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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1No. To begin with, the equality does not even make sense (who are these $x_1$ and $x_2$ guys?). – 2012-03-12
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0@DidierPiau: I thought if $X$ and $Y$ are independent then $f_{XY}(x,y) = f_{X}(x) \cdot f_{Y}(y)$? – 2012-03-12
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0Yes--and this is unrelated to what your question says. – 2012-03-12
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0@DidierPiau: So to find this pdf, we would have to do some sort of transformation? – 2012-03-12
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0@alexm Distinguish between $f_{X,Y}$, the _joint_ density of random variables $X$ and $Y$, and $f_{XY}$, the density of the single random variable $XY$. It is true that $f_{X,Y}(x,y) = f_X(x)f_Y(y)$, but $f_{XY}(x,y)$ makes no sense (the density of a single random variable should have a single argument, not two arguments) and thus $f_{XY}(x,y) = f_{X}(x) \cdot f_{Y}(y)$ does not make sense since the left side is a function of one variable, not two. – 2012-03-12
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0@DilipSarwate: So to find the pdf of $XY$, how would you do that? – 2012-03-12
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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 seems like a lot of effort. Can't you just do Jacobian stuff to the joint density? Take $Z = X_1/X_2$ and $X_1 = X_1$ then the absolute value of the jacobian of $(X_1, X_2) \to (X_1, Z)$ is $|x_1|^{-1}$ so we get $f_{X_1, Z} (x, z) = |x|^{-1} f(x) f(z/x)$ by change of variables; then integrate out. – 2012-03-12
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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