# Help getting the variance of $Y$

We have $Z \sim N(0,1)$ and $Y=a+bZ+cZ^2$. What is the variance of $Y$?

This is what I did:

$$\mbox{Var} (Y)=0+b^2 \, \mbox{Var} (Z)+c^2 \, \mbox{Var} (Z^2) = b^2 + c^2 \, \mbox{Var} (Z^2)$$

To get $\mbox{Var} (Z^2)$, I tried to use the definition $\mbox{Var} (Z^2)= \mathbb E(Z^4)-\mathbb E^2(Z^2)$ But I'm having trouble with this part. If this was a odd for example $\mathbb E(Z^3)$ you can say that because of the symmetry of the normal distribution $\mathbb E(Z^3)=0$, but in this is pair.

• If you have a browser and an internet connection, which apparently you do, it's not terribly difficult to find the value of $E(z^4)$. – Mark L. Stone Jul 20 '16 at 18:04
• $Z^2$ will be a chi-squared random variable. – Patrick Li Jul 20 '16 at 18:13
• You overlooked one term in your calculation: $\operatorname{Cov}(bz,cz^2)$. Although it turns out to be zero, it does need to be considered for your derivation to be complete and valid. – whuber Jul 20 '16 at 19:24
• Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. – gung - Reinstate Monica Jul 20 '16 at 20:05
• Also asked by the OP on math.SE where it has received three answers including one that the OP has accepted. – Dilip Sarwate Jul 21 '16 at 1:32

### Solution 1: look it up.

Write

$$Y = c\left(Z - \frac{b}{2c}\right)^2 + a - \frac{b^2}{4c}.$$

Since $Z-\frac{b}{c}$ has a Normal$(-b/c, 1)$ distribution, its square has a non-central chi-squared distribution with one degree of freedom and noncentrality parameter

$$\lambda = \left(-\frac{b}{2c}\right)^2.$$

Its variance therefore is $2(1 + 2\lambda)$. Consequently the variance of $Y$ is $c^2$ times that, reducing to

$$\operatorname{Var}(Y) = c^2 2(1 + 2\lambda) = b^2 + 2c^2.$$

### Solution 2: compute it.

Use the well-known (and easily derived) result that $\mathbb{E}(Z^4) = 3$ as well as the obvious values $\mathbb{E}(Z^3)=\mathbb{E}(Z) = 0, \mathbb{E}(Z^2)=1$. Starting with the definition of variance, expand the expressions into sums of simpler terms and use the linearity of expectation to simplify them:

\eqalign{ \operatorname{Var}(Y) &= \operatorname{Var}(cZ^2 + bZ + a) \\ &= \mathbb{E}((cZ^2 + bZ + a)^2) - (\mathbb{E}(cZ^2 + bZ + a))^2 \\ &= \mathbb{E}((c^2Z^4 + 2bcZ^3 + (2ac+b^2)Z^2 + 2abZ + a^2) - (c + 0 + a)^2 \\ &= 3c^2 + 0 + (2ac+b^2) + 0 + a^2 - (a^2 + c^2 + 2ac) \\ &= b^2 + 2c^2. }

### Solution 3: exploit basic identities.

This answer uses only the basic properties of covariance and independence, the values of the given two moments of $Z$ (its mean and variance), and the fact that uncorrelated jointly Normal distributions are independent.

Let $Z$ and $W$ independently have standard Normal distributions. Then the two variables $Z\pm W$ have jointly Normal distributions with zero mean and variance $2$. This is equivalent to having the same distribution as $Z$, multiplied by $\sqrt{2}$. Thus

$$\operatorname{Var}((Z\pm W)^2) = \operatorname{Var}((\sqrt{2}Z)^2) = \operatorname{Var}(2Z^2) = 4\operatorname{Var}(Z^2).$$

Moreover, since

$$\operatorname{Cov}(Z+W, Z-W) = \operatorname{Var}(Z) - \operatorname{Var}(W) = 1-1=0,$$

$Z\pm W$ are uncorrelated, implying they are independent. Consequently $(Z\pm W)^2$, since they are functions of independent variables, are also independent. Their covariance must be zero. Therefore

$$8\operatorname{Var}(Z^2) = \operatorname{Var}((Z+W)^2 - (Z-W)^2) = \operatorname{Var}(4ZW) = 16 \operatorname{Var}(Z)\operatorname{Var}(W)=16,$$

showing that

$$\operatorname{Var}(Z^2) = \frac{16}{8} = 2.$$

Finally, note that because $(Z, Z^2)$ and $(-Z, Z^2)$ have identical distributions, they have equal covariances; but since their covariances must be negatives of each other, their covariances are both zero:

$$\operatorname{Cov}(Z^2, Z) = -\operatorname{Cov}(Z^2, Z) = 0.$$

Now

$$\operatorname{Var}(cZ^2 + bZ + a) = c^2\operatorname{Var}(Z^2) + b^2 \operatorname{Var}(Z) + 2bc \operatorname{Cov}(Z^2, Z) = 2c^2 + b^2.$$