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?

  • $\begingroup$ For $Z=X/Y$, see the answers to this question. The result is $$\begin{align*} F_Z(z) &= \int_{x=0}^{\infty}\int_{y=-\infty}^{zx} f_{X,Y}(x,y) \mathrm dy\ \mathrm dx + \int_{x=-\infty}^{0}\int_{y=zx}^{\infty} f_{X,Y}(x,y) \mathrm dy\ \mathrm dx,\\ f_Z(z) = \frac{\mathrm d}{\mathrm dz}F_Z(z) &= \int_{0}^{\infty} x\cdot f_{X,Y}(x,zx) \mathrm dx - \int_{-\infty}^{0} x\cdot f_{X,Y}(x,zx) \mathrm dx, \end{align*}$$ $\endgroup$ – Dilip Sarwate Jan 15 '13 at 16:06
  • $\begingroup$ 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. $\endgroup$ – Dilip Sarwate Jan 15 '13 at 16:14

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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  • $\begingroup$ 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 $f_Z(z)=\int f_X(zt)f_Y(t)tdt$. $\endgroup$ – Did Jan 15 '13 at 22:59
  • $\begingroup$ 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 $u(Z)$)? $\endgroup$ – NumberFour Jan 15 '13 at 23:03
  • $\begingroup$ Great, I didnt know about this approach. Thank you! $\endgroup$ – NumberFour Jan 15 '13 at 23:05

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