# Find the variance of $Y = X_1 X_2 / X_3$ where $X_i \sim$ Uniform(1,2) for $i = 1,2,3$

The problem is very straightforward and it says:

Let $$X_1,X_2,X_3$$ independent random variables with a common density $$f_{x_i} \sim Uniform(1,2)$$. Define $$Y = X_1 X_2 / X_3$$ Find the numeric value of $$var(Y)$$.

I propose the following transformation in order to compute the density of $$Y_1$$ through the margin distribution, and then compute the variance of $$Y_1$$:

$$Y_1 = X_1 X_2 / X_3$$ $$Y_2 = X_2/X_3$$ $$Y_3 = X_3$$

But I'm not getting any results. Is there an easy path to follow?

Disclaimer: I tried to solve the problem with this technique because I wanted to practice the topic of transformation of random variables.

• You have asked two different questions: the question in the title is to find the distribution of $Y$; the question in the text is to find $\text{Var}(Y)$. It is quite easy to find the latter without the former; though if you need the former, then you need the former. Oct 25, 2021 at 5:44
• As $X_1,X_2,X_3$ are independent, $E\left[Y\right]=E\left[\frac{X_1X_2}{X_3}\right]=E\left[X_1\right]E\left[X_2\right]E\left[\frac1{X_3}\right]$. Similarly find $E\left[Y^2\right]$ and hence the variance. Please add the self-study tag and read its wiki. Oct 25, 2021 at 9:55
• Inspirationally a la Ramanujan, looks like the variance should be approximately half of Euler's constant. Oct 25, 2021 at 11:40
• Taking logarithms of everything ought to be revealing ;-). However, your title question and post ask two different things: finding the variance is a matter of straightforward integration (because the uniform distribution has such a simple PDF). Which question are you trying to ask?
– whuber
Oct 25, 2021 at 13:41
• Thanks for all your feedback. I already have edited my question in order to get more accurate answers. I tried to find the variance of Y by proposing a transformation of random variables since i wanted to practice that topic. Oct 25, 2021 at 18:30

One easy path is to apply standard definitions. In particular, when the $$X_i$$ are independent and identically distributed with a density $$f,$$ the moment of $$Y=X_1X_2/X_3 = y(X_1,X_2,X_3)$$ of order $$k$$ is given by

$$\mu_k(Y) = \iiint y(x_1,x_2,x_3)^kf(x_1)f(x_2)f(x_3)\,\mathrm{d}x_1 \mathrm{d}x_2 \mathrm{d}x_3.$$

For the Uniform$$[1,2]$$ distribution, $$f(x) = 1$$ for $$1\le x \le 2$$ (and otherwise is zero). The domain of integration thereby reduces to the cube $$[1,2]^3,$$ giving

$$\mu_k(Y) = \int_1^2\int_1^2\int_1^2 \left(\frac{x_1x_2}{x_3}\right)^k\,\mathrm{d}x_1 \mathrm{d}x_2 \mathrm{d}x_3.$$

The integrand factors, enabling this triple integral to be computed as a product of three elementary univariate integrals,

$$\mu_k(Y) = \int_1^2x_1^k\,\mathrm{d}x_1 \int_1^2x_2^k\,\mathrm{d}x_2\int_1^2 x_3^{-k}\, \mathrm{d}x_3 = \left(\frac{x_1^{k+1}}{k+1}\bigg|^2_1\right) \left(\frac{x_2^{k+1}}{k+1}\bigg|^2_1\right) \left(\frac{x_3^{1-k}}{1-k}\bigg|^2_1\right)$$

(understanding that any expression evaluating to "$$x^0/0$$" means $$\log(x)$$).

Apply this result to the cases $$k=1,2$$ and plug them into the formula

$$\operatorname{Var}(Y) = \mu_2(Y) - \mu_1(Y)^2.$$

Graphical comment: A simulation (in R) may give you some clues.

set.seed(2021);  m = 10^5
x1 = runif(m, 1,2)
x2 = runif(m, 1,2)
x3 = runif(m, 1,2)
y1 = x1*x2/x3
summary(y1);  sd(y1)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
0.5134  1.1573  1.4795  1.5611  1.8731  3.8708
 0.5396972  # aprx SD(X1), about 2 place accuracy
y2 = x2/x3
summary(y2)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
0.5026  0.8214  1.0018  1.0407  1.2195  1.9966 par(mfrow=c(2,2))  # enable 4 panel plot
hist(y1, prob=T, col="skyblue2")
hist(y2, prob=T, col="skyblue2")
hist(x1*x2, prob=T, col="skyblue2")
plot(y1,y2, pch=".")
abline(v=1:2, col="red")