$Y = \frac{X_1 X_2}{X_3}$ where $X_i\sim U(0,1)$ and $X_1,X_2,X_3$ are i.i.d

I need to calculate $Var(Y)$ and $Var[Y|X_3=1.7]$

I know that for each $X_i$,



But I'm not shure how to proceed, neither do I know how to calculate the PDF of Y.


I know that $f_{X_1,X_2, X_3}(x_1, x_2, x_3) = 1 \Bbb1_{(0,1) x (0,1), (0,1)} (x_1, x_2, x_3)$

I thought that since they are independen maybe there were some tricks derived from the expected value, since $E[X_1 X_2] = E[X_1] E[X_2]$

  • $\begingroup$ @SaucyO'Path Well, since the three are i.i.d., $f_{X_1, X_2, X_3}(x_1, x_2, x_3) = 1 for(0,1) x (0,1) x (0,1).$ $\endgroup$ – Leslie Brenes May 21 at 22:45
  • $\begingroup$ As long as you state it. $\endgroup$ – Saucy O'Path May 21 at 22:46
  • $\begingroup$ @LeslieBrenes when did you mention independence ? $\endgroup$ – Graham Kemp May 21 at 22:46
  • $\begingroup$ $X_3=1.7$ is a null event, especially since $X_3$ is supposed to be uniform(0,1). $\endgroup$ – user10354138 May 21 at 22:49
  • $\begingroup$ $\frac{X_1X_2}{X_3}$ is not $L^1$, therefore $\operatorname{Var}(X)$ is not defined. $\endgroup$ – Saucy O'Path May 21 at 23:08

Please see this answer for the variance of the product of independent random variables.

With that result, you can easily answer your second question, which is basically $\frac{1}{1.7^2} \mathrm{Var}[X_1 \, X_2]$.

Again using the result in the linked answer, the first question boils down to $\mathrm{Var}[X_1 \, X_2 \, Z_3]$ where $Z_3$ follows an Inverse uniform distribution. Please note that the variance is infinite in your case (while it would be ok if you had $X_3 \sim U(a,b)$ with $a>0$).


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