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Having two independent random variables $X$ with pdf $f_X$ and $Y$ with pdf $f_Y$ what is the correct way to derive the formula for $f_Z$ where $Z = X/Y$ and $f_W$ where $W = XY$ ?

I know that the classical convolution formula for (V = X+Y) is derived in the following way:

$$F_V(x) = P(V \lt x) = P(X+Y \lt x) = \iint_{X+Y \lt x} df_x df_y = \int_{-\infty}^{\infty}df_x\int_{-\infty}^{z-x}df_y = \int_{-\infty}^{\infty}f_x(z)f_y(z-x) dz$$

pdf is then given by $f_V = \frac{dF_V}{dx}$

If I go through the same process with $Z = X/Y$ and $W = XY$ what is the correct domain of integration in the last step? What would be the approach using characteristic functions?

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For $Z=X/Y$, see the answers to this question. The result is $$\begin{align*} F_Z(z) &= \int_{x=0}^{\infty}\int_{y=-\infty}^{zx} f_{X,Y}(x,y) \mathrm dy\ \mathrm dx + \int_{x=-\infty}^{0}\int_{y=zx}^{\infty} f_{X,Y}(x,y) \mathrm dy\ \mathrm dx,\\ f_Z(z) = \frac{\mathrm d}{\mathrm dz}F_Z(z) &= \int_{0}^{\infty} x\cdot f_{X,Y}(x,zx) \mathrm dx - \int_{-\infty}^{0} x\cdot f_{X,Y}(x,zx) \mathrm dx, \end{align*}$$ –  Dilip Sarwate Jan 15 '13 at 16:06
Incidentally, your "classical convolution formula" needs some work. The integral on the right gives the probability density function $f_{X+Y}(z)$, not the cumulative probability distribution function $F_{X+Y}(z)$ or $F_{X+Y}(x)$ as you write it. Note that the integral on the right is a function of $z$, not $x$, since the latter is the variable of integration and disappears when the limits are substituted. –  Dilip Sarwate Jan 15 '13 at 16:14

1 Answer 1

up vote 2 down vote accepted

Try this systematic approach which avoids the unnecessary complications which arise when one uses the détour through CDFs you seem to be mentioning.

For example, to compute the density $f_Z$ of $Z$, note that, for every bounded measurable function $u$, $$ \mathbb E(u(Z))=\iint u(x/y)f_X(x)f_Y(y)\mathrm dx\mathrm dy. $$ The change of variables $x=zt$, $y=t$, yields $\mathrm dx\mathrm dy=t\mathrm dz\mathrm dt$ hence $$ \mathbb E(u(Z))=\iint u(z)f_X(zt)f_Y(t)t\mathrm dz\mathrm dt. $$ This identity holds for every test function $u$ hence $$ f_Z(z)=\int f_X(zt)f_Y(t)t\mathrm dt. $$ If one writes correctly the density functions $f_X$ and $f_Y$, there is no problem of domain of integration. For example, if $X$ and $Y$ are i.i.d. uniform on $(0,1)$, then $f_X=f_Y=\mathbf 1_{(0,1)}$ hence, for every $z\gt0$, $$ f_Z(z)=\int \mathbf 1_{0\leqslant zt\leqslant1}\mathbf 1_{0\leqslant t\leqslant1}t\mathrm dt=\int_0^{\min(1,1/z)}t\mathrm dt=\frac{\min(1,1/z)^2}2, $$ that is, $$ f_Z(z)=\begin{cases}\frac12&\text{if}\ z\leqslant1,\\ \frac1{2z^2}&\text{if}\ z\geqslant1.\end{cases} $$

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Almost. You know that $E(u(Z))=\int u(z)f_Z(z)dz=\iint u(z)f_X(zt)f_Y(t)tdtdz$ for every $u$ hence $f_Z(z)=\int f_X(zt)f_Y(t)tdt$. –  Did Jan 15 '13 at 22:59
Do I understand correctly (3rd step) that since $\mathbb E(u(Z)) = \int u(Z)f_Z(t)\mathrm dt$ and the other equation says that $\mathbb E(u(Z))=\iint u(z)f_X(zt)f_Y(t)t\mathrm dz\mathrm dt$ I can just compare the integrands (getting rid of $u(Z)$)? –  NumberFour Jan 15 '13 at 23:03
Great, I didnt know about this approach. Thank you! –  NumberFour Jan 15 '13 at 23:05

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