Deriving the Characteristic Function of the Gaussian Distribution

[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

 

[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 (probability 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

 

[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.

 

[tuanle007]

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

 

[uart]

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

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.

 

[HallsofIvy]

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

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]<br /> and them immediately uses the fact that the Gaussian distribution (with mean 0) itself can be written as <br /> e^{\frac{-x^2}{2\sigma^2}<br /> <br /> Those are two completely different functions.