf $f_{Y}(y)$?

Solution

$$f_{XH}(x, h) = f_X(x) f_H(h) = \frac{1}{x} e^{-x^2} \cdot \frac{1}{h} e^{-3h^2} = f_{X\frac{Y}{X}}(x, \frac{y}{x}).$$ $$f_{XY}(x, y) = f_{X\frac{Y}{X}}(x, \frac{y}{x}) \cdot |J|,$$

where

$$|J| = \left| \begin{array}{cc} \frac{\partial x}{\partial x} & \frac{\partial x}{\partial y} \\ \frac{\partial \left(\frac{y}{x}\right)}{\partial x} & \frac{\partial \left(\frac{y}{x}\right)}{\partial y} \end{array} \right| = \left| \begin{array}{cc} 1 & 0 \\ -\frac{y}{x^2} & \frac{1}{x} \end{array} \right| = \frac{1}{x}.$$

So

$$f_{XY}(x, y) = \frac{1}{x} \cdot \frac{1}{x} e^{-x^2} \cdot \frac{1}{\left(\frac{y}{x}\right)} e^{-3\left(\frac{y}{x}\right)^2} = \frac{1}{xy} e^{-x^2 - 3\left(\frac{y}{x}\right)^2}.$$

Then

$$f_{Y|X}(y|x) = \frac{f_{XY}(x, y)}{f_X(x)} = \frac{\frac{1}{xy} e^{-x^2 - 3\left(\frac{y}{x}\right)^2}}{\frac{1}{x} e^{-x^2}} = \frac{1}{y} e^{-3\left(\frac{y}{x}\right)^2}.$$ $$f_Y(y) = \int_0^\infty f_{XY}(x, y) \, dx = \frac{1}{y} \int_0^\infty \frac{1}{x} e^{-x^2 - 3\left(\frac{y}{x}\right)^2} \, dx = \frac{1}{y} K_0\left(2\sqrt{3}|y|\right),$$

where $K_0\left(2\sqrt{3}|y|\right)$ is the modified Bessel function of the second kind.

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