$Z$ ~ $N(0, 1)$. Find $\operatorname{Var}(Z)$

Question

$Z$ ~ $N(0, 1)$. Find $\operatorname{Var}(Z)$

We know:

$E(Z) = 0$

$\operatorname{Var}(Z) = E(Z^2) - E(Z)^2 = E(Z^2) - 0^2 = E(Z^2)$

Thus, we need:

$E(Z^2) = \int_{-\infty}^{\infty} z^2 \frac{1}{\sqrt{2\pi}}e^{-1/2(z^2)}dz$

Using int by parts, $u = z, du = dz$, then $dv = z \frac{1}{\sqrt{2\pi}}e^{-1/2z^2}dz$, $v = -\frac{1}{\sqrt{2\pi}}e^{-1/2z^2}$

Then we have

$$\left[-\frac{z}{\sqrt{2\pi}}e^{-1/2z^2} \right]_{-\infty}^{\infty} + \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-1/2z^2}dz$$

I'm aware that $\left[-\frac{z}{\sqrt{2\pi}}e^{-1/2z^2} \right]_{-\infty}^{\infty} = (0-0) = 0$

No clue how to proceed.

Answer 1

The standard method for evaluating $\int_{-\infty}^{\infty}e^{-\frac{1}{2}z^2}dz$ isn't an obvious one, but one worth remembering. It involves squaring it and turning it into a double integral.

First note that

$$\int_{-\infty}^{\infty}e^{-\frac{1}{2}z^2}dz=2\int_{0}^{\infty}e^{-\frac{1}{2}z^2}dz$$

as $e^{-\frac{1}{2}z^2}$ is an even function.

Let $I:=\int_{0}^{\infty}e^{-\frac{1}{2}z^2}dz$. Also note that "$z$" is just a dummy variable, so $I=\int_{0}^{\infty}e^{-\frac{1}{2}x^2}dx=\int_{0}^{\infty}e^{-\frac{1}{2}y^2}dy$.

$$ \begin{align} I^2 &= \int_{0}^{\infty}e^{-\frac{1}{2}x^2}dx \int_{0}^{\infty}e^{-\frac{1}{2}y^2}dy \\ &= \int_{0}^{\infty}\int_{0}^{\infty}e^{-\frac{1}{2}x^2}e^{-\frac{1}{2}y^2}dxdy \\ &= \int_{0}^{\infty}\int_{0}^{\infty}e^{-\frac{1}{2}(x^2+y^2)}dxdy \\ &= \int_{0}^{\frac{\pi}{2}}\int_{0}^{\infty}e^{-\frac{1}{2}r^2}rdrd\theta \\ &= \frac{\pi}{2}\int_{0}^{\infty}re^{-\frac{1}{2}r^2}dr \\ &= \frac{\pi}{2} \left[ -e^{-\frac{1}{2}r^2} \right]_{0}^{\infty} \\ &= \frac{\pi}{2} \end{align} $$

... and hence

$$ I = \frac{\sqrt{2\pi}}{2} $$

which, together with your working, gives:

$$ \begin{align} \mathrm{Var}(Z) &= E(Z^2) \\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}z^2}dz \\ &= \frac{2}{\sqrt{2\pi}} I \\ &= 1 \end{align} $$

as required!

Answer 2

The answer is by definition of the law 1.

Moreover, you are trying to find the variance of a normal law knowing only the mean. Do you think it is even possible ? You could have any positive value for the variance.

Answer 3

Using Fubini's theorem you can compute $(\int e^{-x^2})^2$=$\pi$ and in the same way that $\int e^{-\frac{x^2}{\alpha}}=\sqrt{\alpha\pi}$, which gives you the result you were looking for.