Trying to approximate $E[f(X)]$ - Woflram Alpha gives $E[f(X)] \approx \frac{1}{\sqrt{3}}$ but I get $E[f(X)] \approx 0$?

Question

Let $X \sim \mathcal{N}(\mu_X,\sigma_X^2) = \mathcal{N}(0,1)$. Let $f(x) = e^{-x^2}$. I want to approximate $E[f(X)]$.

Wolfram Alpha gives \begin{align} E[f(X)] \approx \frac{1}{\sqrt{3}}. \end{align}

Using a Taylor expansion approach, and noting that $f''(0) = -2$, I get \begin{align} E[f(X)] &\approx f(\mu_X) + \frac{f''(\mu_X)}{2} \sigma_X^2 \\ & = f(0) + \frac{f''(0)}{2} \\ & = 1 + \frac{-2}{2} \\ & = 0. \end{align}

Why does my approximation fail to match the Wolfram Alpha result? What can be done to fix it?

Answer 1

You does not need an approximation here. Use properties of moment generating functions, $X$ is standard normal so $X^2$ is chisquared with one df, with moment generating function $M_{X^2}(t)=\frac1{\sqrt{1-2t}}$ (for $t<1/2$.) Then note that $$\DeclareMathOperator{\E}{\mathbb{E}}M_X(t)=\E e^{t X} $$ is the definition, so that $$\E e^{-X^2}=M_{X²}(-1)=\frac1{\sqrt{1-2\cdot (-1)}}=\frac1{\sqrt{3}} $$ We can check that in R with a fast simulation (always a good idea to do a simulation check):

 mean( exp(-rnorm(1E6)^2) )
[1] 0.5774847
 1/sqrt(3)
[1] 0.5773503

Answer in comments:

What about if 𝑋 was not standard normal but normal with mean $𝜇_𝑋$ and variance $𝜎^2_𝑋$. Can your approach still be used or is it specific to the case of a standard normal distribution?

It can still be used. I will not give full details. First, the easy case $X \sim \mathcal{N}(0,\sigma^2)$. Then $X=(\sigma Z)^2$ with $Z$ standard normal, so in the above argument you get the argument $-\sigma^2$ in place of $-1$ for the mgf (moment generating function.) For the fully general case, see for instance Moment-generating function (MGF) of non-central chi-squared distribution and work from there.

Answer 2

There's no need to "approximate" when you can derive the exact value of $\mathbb{E}[f(X)]$ . Let us apply the Law of the Unconscious Statistician (LoTUS) to obtain : \begin{align*} \mathbb{E}[f(X)] &= \int_{-\infty}^{+\infty} e^{-x^2} \cdot \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{x^2}{2}\right)~dx\\ &= 2\int_0^{+\infty} \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{3x^2}{2}\right)~dx\\ &= 2\int_0^{+\infty} \frac{1}{\sqrt{2\pi}} \cdot\frac{1}{\sqrt{6z}} e^{-z}~dz\\ &= \frac{2}{2\sqrt{3\pi}} \int_0^{+\infty} e^{-z} z^{\frac{1}{2} -1} ~dz\\ &=\frac{1}{\sqrt{3\pi}}\cdot \Gamma\left(\frac{1}{2}\right)\\ &= \frac{1}{\sqrt{3\pi}}\cdot \sqrt{\pi}\\ &= \frac{1}{\sqrt{3}} \end{align*}

Hope this helps. :)