# Expected value and variance of given probability density function

A continuous random variable x has a probability density function:

$\mathrm{pdf}(x) = \frac{\exp\left(-\frac{(x-3)^2}{16}\right)}{\sqrt{16\pi}}, \qquad x\in (-\infty, \infty)$

a) Find the expected value, variance and the standard deviation of continuous random variable.

b) Find the expected value of a random variable $\exp(-x/2)$

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Hint: (a) Do you know about normal random variables? (b) Complete the square. –  Dilip Sarwate Nov 9 '12 at 16:13

Continuing on Dilip's suggestion. PDF or a Normal RV is $$f(x)=\frac{1}{\sqrt{2 \pi} \sigma}e^{-\big( \frac{x- \mu}{\sigma \sqrt{2}} \big)^2}$$ Using the values, you get $\mu=3, \sigma^2=8$, hence $X \sim N(3,8)$.
EDIT I probably misunderstood the OP's second question, as Dilip pointed out. Here's what I'd do. If $X \sim N(0,1)$, then: $$\mathbf{E}e^{-\frac{x}{2}}=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-\frac{x}{2}}e^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-\big(\frac{x^2}{2}+\frac{x}{2})}dx=\frac{e^{\frac{1}{8}}}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-\frac{1}{2} \big(x+ \frac{1}{2} \big)^2}dx = e^{\frac{1}{8}} \cdot 1\\$$ The last step is since the integral is $\mathbf{P}(-\infty < X<\infty)$=1, where $X \sim N(-\frac{1}{2},1)$
I think the second question asks for $E[\exp(-X/2)]$ and $X\sim N(0,1)$ has nothing to do with the matter. –  Dilip Sarwate Nov 10 '12 at 5:24