Avg. of sum of independent variables

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

The probability that $X_1$ is between $X_1$ and $X_1 + dX_1$ and $X_2$ is between $X_2$ and $X_2 + dX_2$ and so on till the nth variable is
dP(##X_1,
X_2, ..., X_n) = p ( X_1) p( x_2) p(X_3)...p(X_n) dX_1 dX_2 ...dX_n
\\ dP(Y) = p(Y) dY
\\<Y>= \int Yp(Y) dY
\\ assuming dP(y) =
dP(X_1,
X_2, ..., X_n)
\\= \int ( X_1 +X_2 + ... + X_n) p ( X_1) p( x_2) p(X_3)...p(X_n) dX_1 dX_2 ...dX_n
\\ = \int X_1 p ( X_1)dX_1+ \int X_2 p ( X_2)dX_2+ ...+\int X_n p ( X_n)dX_n
\\= <X_1> + <X_2>+...+<X_n>
\\= n <X>##
Is the assumption o.k?

 

[Orodruin]

You are never really using that assumption, you are using
$$
\langle Y \rangle = \int Y(x_1,\ldots,x_n) P(x_1,\ldots, x_n) dx_1 \ldots dx_n.
$$
If you want to express $P(Y)$ as a one-variable function you need to integrate out the $X_i$ variables, essentially
$$
P(Y) = \int \delta(Y-x_1-\ldots-x_n) P(x_1,\ldots,x_n) dx_1 \ldots dx_n.
$$
Note that this leads to
$$
\langle f(Y)\rangle = \int f(y) P(y) dy = \int f(y) \delta(y-x_1-\ldots-x_n) P(x_1,\ldots,x_n) dx_1 \ldots dx_n dy
= \int f(x_1+\ldots+x_n) P(x_1,\ldots,x_n) dx_1 \ldots dx_n,
$$
so it is compatible with the original probability distribution.

 

[Pushoam]

Thank you for this insight.