$Y = \frac{X_1 X_2}{X_3}$ where $X_i$ is a uniform random variable

$$Y = \frac{X_1 X_2}{X_3}$$ 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 $$X_i$$,

$$E[X_i]=\frac{1}{2}$$

$$Var[X_i]=\frac{1}{12}$$

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

¿Tips?

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 $$E[X_1 X_2] = E[X_1] E[X_2]$$

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

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 $$a>0$$).