# Normal distribution $var(X^2+Y^2)$

There is a statistics problem I unfortunately have no idea where to start (I’m studying on my own so there is nobody I can ask, if I don’t understand something.

The question is

$X,Y$ i.i.d. $N(a,b^2); a=0; b^2=6; var(X^2+Y^2)=?$

Since you are dealing with IID normal data, it is worth generalising your problem slightly to look at the case where you have $X_1, ..., X_n \sim \text{IID N}(a, b^2)$ and you want $Q_n \equiv \mathbb{V}(\sum_{i=1}^n X_i^2)$. (Your question corresponds to the case where $n=2$.) As other users have pointed out, the sum of squares of IID normal random variables is a scaled non-central chi-squared random variable, and so the variance of interest can be obtained from knowledge of that distribution. However, it is also possible to obtain the required variance using ordinary moment rules, combined with knowledge of the moments of the normal distribution. I will show you how to do this below, in steps.

Finding the variance using moments of the normal distribution: Since the values $X_1, ..., X_n$ are IID (and taking $X$ to be a generic value from this distribution) you have:\begin{aligned}Q_n \equiv \mathbb{V}\Big( \sum_{i=1}^n X_i^2 \Big) &= \sum_{i=1}^n \mathbb{V}(X_i^2) \\[6pt]&= n\mathbb{V}(X^2) \\[6pt]&= n(\mathbb{E}(X^4) - \mathbb{E}(X^2)^2) \\[6pt]&= n(\mu_4' - \mu_2'^2), \\[6pt]\end{aligned} where we are denoting the raw moments as $\mu_k' \equiv \mathbb{E}(X^k)$. These raw moments can be written in terms of the central moments $\mu_k \equiv \mathbb{E}((X-\mathbb{E}(X))^k)$ and the mean $\mu_1' = \mathbb{E}(X)$ using standard conversion formulae, and we can then look up the central moments of the normal distribution and substitute them in.

Using the moment conversion formulae you should get:\begin{aligned}\mu_2' &= \mu_2 + \mu_1'^2, \\[6pt]\mu_3' &= \mu_3 + 3 \mu_1' \mu_2 + \mu_1'^3, \\[6pt]\mu_4' &= \mu_4 + 4 \mu_1' \mu_3 + 6 \mu_1'^2 \mu_2 + \mu_1'^4. \\[6pt]\end{aligned}For the distribution $X \sim \text{N}(a,b^2)$ we have mean $\mu_1' = a$ and higher-order central moments $\mu_2 = b^2$, $\mu_3 = 0$ and $\mu_4 = 3 b^4$. This gives us the raw moments:\begin{aligned}\mu_2' &= b^2 + a^2, \\[6pt]\mu_3' &= 3 a b^2 + a^3, \\[6pt]\mu_4' &= 3 b^4 + 6 a^2 b^2 + a^4. \\[6pt]\end{aligned} Now, try substituting these back into the original expression to find the variance of interest.

Substituting back into the first expression gives:\begin{aligned}Q_n &= n(\mu_4' - \mu_2'^2) \\[6pt]&= n[(3 b^4 + 6 a^2 b^2 + a^4) - (b^2 + a^2)^2] \\[6pt]&= n[(3 b^4 + 6 a^2 b^2 + a^4) - (b^4 + 2 a^2 b^2 + a^4)] \\[6pt]&= n[2 b^4 + 4 a^2 b^2] \\[6pt]&= 2nb^2 (b^2 + 2a^2). \\[6pt]\end{aligned}For the special case where $n=2$ you have $Q_2 = 4b^2 (b^2 + 2a^2)$. It can be shown that this result accords with the solution you would get if you used the alternative method of deriving your result from the scaled non-central chi-squared distribution.

Alternative working based on use of the non-central chi-squared distribution: Since $X_i / b \sim \text{N}(a/b, 1)$ we have:$$\sum_{i=1}^n \Big(\frac{X_i}{b} \Big)^2 \sim \text{Non-central Chi-Sq}\Big( k=n, \lambda= \frac{n a^2}{b^2} \Big).$$Using the known variance of this distribution we have:\begin{aligned}Q_n \equiv \mathbb{V}\Big( \sum_{i=1}^n X_i^2 \Big) &= b^4 \cdot \mathbb{V}\Big( \sum_{i=1}^n \Big(\frac{X_i}{b} \Big)^2 \Big) \\[6pt]&= b^4 \cdot 2(k+2 \lambda) \\[6pt]&= 2 b^4 \Big( n + 2 \frac{na^2}{b^2} \Big) \\[6pt]&= 2 n b^2 (b^2 + 2a^2). \\[6pt]\end{aligned} This result matches with the result above.

• Spoiler tags are unnecessary and distracting. – Alexis Jul 5 '18 at 16:15

If $X$ and $Y$ are $\text{N} (a, b^2)$ independent random variables, then $\left( \frac{X - a}{b} \right)^2 + \left( \frac{Y - a}{b} \right)^2$ is a $\chi ^2(2)$ random variable.

Do you think you can take it from there?

The answer is in the non-central Chi-squared distribution.

For example, if b=1, the answer to your question is: $2(k + 2(a^2))$, where $k=2$ is the number of components ($X$ and $Y$).