How to Compute the Ensemble Average of a Product of Integrals?

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quasar_4
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Homework Statement



I'm trying to compute something of the form [tex]\langle \int_a^b{f(x) dx} \int_a^b{f(x)^{\dagger}dx} \rangle[/tex] where the dagger means complex conjugate and the brackets are ensemble average (f(x) is a statistical quantity). I'm supposed to use the relation that [tex]\langle f(x) f(x')^{\dagger} \rangle = c*\delta(f-f')[/tex] where c is some constant.


Homework Equations



[tex]\langle f(x) f(x')^{\dagger} \rangle = c*\delta(f-f')[/tex]

The Attempt at a Solution



I'm a bit perplexed. I have the function and its complex conjugate, but inside different integrals, which are being multiplied. And the ensemble average of a product isn't the same as the product of ensemble averages, either... is it? I'd be surprised.

I thought maybe I could multiply the entire quantity by an extra f dagger, then somehow use the relation, but it didn't really get me anywhere.

So I have no idea how to use the given relation. Can anyone help??
 
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quasar_4 said:

Homework Statement



I'm trying to compute something of the form [tex]\langle \int_a^b{f(x) dx} \int_a^b{f(x)^{\dagger}dx} \rangle[/tex] where the dagger means complex conjugate and the brackets are ensemble average (f(x) is a statistical quantity).
Are you sure that the x appearing in the second integral should not all have a prime on them?

I'm supposed to use the relation that [tex]\langle f(x) f(x')^{\dagger} \rangle = c*\delta(f-f')[/tex] where c is some constant.
Are you sure that it is not [itex]\delta(x-x')[/itex] ??


Check these two things and let us know. If I am correct about the two corrections, the problem becomes very easy.
 
Oops, you're right. I've been working with functions of frequency and forgot. So yes, should be f(x) and f(x'), and the delta function should then be delta(x-x').
 
quasar_4 said:
Oops, you're right. I've been working with functions of frequency and forgot. So yes, should be f(x) and f(x'), and the delta function should then be delta(x-x').

Great. Then are you all set? Replacing the product of the functions by a delta function makes the two integrations trivial.