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Let the joint PDF of $(X, Y)$ be defined by

$$f(x, y) = \frac{c}{y} \text{exp}\left(-y -\frac{x}{y}\right) $$

for $x > 0$ and $y > 0 $

(a) Determine the constant $c$

(b) Find the marginal PDF of $Y$

(c) Find the conditional pdf of $X$ given $Y = y$

(d) Compute $\mathbb{E}[X^2 \mid Y = y]$

(a)

Ok, so using the fact that the sum of probability density functions equals $1$, I computed the value of $c$ as follows:

$$ \begin{align*} 1 = c \int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{y} \text{exp}\left(-y - \frac{x}{y}\right) \mathop{dx} \mathop{dy} \\ = c\int_{0}^{\infty} \frac{1}{y} \int_{0}^{\infty}\text{exp}\left(-y - \frac{x}{y} \right) \mathop{dx} \mathop{dy} \\ c\int_{0}^{\infty} \text{exp}(-y) \mathop{dy} \\ = c \\ \Longleftrightarrow \boxed{c = 1} \end{align*} $$

(b) I compute the marginal PDF of $Y$ as follows:

$$ \begin{align*} f_{Y}(y) = \int_{0}^{\infty} f(x, y) \mathop{dx} \\ = \int_{0}^{\infty} \frac{1}{y} \text{exp} \left(-y - \frac{x}{y}\right) \mathop{dx} \\ = \frac{1}{y} \int_{0}^{\infty} \text{exp}\left(-y - \frac{x}{y} \right) \mathop{dx} \\ = \frac{1}{y} \cdot y = 1, \end{align*} $$

but I'm pretty sure this is wrong. Can someone please check this part of the work for me? If it's right, then I'd have

$$ \begin{align*} f_{X|Y} = \frac{f(x, y)}{f_{Y}(x, y)} = \frac{1}{y} \text{exp} \left(-y - \frac{x}{y}\right) \end{align*} $$ for $x, y > 0$ for part $(c)$.

(d): To compute $\mathbb{E}[X^2 \mid Y = y]$, we use the conditional pdf of $x$ given $Y = y$. So,

$$E[X^2 \mid Y = y] = \int_{0}^{\infty}x^{2}\cdot \frac{\text{exp}(-x/y)}{y} \mathop{dx} = 2y^{2}$$

(I omitted the steps for integration).

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  • $\begingroup$ When computing the marginal PDF you seem to have lost a factor of $\exp(-y)$. $\endgroup$
    – angryavian
    Oct 29, 2018 at 1:03

1 Answer 1

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Part $(a)$ is correct.

In part $(b)$ you have made a mistake whilst calculating the integral.

\begin{align*} f_{Y}(y) &= \int_{0}^{\infty} f(x, y) \mathop{dx} \\ &= \int_{0}^{\infty} \frac{1}{y} \text{exp} \left(-y - \frac{x}{y}\right) \mathop{dx} \\ &= \frac{1}{y} \int_{0}^{\infty} \text{exp}\left(-y\right) \text{exp}\left(-\frac{x}{y} \right) \mathop{dx} \\ &= \frac{\text{exp}(-y)}{y} \int_{0}^{\infty} \text{exp}\left( - \frac{x}{y} \right) \mathop{dx} \\ &= \frac{\text{exp}(-y)}{y} \cdot y \\ &= \text{exp}(-y). \end{align*}

For part $(c)$ you would then have that

$$ f_{X|Y}(x) = \frac{f(x, y)}{f_{Y}( y)} = \frac{\frac{1}{y} \text{exp} \left(-y - \frac{x}{y}\right)}{\text{exp}(-y)} $$ which reduces to (I'll leave the details up to you) $$f_{X|Y}(x) = \frac{\text{exp}(-x/y)}{y}.$$

Try and do part $(d)$.

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  • $\begingroup$ Can you please check? I've updated my post. $\endgroup$
    – user400359
    Oct 30, 2018 at 0:47
  • $\begingroup$ Indeed, looks right to me! $\endgroup$
    – Bergson
    Oct 30, 2018 at 2:54

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