normal distribution expected value

108 Views Asked by At

In this derivation:

http://www.sonoma.edu/users/w/wilsonst/Papers/Normal/default.html

$$f(x) = \sqrt{\frac{k}{2\pi}}e^{-\frac{k(x-\mu)^2}{2}}$$

they let

$x-\mu = v$

$dx = dv$

and conclude that

$$E(v) = \sqrt{\frac{k}{2\pi}}\int_{-\infty}^\infty ve^{-\frac{kv^2}{2}}dv$$

Why? Shouldn't it be

$$E(v) = \sqrt{\frac{k}{2\pi}}\int_{-\infty}^\infty (v+\mu)e^{-\frac{kv^2}{2}}dv$$

1

There are 1 best solutions below

0
On BEST ANSWER

Look further down the web page. You will see $$ E(x) = E(v) + \mu = \mu .$$ In other words, the problem is with awkward notation rather than any genuine mistake. The web page really shouldn't use the same letter for the random variable and the variable being integrated over. A common notation is to use capital letters for the random variable - then it would have said let $V = X-\mu$ and $v=x-\mu$, then $$ E(V) = \text{(what the web page wrote on R.H.S.)} = 0$$ and I think it would have been clearer.