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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..)

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1 Answer 1

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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.

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