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Suppose that $X1$ and $X2$ have a continuous joint distribution for which the joint PDF is as follows:

\begin{equation*} f(x_1,x_2) = \begin{cases} x_1 + x_2 & \text {for $0 < x_1 < 1,$ and $0 < x_2 < 1$}\\ 0 & \text {otherwise}\\ \end{cases} \end{equation*}

Find the PDF of $Y = X_1 X_2 $.

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  • $\begingroup$ Looked up wikipedia? $\endgroup$ – Karl Feb 1 '15 at 14:24
  • $\begingroup$ Can you please guide me through the steps ? $\endgroup$ – mike111 Feb 1 '15 at 15:21
  • $\begingroup$ What are the definitions and what haev you tried? $\endgroup$ – Alec Teal Feb 2 '15 at 10:35
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Let's do it. Let $F_Y$ the CDF of $Y$.

$$F_Y(t) = P(Y \leq t) = P \left(X_1 \, X_2 \leq t \right) = P \left(X_1 \leq \dfrac{t}{X_2} \right).$$

The question here is: when does the inequality happen?

$$X_1 \leq \dfrac{t}{X_2}, \quad \text{when?}$$

For that, we need to find the marginal PDF of $X_1$ and $X_2$. How? With their joint distribution.

$$f_{X_1}(x_1) = \int_0^1 (x_1 + x_2)\, dx_2 = 1/2 + x_1$$ $$f_{X_2}(x_2) = \int_0^1 (x_1 + x_2)\, dx_1 = 1/2 + x_2$$

Ok, now... when does this happen?

$$1/2 + x_1 \leq \dfrac{t}{1/2 + x_2}$$

With $x_1 = x_2$, find $x_1$ (or $x_2$) in terms of $t$.

$$x_1 = 1/2 \cdot (-1 + 2 \sqrt t). \tag{1} \label{exacto}$$

If you draw $1/2 \, + x_1$ and $t /( 1/2 + x_2)$ between $0$ and $1$ (do it!), you see that the inequality is true for some values of $t$.

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So the CDF of $Y$ is:

$$F_Y(t) = \int_0^1 \left( \int_0^{1/2 \, (-1+ 2\sqrt t)} (x_1 + x_2) \, dx_1 \right) dx_2 = \dfrac{1}{8} (4 t-1).$$

We need to find an interval for $t$. What is it? Is $t$ in $[0,1]$ or in $[1, 10]$?. Refer to $\eqref{exacto}$ to find $t$ when $x_1=0$ and $x_1=1$, so $t \in [1/4, 9/4]$. The PDF is:

$$f_Y(t) = \dfrac{d}{dt}F_Y (t) = \dfrac{1}{2}.$$

Check that:

$$\int_{1/4}^{9/4} \dfrac{1}{2}\, dt = 1.$$

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