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Let be $X,Y$ two independent random variables having the same distribution (the following is the density of this distribution)

$$f(t)= \frac{1}{t^2} \,\,\, \text{for $t>1$}$$

Calculate the density of $Z=(XY)^{1/2}$. Thanks

afedder
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White_noyze
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  • Can you add some of your own thoughts and the attempts that you have made to solve the problem? – afedder May 28 '14 at 22:03
  • What have you tried? I've already shown you how to do the other question. Can you try to apply the same idea to do this one? – heropup May 28 '14 at 22:04
  • Draw a picture of the plane and identify where the joint density "lives." Let $Z=\sqrt{XY}$. We find the cdf of $Z$. So we want $\Pr(XY\le z^2)$. For some fixed $z$, draw the hyperbola $xy=z^2$. We want the probability of landing below the hyperbola (but of course to the right of $x=1$, and above $y=1$). A not hard integral. – André Nicolas May 28 '14 at 22:29
  • @AndréNicolas i got that point, i mean the hyperbola, but still have problems difining the integral – White_noyze May 28 '14 at 22:40
  • @heropup do you mean "by conditioning" or by "applying the convolution formula"? – White_noyze May 28 '14 at 22:41
  • Integrate the joint density, first with respect to $y$, then with respect to $x$. The integral with respect to $y$ goes $y=1$ to $y=z^2/x$. Then $x$ goes from $x=1$ to $x=z^2$. – André Nicolas May 28 '14 at 22:44
  • sorry @AndréNicolas but i dont get you... is it correct the joint i posted below $f(z,w)=2(z^3w)$^$(-1)$? why do i have to integrate w.r.t. $y$ and wrt $x$ and why does $x$ go from 1 to $z^2$? – White_noyze May 28 '14 at 22:59
  • @White_noyze the joint density of $X,Y$ is $$f_{X,Y}(x,y) = \frac{1}{x^2y^2}$$ for $x,y>1$. – afedder May 28 '14 at 23:02
  • The joint density is just the product of the individual densities. And the limits of integration come from a careful picture. The hyperbola meets $y=1$ and $x=z^2$. – André Nicolas May 28 '14 at 23:20

2 Answers2

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Refer to the following post for the strategy to find the density of $XY$: PDF of product of variables? Let $Z=XY$ and we want to find the density of $W=Z^{1/2}$. To this end, apply the technique of computing the cumulative distribution function and then differentiating this to find the pdf. To start, suppose we know the density function of $Z$, which we'll call $f_Z$, and also we will label the cumulative distribution functions of $Z$ and $W$ as $F_Z$ and $F_W$, respectively (note that we will not actually need to know $F_W$ in the end, but we will need $F_Z$ to determine $f_Z$). Additionally, we will assume $Z>0$ (if you want to include $Z < 0$, this would be including complex random variables). Then, $$\begin{align} F_W(w) &= P(W \leq w) \\&= P(Z^{1/2} \leq w) \\&= P(Z \leq w^2) \\&= F_Z(w^2) \,\,,\end{align}$$ where the third equality uses the assumption $Z > 0$. Differentiating both sides of the above, we obtain $$f_W(w) = \frac{\partial}{\partial w}F_Z(w^2)=2w \, f_Z(w^2)\,\,.$$

In regards to finding the density $f_Z$, here is the approach we will exploit. With the assumption that $X,Y$ are nonnegative, we have $$\begin{align}F_Z(z)= P(Z \leq z) &=\int\int_{z \,|\, z \geq xy} f_{X,Y}(x,y)\,\text{d}x\text{d}y \\&= \int\int_{z \,|\, z \geq xy} f_{X}(x)f_Y(y)\,\text{d}x\text{d}y \\&= \int\int_{z \,|\, z \geq xy} \frac{1}{x^2y^2} \text{d}x\text{d}y \\&= \int_{1}^z\int_{1}^{z/x} \frac{1}{x^2y^2} \text{d}y\text{d}x \,\,.\end{align}$$ Then, once we compute this integral, $f_Z(z) = F_Z'(z)$ and we can apply the above to find the density of $W$, which is what we wanted.

Community
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afedder
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$$\begin{align*} F_Z(z) &= \Pr[Z \le z] = \Pr[XY \le z^2] \\ &= \int_{y=1}^{z^2} \Pr[X \le z^2/y \mid Y = y]f_Y(y) \, dy \\ &= \int_{y=1}^{z^2} \int_{x=1}^{z^2/y} \frac{1}{x^2} \cdot \frac{1}{y^2} \, dx \, dy \\ &= \int_{y=1}^{z^2} \frac{1}{y^2} \biggl(1 - \frac{y}{z^2}\biggr) \, dy \\ &= \biggl[ -\frac{1}{y} - \frac{1}{z^2} \log y \biggr]_{y=1}^{z^2} \\ &= 1 - \frac{1 + 2 \log z}{z^2}. \end{align*}$$

Therefore, $$f_Z(z) = \frac{4 \log z}{z^3}, \quad z > 1.$$

heropup
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