Suppose $X_{1}$ and $X_{2}$ and i.i.d random variables. Consider $K = X_{1}X_{2}$. Then does $f_{K}(k) = f_{X_{1}}(x_1) \cdot f_{X_{2}}(x_2)$?
Pdf of a product
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1How about reading your probability textbook for a while and trying to understand what it says about solving such questions? Your several questions in the past few hours reveal wild misconceptions about standard basic notions, and seem to indicate that some serious self-study and trying to understand some basic ideas before asking another question on math.SE will probably serve you better. – 2012-03-12
1 Answers
Let $X_1$ and $X_2$ be identically distributed independent random variables. Let $f$, $f_{X_1}$ and $f_{X_2}$ be the joint density, density of $X_1$ and that of $X_2$ respectively.
We are interested in the density of $Z=X_1X_2$.
Let $z \in \Bbb R$. Define $A_z :=\{(x_1,x_2) \mid x_1x_2 \le z\}$. Observe the following in succession.
- $P(Z \le z)=P(Z \in A_z)$
- Note that if $x_1>0$, we have that $x_2 \le \dfrac{z}{x_1}$ and if $x_1 \lt 0$ we have that $x_2 \gt \dfrac{z}{x_1}$. Consequently, we can write $A_z$ as disjoint union of two sets.
$A_z=\{(x_1,x_2) \mid x_1 \gt 0~~\mbox{and} x_2 \lt \frac{z}{x_1} \}\bigsqcup \{(x_1,x_2) \mid x_1 \lt 0~~\mbox{and} x_2 \gt \frac{z}{x_1} \}$
$\begin{align} P(Z \le z) &= \iint_{A_z} f(x_1,x_2) dx_1 dx_2 \\&= \int _0 ^\infty \left(\int_{-\infty}^{z/x_1} f(x_1,x_2) dx_2\right)dx_1+\int_{-\infty}^0 \left(\int_{z/x_1}^\infty f(x_1,x_2) dx_2\right)dx_1\\&\overset{(1)}{=}\int _0 ^\infty \left(\int_{-\infty}^{z} \frac 1 x_1 f(x_1,\frac v x_1) dv\right)dx_1+\int_{-\infty}^0 \left(\int_{z}^{-\infty} \frac 1 x_1 f(x_1,\frac v x_1) dv\right)dx_1\\&=\int _0 ^\infty \left(\int_{-\infty}^{z} \frac 1 x_1 f(x_1,\frac v x_1) dv\right)dx_1+\int_{-\infty}^0 \left(\int_{-\infty}^z \left(-\frac 1 x_1\right) f(x_1,\frac v x_1) dv\right)dx_1 \\ &=\int_{-\infty}^\infty \left(\int_{-\infty}^z\left| \dfrac{1}{x_1} \right| f(x_1, \frac v x_1)dv \right) dx_1\\&\overset{(2)}{=}\int_{-\infty}^z \left(\int_{-\infty}^\infty\left| \dfrac{1}{x_1} \right| f(x_1, \frac v x_1)dx_1\right) dv \end{align}$
From the definition it follows that $f_{X_1X_2}(z)=\int_{-\infty}^{\infty}\left| \dfrac{1}{x}\right|f(x, \frac z x) dx$
Now use the fact that they are independent random variables to see:
$f_{X_1X_2}(z)=\int_{-\infty}^\infty \left| \frac 1 x\right| f_{X_1}(x) f_{X_2}(\frac z x) dx$
$(1)$ Make a change of variables in the inner integrals as follows: $x_2=\dfrac{v}{x_1}$
$(2)$ Change the order of integration by appealing to Fubini's Theorem.
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0This is from the first principles and you don't need to rely on any extra machinery. I personally prefer this because OP seems relatively new to the topic. He should get used to this kind of an integration before having to make use of shorter routes. – 2012-03-12