How can I find expection value?

Question

Let suppose that $X$ is gaussian RV whose mean value is $0$ and variance is $\sigma^2$ and $Y=10^\frac{X}{10}$ .

What I like to know is $E[Y]$.

First of all I search pdf of $Y$ like belows.

$$P(Y<y)=P(X\lt10 \log_{10}(y))=\int_{-\infty}^{10 \log_{10}y}\frac{1}{\sigma \sqrt{2\pi}}\exp(-\frac{x^2}{2\sigma^2})\,\text dx$$ $$P_{Y}(y)=\frac{\text d}{\text dY}P(Y\lt y)=\frac{10}{\sigma \ln10y\sqrt{2\pi}}\exp(-\frac{(10\log_{10}y)^2}{2\sigma^2})$$

And then to calculate mean value, I use moment generating function $$M_{Y}(t)=E[e^{tY}]$$ $$M'_{Y}(0)= \frac{10}{\sigma \ln10\sqrt{2\pi}}\int_{0}^{\infty}\exp(-\frac{(10\log_{10}y)^2}{2\sigma^2})\,\text dy$$

But I can't obtain value of $\int_{0}^{\infty}\exp(-\frac{(10\log_{10}y)^2}{2\sigma^2})\,\text dy$.

How can I solve this problem?

Answer 1

There are two ways that are relatively straightforward.

1) $10^{x/10} = e^{x/10 \log 10}$, so $E[10^{X/10}] = E[e^{X \frac{\log 10}{10}}] = M_X(\frac{\log 10}{10})$ where $M_X$ is the moment generating function of a $N(0,\sigma^2)$ r.v.

2) $E[10^{X/10}] = E[e^{X \frac{\log 10}{10}}] = \int_{-\infty}^\infty e^{x \frac{\log 10}{10}} \left( \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{x^2}{2 \sigma^2}} \right) dx$ and then evaluate this integral by completing the square in the exponent.