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Given two independent random variables $X\sim N(0, \sigma^2), Y\sim N(0, \sigma^2)$, what is the distribution of a variable $Q = (X-Y)^2/4$ ? What would be the expected value and variance of $Q$?

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    $\begingroup$ The distribution of $Q$ depends on the joint distribution of $(X,Y)$, not just their marginal distributions. $\endgroup$
    – Glen_b
    Jul 7, 2022 at 2:59
  • $\begingroup$ Sorry for the confusing notation, hopefully my edit makes things more clear? $Q$ is proportional to the difference of two independent instances of the same distribution, not two separate distributions $\endgroup$
    – Galaxy
    Jul 7, 2022 at 3:22
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    $\begingroup$ That's surely not what you intended at all. $X - X$ is identically zero, and so $P(Q=0)=1$. I wasn't asking you to change notation. I was asking you to talk about the joint distribution of the two underlying variables in your question. Perhaps you intended that they be independent? $\endgroup$
    – Glen_b
    Jul 7, 2022 at 3:25
  • $\begingroup$ Ah ok sorry, maybe now it's cleared up? I did intend them to be independent $\endgroup$
    – Galaxy
    Jul 7, 2022 at 3:38
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    $\begingroup$ @Galaxy Usually, a normal distribution with variance $\sigma^2$ is written $N(0, \sigma^2)$. If you write $N(0,\sigma)$, do you really mean that the variance is $\sigma$ and not $\sigma^2$, i.e. you use the symbol $\sigma$, that is usually used for the standard deviation, as the symbol for the variance? $\endgroup$
    – frank
    Jul 7, 2022 at 5:54

1 Answer 1

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Below, I presume that you actually want X and Y to have variance $\sigma^2$, and not $\sigma$.

Since $N(0, \sigma^2)$ is symmetric over zero, the random variable $Y$ is equal to the random variable $-Y$. Thus $X-Y$ is just the sum of two independent normal distributions $N(0, \sigma^2)$. Using the general formula for the sum of two independent normal distributions $A\sim N(\mu_A, \sigma_A^2), B\sim N(\mu_B, \sigma_B^2)$: $$ A+B \sim N(\mu_A + \mu_B, \; \sigma^2_A + \sigma_B^2) $$

we get for $X-Y$: $$ \begin{align} X-Y &\sim N(0, 2\sigma^2),\\ \frac{X-Y}{\sqrt{2}\sigma} &\sim N(0, 1). \end{align} $$ Next, the square of a $N(0, 1)$ variable has a $\chi^2$ distribution of degree one, i.e. $\left(\frac{X-Y}{\sqrt 2\sigma}\right)^2$ is $\chi^2_1$-distributed. Furthermore $$ \begin{align} \frac{(X-Y)^2}{4} &= \frac{1}{4}\left(\frac{\sqrt 2 \sigma (X-Y)}{\sqrt 2 \sigma}\right)^2\\ &= \frac{\sigma^2}{2}\left(\frac{X-Y}{\sqrt 2 \sigma}\right)^2.\\ \end{align} $$ Thus, it is a scaled $\chi^2_1$-distribution, and the scaling factor is $\frac{\sigma^2}{2}$. Now, the $\chi^2_1$ distribution has mean equal to one, and variance equal to two. Since the expectation is linear, we have: $$ \begin{align} E\left[\frac{(X-Y)^2}{4}\right] &= \frac{\sigma^2}{2}E\left[\left(\frac{X-Y}{\sqrt 2 \sigma}\right)^2\right]\\ &=\frac{\sigma^2}{2}. \end{align} $$ The variance is quadratic, thus $$ \begin{align} Var\left[\frac{(X-Y)^2}{4}\right] &= \left(\frac{\sigma^2}{2}\right)^2 Var\left[\left(\frac{X-Y}{\sqrt 2 \sigma}\right)^2\right]\\ &=\frac{\sigma^4}{4}\cdot 2\\ &= \frac{\sigma^4}{2}. \end{align} $$

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  • $\begingroup$ Thank you! If you have the time, would you mind explaining why $Q$ only has one degree of freedom? My intuition tells me there should be either two since $Q$ is a function of two independent variables $X$ and $Y$, or three since if you multiply out the squared sum you would get $Q=(X^2+Y^2-2XY)/4$ which is a sum of three chi-squared distributions $\endgroup$
    – Galaxy
    Jul 7, 2022 at 8:58
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    $\begingroup$ @Galaxy $Z = (X-Y)/(\sqrt 2\sigma)$ is a single $N(0,1)$ random variable, so its square is $\chi^2_1$, i.e. of degree 1 (by definition, the degree is the number of random variables you sum the squares of). Note, that your $Q$ is a square of a sum, not a sum of squares. And if you multiply $Q$ out, this doesn't give a sum of squares of independent random variables. $\endgroup$
    – frank
    Jul 7, 2022 at 9:13

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