The random vector X = $(X_1,X_2,X_3)'$ has density function

$f_{X}(x)=\begin{cases} \frac{2}{2e-5}.x_1^2.x_2.e^{x_1.x_2.x_3}, & for & 0 < x_1,x_2,x_3<1 \\[2ex] 0, & \text{otherwise} \end{cases}$

Determine the distribution of $X_1.X_2.X_3$:

I've tried getting the marginals and multiply them to get the pdf of


but I'm not sure if that could be done since I'm not sure if I can assume that $X_i$'s are all independent.

My main issue is with the bounds with min. I can't figure those out.

  • 1
    $\begingroup$ They are not independent. Even if they were, multiplying the marginal densities would not give you the density of the product $\endgroup$ – Henry Jan 27 at 10:45

First we write the CDF of $Y=X_1X_2X_3$, $F_Y(y)=P(Y\leq y)$ when $0\leq y \leq 1$: $$F_Y(y)=\int_0^{1}\int_0^1\int_0^1f(x_1,x_2,x_3)\mathbf{I}(x_1x_2x_3\leq y)dx_1dx_2dx_3$$

This can be written in three regions: (simplify the notation as $f(\mathbf{x})$ and $d\mathbf{x}$ to denote $f(x_1,x_2,x_3)$ and $dx_1dx_2dx_3$, in the same order. $$F_Y(y)=\int_0^y\int_0^1\int_0^1f(\mathbf{x})d\mathbf{x}+\int_y^1\int_0^{y/x_1}\int_0^1f(\mathbf{x})d\mathbf{x}+\int_y^1\int_{y/x_1}^1\int_0^{y/x_1x_2}f(\mathbf{x})d\mathbf{x}$$ Performing the integrations, if I don't have any computational mistakes, this CDF can be found as $$F_Y(y)=\frac{e^y(y^2-4y+5)-5}{2e-5}$$ which yields (for $0\leq y \leq 1$) $$f_Y(y)=\frac{dF_Y(y)}{dy}=\frac{e^y(y-1)^2}{2e-5}$$

  • $\begingroup$ my main issue is with the bounds that you have with min. I can't figure those out $\endgroup$ – Mahamad A. Kanouté Jan 28 at 3:05
  • $\begingroup$ like going from the first line to the second $\endgroup$ – Mahamad A. Kanouté Jan 28 at 3:06
  • $\begingroup$ I actually went from second to first :) while trying to represent it more compact, introduced a mistake. Now edited. However, the second integral is correct. Do you have problems with it? $\endgroup$ – gunes Jan 28 at 4:28
  • $\begingroup$ I now have a problem with the $F_Y(y)=\int_0^y\int_0^1\int_0^1f(\mathbf{x})d\mathbf{x}+\int_y^1\int_0^{y/x_1}\int_0^1f(\mathbf{x})d\mathbf{x}+\int_y^1\int_{y/x_1}^1\int_0^{y/x_1x_2}f(\mathbf{x})d\mathbf{x}$ $\endgroup$ – Mahamad A. Kanouté Jan 28 at 4:38
  • 1
    $\begingroup$ The first integral: When $x_1<y$, $x_2,x_3$ can be anything in $[0,1]$, still $x_1x_2x_3<y$. When $x_1>y$, there are two options. Second integral: If $x_2<y/x_1$, $x_3$ can be anything since $x_1x_2<y$. Third integral: When $x_2>y/x_1$, $x_1x_2$ becomes $>y$, and we need to consider $x_3$ 's where $x_1x_2x_3<y$. To be able to do that, we need to choose $x_3<y/x_1x_2$. $\endgroup$ – gunes Jan 28 at 4:44

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