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If I define

$$\phi (k) = \int_{-\infty}^\infty e^{ikx} p(x) dx$$

but I want to define $x=x_1+x_2+x_3+...x_n$, how would I change my variables of integration to do that integral? I ultimately want to show that if

$$\phi_i (k) = \int_{-\infty}^\infty e^{ikx_i} p(x_i) dx$$

then

$$\phi = \phi_i^n.$$

I know I can appeal to the definition of the expectation value to do this if $p(x)$ is interpreted as a probability distribution, but I want to see this done from the definition of $\phi$ directly, and it's unclear to me how you would change that first integral to be integrated over multiple variables.

Edit: See comment below regarding what I mean that this can be shown by appealing to the definition of expectation values. Sorry if that was unclear.

Dargscisyhp
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  • As it stands, what you are asking doesn't seem to make much sense. Could you explain what you mean about using the expectation to do it, so we can see what you are thinking? – Chappers Mar 17 '15 at 01:24
  • If $p(x)$ is a probability distribution then $\phi (k) = E \left[ e^{ikx} \right] = e^{ik E[x]}.$ But if $x=\sum x_i$ then $\phi (k) = E \left[ e^{ik x_1 + x_2 + ...} \right] = e^{ik E[x_1] + E[x_2] + ... } = e^{ik E[x_1]} e^{ik E[x_2]} ... = \phi_1 \phi_2 ...$ – Dargscisyhp Mar 17 '15 at 01:42
  • This proves that the characteristic function, $\phi$ of the sum is the product of the characteristic function of each individual $x_i$, but I would like to see this done without appealing to algebra of expectation values. It would be nice to see it done within the integral itself. – Dargscisyhp Mar 17 '15 at 01:44

1 Answers1

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I think this answer covers most of what you are asking about. What it essentially boils down to is that you have to integrate over $x_1$, $x_2$, . . . and $x_n$. What you haven't mentioned is that your random variables are independent, which means that the joint probability density function factorises into a product: $$ f(x_1,x_2,\dotsc,x_n) = p(x_1) p(x_2) \dotsm p(x_n). $$ (And if you don't have independence, you can't do anything to split this up.)

Then in your notation you get the simple factorisation $$ \phi(k) = \idotsint e^{ik(x_1+\dotsb + x_n)} f(x_1,x_2,\dotsc,x_n) dx_1 \dotsm dx_n \\ = \left( \int e^{ikx_1} p(x_1) dx_1 \right) \dotsm \left( \int e^{ikx_1} p(x_n) dx_n \right) = \phi_i^n $$

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Chappers
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  • It's not clear to me why $dx=dx_1...dx_..$ I think that is what is tripping me up the most. They don't typically cover that kind of manipulation in an introductory calculus course. – Dargscisyhp Mar 17 '15 at 19:48