Proving that integral of transfer function equals 1

[whatsoever]

Homework Statement

Given that the expected value of a function is equal to the expected value of the convolution of that function f with a transfer function g
<br /> E(f) = E(\widetilde{f})<br />
prove that the following holds
<br /> \int^{+\infty}_{-\infty} g(t) \, dt = 1 <br />
Here f is the function and g is the transfer function, and \widetilde{f}=f*g is the convolution

Homework Equations

<br /> \widetilde{f} = \int^{+\infty}_{-\infty} f(x-t)g(t) \, dt<br />

<br /> E(f) = \int^{+\infty}_{-\infty} xf(x) \, dx<br />

The Attempt at a Solution

I have no idea how to approach this, my guess is that by replacing E(f) on both sides would lead to the proof:
<br /> \int^{+\infty}_{-\infty} xf(x) \, dx = \int^{+\infty}_{-\infty}\int^{+\infty}_{-\infty} xf(x-t)g(t) \, dt\, dx<br />

 

[blue_leaf77]

$$
E(\tilde{f}) = \int^{+\infty}_{-\infty}\int^{+\infty}_{-\infty} xf(x-t)g(t) \, dt\, dx = \int^{+\infty}_{-\infty} \left( \int^{+\infty}_{-\infty} xf(x-t)\,dx \right) g(t) \, dt
$$
How do you recognize the integral inside the parentheses?
EDIT: On a second thought, are you sure you don't miss any given information?

 

[whatsoever]

$$
E(\tilde{f}) = \int^{+\infty}_{-\infty}\int^{+\infty}_{-\infty} xf(x-t)g(t) \, dt\, dx = \int^{+\infty}_{-\infty} \left( \int^{+\infty}_{-\infty} xf(x-t)\,dx \right) g(t) \, dt
$$
How do you recognize the integral inside the parentheses?
EDIT: On a second thought, are you sure you don't miss any given information?

No, I am not sure I am not missing anything. It's very hard to understand what the professor is saying or writing on the board, most of the time we try to decode what he meant to say.
How can I relate the integral in the parentheses to E(f), my calculus-fu is weak?

 

[blue_leaf77]

What about the expected value of $g(t)$? Is it not given?

 

[whatsoever]

What about the expected value of $g(t)$? Is it not given?

It's not given for sure.

 

[Ray Vickson]

Homework Statement

Given that the expected value of a function is equal to the expected value of the convolution of that function f with a transfer function g
<br /> E(f) = E(\widetilde{f})<br />
prove that the following holds
<br /> \int^{+\infty}_{-\infty} g(t) \, dt = 1<br />
Here f is the function and g is the transfer function, and \widetilde{f}=f*g is the convolution

Homework Equations

<br /> \widetilde{f} = \int^{+\infty}_{-\infty} f(x-t)g(t) \, dt<br />

<br /> E(f) = \int^{+\infty}_{-\infty} xf(x) \, dx<br />

The Attempt at a Solution

I have no idea how to approach this, my guess is that by replacing E(f) on both sides would lead to the proof:
<br /> \int^{+\infty}_{-\infty} xf(x) \, dx = \int^{+\infty}_{-\infty}\int^{+\infty}_{-\infty} xf(x-t)g(t) \, dt\, dx<br />

Are you sure about the definition of $E(f)$? IF $f(x)$ is a probability density function (that is, $f \geq 0$ and $\int f = 1$) then the quantity $\int x f(x) \, dx$ is, indeed, the expected value of the random variable associated with $f$. However, if your problem is not in the context of probability and statistics, there is no reason to look at $\int x f(x) \, dx$. Are you sure you are not supposed to take $E(f) = \int_{-\infty}^{\infty} f(x) \, dx$?

 

[whatsoever]

Are you sure about the definition of $E(f)$? IF $f(x)$ is a probability density function (that is, $f \geq 0$ and $\int f = 1$) then the quantity $\int x f(x) \, dx$ is, indeed, the expected value of the random variable associated with $f$. However, if your problem is not in the context of probability and statistics, there is no reason to look at $\int x f(x) \, dx$. Are you sure you are not supposed to take $E(f) = \int_{-\infty}^{\infty} f(x) \, dx$?

This would make more sense, as I said it's kind of hard to make out what is written on the board and I didn't check the equation for the expected value.