# Proving the Variance of the standard normal distribution

I am trying to prove the variance of the standard normal distribution $$\phi(z)=\frac{e^{-\frac{1}{2}z^2}}{\sqrt{2\pi}}$$ using high school level mathematics only. The proof given in my textbook seems wrong to me. Here is what it says:

Because the mean is zero, the variance is given by the integral

$$\operatorname{Var}(Z)=\int_{-\infty}^{\infty} z^{2} \phi(z) d z$$

We showed above while finding the second derivative of $$\phi(z)$$ that

$$\phi^{\prime \prime}(z)=\phi(z)\left(z^{2}-1\right)$$

and rearranging, $$z^{2} \phi(z)=\phi^{\prime \prime}(z)+\phi(z)$$

Hence $$\operatorname{Var}(Z)=\int_{-\infty}^{\infty} z^{2} \phi(z) d z$$

$$=\int_{-\infty}^{\infty} \phi^{\prime \prime}(z) d z+\int_{-\infty}^{\infty} \phi(z) d z$$

$$=\left[\phi^{\prime}(z)\right]_{-\infty}^{\infty}+\int_{-\infty}^{\infty} \phi(z) d z$$

$$=0+1$$

$$=1$$

The first integral above is zero, because the integrand $$\phi^{\prime}(z)=-z \phi(z)$$ is odd, as we saw before. The second integral above is 1 because $$\phi(z)$$ is a probability density function.

The only part that I don't understand is the 3rd last line, where they say that $$\left[\phi^{\prime}(z)\right]_{-\infty}^{\infty}=0$$

Because indeed $$\phi^{\prime}(z)=-z \phi(z)$$, so

$$\left[\phi^{\prime}(z)\right]_{-\infty}^{\infty}$$

$$=\left[-z \phi(z)\right]_{-\infty}^{\infty}$$

$$=-\{\infty \phi(\infty)+\infty \phi(-\infty)\}$$

$$=-(\infty \times 0+\infty \times 0)$$ which we can't say is equal to zero, right?

I think what the textbook is trying to say is that since the integrand $$\phi^{\prime}(z)=-z \phi(z)$$ is odd, $$\int_{-\infty}^{\infty} \phi^{\prime}(z) d z=0$$, but clearly the integrand is $$\phi^{\prime \prime}(z)$$ not $$\phi^{ \prime}(z)$$, and $$\phi^{\prime \prime}(z)$$ which is equal to $$(z^2-1)\phi(z)$$ is not odd.

Please let me know if I am missing something here.. And if the textbook is wrong, is there another way to prove the variance of the standard normal distribution using high school level mathematics only? (I had a look around here and saw the use of Gamma function, which I don't know what it is..)

Actually, $$\lim_{z\to\pm\infty}z\phi(z)=0$$ because $$\phi(z)$$ decays exponentially at infinity. You can use L'Hopital's rule to see that the limit is zero. Therefore, $$[\phi'(z)]|_{-\infty}^\infty=0$$ and the argument in your book is correct.