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Let $X_1,...,X_n$ be independent variables, each of them has a Discrete uniform distribution between $0$ and $m$, $m= \left( 2,3,4,,... \right)$.

Let $Y$ be a random variable which is defined by $Y = X_1 + X_2 +...+X_n$.

What is the Moment-generating function of $Y$?

I do know that since the $X_i$s are independent, the function is a sum of all the moment generating functions of the $X_i$s. I also know that each function is essentially the expected value of $e^{tx}$. However, here I'm lost.

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closed as unclear what you're asking by Did, Pierre-Guy Plamondon, José Carlos Santos, mrtaurho, max_zorn Jan 13 at 22:13

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  • $\begingroup$ "since the $X_i$s are independent, the function is a sum of all the moment generating functions of the $X_i$" What you mean by that is unclear but almost certainly wrong. $\endgroup$ – Did Jan 13 at 9:06
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The moment-generating function of a sum of independent random variables is the product of their moment generating functions.

To a large extent, that's the point of using the moment-generating function at all; the transform method turns the convolution that is the density/probability function of the sum into a pointwise product.

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  • $\begingroup$ Thank you, but what is the actual function? $\endgroup$ – Alan Jan 13 at 9:20
  • $\begingroup$ Calculating the moment-generating function for a discrete uniform distribution is just a finite geometric series. You should be able to do that yourself. $\endgroup$ – jmerry Jan 13 at 19:44
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$Ee^{tY}=[\frac 1 {m+1} \frac {e^{t(m+1)}-1} {e^{t-1}}]^{n}$ for $t \neq 0$, $1$ for $t=1$. I have used the formula for a geometric sum as well as the fact that the MGF of a sum of independent random variables is the product of the individual MGF's.

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  • $\begingroup$ Thank you. Why $\frac{1}{m+1}$ and not $\frac{1}{m}$? $\endgroup$ – Alan Jan 13 at 16:43
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    $\begingroup$ @Alan How many numbers are there from $0$ to $m$ (both included)? There are $m+1$ of them not $m$. $\endgroup$ – Kabo Murphy Jan 13 at 23:31
  • $\begingroup$ Thank you. Just to be fair, I HAVE NOT down graded your answer, but rather upgraded it. :-) $\endgroup$ – Alan Jan 14 at 7:19

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