Say $X_m$ is a normal distribution with mean $0$ and variance $2m$. Now suppose we have an alternating series $-X_1+X_2-X_3+...$ up to $X_n$. Find the distribution of this series.

My working so far: I know that the moment generating function of a series of random variables is the product of each random variable's MGF. Could this lead me to the distribution?

I know the MGF of all positive $X_i$ but what is the MGF of a negative $X_i$? Is it simply the negative of the MGF of positive $X_i$?

If I'm on the wrong track, how do I find this distribution?

Are they independent? If so, the fact that $-X_i$ has the same distribution implies the signs have no effect.

Since you ask, $M_{aX}(t)=M_X(at)$.

  • They are indeed independent; why does that mean the sign has no effect? – mathenthusiast May 21 at 15:50
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    @mathenthusiast Technically, the individual variables having a symmetric distribution is why the signs don't matter, so we may as well take a sum; the independence is just needed to compute the sum's distribution. – J.G. May 21 at 15:53

Hint: Have a look at how the MGF behaves under linear transformations - that will answer your question about the change in the MGF under the mapping $X_i \mapsto -X_i$.

More generally let $\mathbf{X} = (X_1,\ldots,X_n)$ be your vector of random variables and note that you haven't yet specified the joint distribution which you should do, after doing this then you can write your series in terms of the linear transformation $$ \mathbf{X} \mapsto \langle \mathbf{a}, \mathbf{X} \rangle, $$ for an appropriate choice of vector $\mathbf{a}$ and combine that with what you have learned about the effect of linear transformations on the moment generating function.

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