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More of a general integration question, but I just saw the following proof for the derivation of the expectation of a normal variable:

[tex]

E[X] = \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}{x exp\left( -\frac{1}{2\sigma^2}(x-\mu)^2 \right) dx}

[/tex]

Set z=(x-mu):

[tex]

E[X] = \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}{z exp\left( -\frac{1}{2\sigma^2}z^2 \right) dx} + \mu \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}{exp\left( -\frac{1}{2\sigma^2}z^2 \right)dx}

[/tex]

[tex]=\mu[/tex]

Now, I dont really understand how this works: if z=x-mu, then I would assume that the term inside the integral becomes:

[tex]

(z+\mu) exp\left( -\frac{1}{2\sigma^2}z^2\right)

[/tex]

[tex]

= z \left( exp\left( -\frac{1}{2\sigma^2}z^2 \right) \right) + \mu \left( exp\left( -\frac{1}{2\sigma^2}z^2 \right) \right)

[/tex]

However, I dont see how we get two separate integrals, as in the proof above. Can anyone help shed any light on this?

Cheers

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# Expectation of Normal Variable

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