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- Thread starter tuanle007
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uart

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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,

[tex] E(x) = \int_{-\infty}^{+\infty} x f(x) dx [/tex]

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

[tex] E(\, \phi(x)\, ) = \int_{-\infty}^{+\infty} \phi(x) f(x) dx [/tex]

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,

[tex] E(x) = \int_{-\infty}^{+\infty} x f(x) dx [/tex]

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

[tex] E(\, \phi(x)\, ) = \int_{-\infty}^{+\infty} \phi(x) f(x) dx [/tex]

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uart

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yeah.i understand that part..

but how does exp(-o^2w^2/2) = exp(-x^2/2o^2)?

the arrow....

but how does exp(-o^2w^2/2) = exp(-x^2/2o^2)?

the arrow....

- #5

uart

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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,

[tex]\Phi(\omega) = exp(-\sigma^2 \omega^2 /2)[/tex]

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

- #6

HallsofIvy

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yeah.i understand that part..

but how does exp(-o^2w^2/2) = exp(-x^2/2o^2)?

the arrow....

It

[tex]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}[/tex]

Those are two completely different functions.

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