Expectation with respect to normal distribution

Question

In one of the paper (from IEEE journal) it is shown that $E\left[10^{-\frac{2x}{10\alpha}}\right]$ is equal to $$E\left[10^{-\frac{2x}{10\alpha}}\right]=\exp\left[2\left(\frac{.1\sigma\ln(10)}{\alpha}\right)^2\right]$$ where $x$ is a normal random variable with mean zero and variance $\sigma^2$, $\ln(x)$ is the natural logarithm, $\alpha\geq 2$ is a constant. I tried to obtain this result but I do not get the same result but a different result as shown below.

My Attempt:

We can write $$E\left[10^{-\frac{2x}{10\alpha}}\right]=E\left[\exp\left(-\frac{.2x\ln(10)}{\alpha}\right)\right]$$ Hence we will have $$E\left[10^{-\frac{2x}{10\alpha}}\right]=\frac{1}{\sqrt{2\pi \sigma^2}}\int_0^{\infty} \exp\left(-\frac{.2x\ln(10)}{\alpha}-\frac{x^2}{2\sigma^2}\right)dx$$ which we can solve with the help of following formula (obtained from wolfram alpha)$$\int_0^{\infty}\exp(-mx-nx^2)dx=.5\sqrt{\frac{\pi}{n}}\exp(\frac{m^2}{4n})erfc(\frac{m}{2\sqrt{n}})$$ therefore we will have following result $$E\left[10^{-\frac{2x}{10\alpha}}\right]=.5\exp\left(2\left(\frac{.1\sigma\ln(10)}{\alpha}\right)^2\right)erfc\left(\frac{.1\ln(10)\sqrt{2\sigma^2}}{\alpha}\right)$$ which is not exactly same as the right side of the first equation. Am I doing something terribly wrong here. I will be very thankful to you for your help in clarifying my problem. Thanks in advance.

Answer 1

your first step actually is right

do you know moment generating function? (def $M_X(t)=E(e^{tX})$), then actually your first formula is equal to $M_x\left(-\frac{0.2\ln 10}{\alpha}\right)$, and then use the mgf of $N(0,\sigma^2)$, i hope it is clear enough

Answer 2

First, you should avoid mixing decimal notation and fractions, and if you must use decimals, you should not omit the zero in the units' place.

The difficulty with your calculation is that if $X$ is normal, then the support is always $-\infty < X < \infty$, and you should not skip so many intermediate steps.

The key to avoiding your first mistake here is that if $$Y = g(X) = 10^{-X/(5\alpha)} = \exp(-(X \log 10)/(5 \alpha)),$$ then $$\operatorname{E}[Y] = \int_{x=-\infty}^\infty g(x) \frac{e^{-x^2/(2\sigma^2)}}{\sqrt{2\pi} \sigma} \, dx = \frac{1}{\sqrt{2\pi}\sigma} \int_{x=-\infty}^\infty \exp\left( -\frac{\log 10}{5\alpha} x - \frac{x^2}{2\sigma^2}\right) \, dx ,$$ where we do not integrate over the half-interval, because $g(x) f_X(x)$ is no longer an even function. Then we complete the square of the exponential term: $$\frac{1}{2\sigma^2} x^2 + \frac{\log 10}{5\alpha} x = \frac{1}{2\sigma^2} \left(x + \frac{\sigma \log 10}{5\alpha} \right)^2 - \frac{(\sigma \log 10)^2}{50\alpha^2}.$$ Consequently, $$\operatorname{E}[Y] = \frac{1}{\sqrt{2\pi} \sigma} \exp\left(\frac{(\sigma \log 10)^2}{50\alpha^2}\right) \int_{x=-\infty}^\infty e^{-(x - \mu')^2/(2\sigma^2)} \, dx, $$ where $\mu' = -(\sigma \log 10)/(5\alpha)$ But this last integral is merely a location shift of the PDF of $X$, namely a normal random variable with mean $\mu'$ and standard deviation $\sigma$, hence $$\operatorname{E}[Y] = \exp\left(\frac{1}{2} \left( \frac{\sigma \log 10}{5\alpha} \right)^2\right)$$ as claimed.

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Of course, we can avoid much of this computation by using the probability generating function of $X$: $$P_X(t) = \operatorname{E}[t^X] = \exp((\sigma \log t)^2/2),$$ which gives us the desired result for $t = 10^{-1/(5\alpha)}$.