Variance of the Average of Two Half Normal Distributions

Question

Say I have two independent normal distributions (both with $\mu=0$, $\sigma=\sigma$) one for only positive values and one for only negatives so their pdfs look like:

$p(x, \sigma) = \frac{\sqrt{2}}{\sqrt{\pi} \sigma} exp(-\frac {x^2}{2 \sigma^2}), \forall x>0$ and

$p(y, \sigma) = \frac{\sqrt{2}}{\sqrt{\pi} \sigma} exp(-\frac{y^2}{2 \sigma^2}), \forall y<0$.

If I pluck samples from both and then take the average $ = \frac{x+y}{2}$ I would imagine the expected value of this average to be zero but I would imagine the variance would be less than the variance of the individual distributions because the averaging of a positive and negative number would "squeeze" the final distribution.

I think the correct way to calculate it is using the following integral.

$Var( \frac{x+y}{2}) = \frac{2}{ \pi \sigma^2} \int^{\infty}_{0} \int^{0}_{- \infty} \frac{(x + y)^2}{4}exp(-\frac {x^2}{2 \sigma^2}) exp(-\frac {y^2}{2 \sigma^2}) dx dy$

But I am not sure if I am over-simplifying it. Does that logic seem correct or am I missing something?

Thank you.

Edited to mention independence and correct formulae mistakes.

Answer 1

Your approach is workable (although the $\tfrac{1}{2\pi\sigma^2}$ should be $\tfrac{2}{\pi\sigma^2}$), but there's a much easier way @callculus pointed out. Since $X,\,Y$ are independent, $\operatorname{Var}(aX+bY)=a^2\operatorname{Var}X+b^2\operatorname{Var}Y=(a^2+b^2)^2\operatorname{Var}X$, so $\operatorname{Var}\tfrac{X+Y}{2}=\tfrac12\operatorname{Var}X=\sigma^2(\tfrac12-\tfrac{1}{\pi})$. (I've used a variance from here.)

Answer 2

I think you can use moment generating functions to be able to get $E\big(\big(\frac{X+Y}{2}\big)^2\big)$ and $E\big(\frac{X+Y}{2}\big)$ to calculate the variance.

$$M_\frac{X+Y}{2}(t)= E(e^{t\frac{X+Y}{2}}) = E\big(e^{\frac{t}{2}X + \frac{t}{2}Y}\big) = E\big(e^{\frac{t}{2}X}\big) + E\big(e^{\frac{t}{2}Y}\big)$$

You can calculate both of these independently:

$$E(e^{\frac{t}{2}}X) = \frac{1}{2\sqrt{\pi}}\int_0^\infty e^{\frac{t}{2}x}e^{-\frac{x}{2\sigma^2}}dx = \frac{1}{2\sqrt{\pi}}\int_0^\infty e^{{-\frac{1}{2}}\big(t^2\sigma^2 - tx + \frac{x^2}{\sigma^2}\big) + \frac{t^2\sigma^2}{2}}dx = e^\frac{t^2\sigma^2}{2}\int_0^\infty\frac{1}{2\sqrt{\pi}}e^{-\frac{(x-t\sigma)^2}{2\sigma^2}} $$

Note that the last integral represents a normal distribution that is centred in $t\sigma$ and since the Normal Distribution is symmetric this is half of the value. Since the integral of the PDF on the entire space is 1, it means that the above mentioned integral will integrate to $\frac{1}{2}$.

$$E(e^{\frac{t}{2}}X) = \frac{1}{2}e^\frac{t^2\sigma^2}{2}$$

In similar manner you can compute $E(e^{\frac{t}{2}}X)$ but in that case the integral would be from $-\infty$ to $0$.

After that you can take the first and second derivative and get the values required to calculate the variance.