# How to derive to this point

1. Apr 28, 2008

### tuanle007

hello, can someone look at my problem and tell me how to get to the arrow?
thank you so much..
i am studying for my midterm and i have no clue where it come from.
thanks

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2. Apr 28, 2008

### uart

That first line you point to is basically just the definition of how you find the "expected value" of a function.

That is, if a random variable X has a pdf (probablity density function) of f(x) then the expected value of x can be written as,

$$E(x) = \int_{-\infty}^{+\infty} x f(x) dx$$

And more generally the expected value of a function of x can be written as,

$$E(\, \phi(x)\, ) = \int_{-\infty}^{+\infty} \phi(x) f(x) dx$$

Last edited: Apr 28, 2008
3. Apr 28, 2008

### uart

BTW, there's an error in that derivation. The x^2 term that appears in the first square bracketed term in line 4 (2nd line under the completing the square heading) should be just "x" (not squared). This error is carried all the way through the derivation BTW.

4. Apr 28, 2008

### tuanle007

yeah.i understand that part..
but how does exp(-o^2w^2/2) = exp(-x^2/2o^2)?
the arrow....

5. Apr 30, 2008

### uart

What? It never says that anywhere. The question says to prove that,

$$\Phi(\omega) = exp(-\sigma^2 \omega^2 /2)$$

You must be misreading it because nowhere does it claim the thing you state.

6. Apr 30, 2008

### HallsofIvy

It doesn't say that. It says that you are to prove that the [/b]characteristic function[/b] for the the Gaussian distribution is
$$e^{\frac{-\sigma^2\omega^2}{2}}[/itex] and them immediately uses the fact that the Gaussian distribution (with mean 0) itself can be written as [tex]e^{\frac{-x^2}{2\sigma^2}$$

Those are two completely different functions.