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Suppose $\ X_i \sim Unif (0,m), \ m = 2,3 ,... $

What will be the moment generating function of this distribution?

$$\ E[e^{Xt}] = \sum_{i=0}^m i \cdot e^{ti} \cdot \frac{1}{1+m} = \frac{1}{1+m} \cdot \sum_{i=0}^m e^{ti} \cdot i$$

I couldn't find a way to proceed from here? found this answer Moment Generating Function for a discrete random distribution but why do they ignore the possible values of $\ i $ as it is uniform and can get any value between $\ 0,m $

bm1125
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    You shouldn't multiply by $i$. Just remove the multiplication by $i$ and you'll be good. The values of $X$ are still accounted for by the $i$ in the exponential, which is all we need since we want the expected value of $e^{tX}$ (not of just $X$). – Minus One-Twelfth Mar 21 '19 at 11:57
  • (and, for reference, also the summation with the i, which is the expectation value $E[Xe^{Xt}]$, has a closed form and would not be a reason to stop, since in this case it holds $E[Xe^{Xt}]=\partial_t E[e^{Xt}]$ ) – Thomas Mar 21 '19 at 12:01

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Hint: You shouldn't be multiplying by $i$ in the sum. The expression for $E\left[\color{blue}{e^{tX}}\right]$ is $$\sum\limits _{i =0}^m \color{blue}{e^{ti}} P(X=i).$$ You should find this sum easier to deal with. (It will just be a geometric series.)

(Remember for a general function $\color{blue}{f(X)}$, its expectation here would be given by $\sum\limits _{i=0}^{m}\color{blue}{f(i)}P(X=i)$. In other words, just put in the function (with $X$ replaced by $i$) next to $P(X=i)$ in the sum).)

Minus One-Twelfth
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    You can find $1+e^t+e^{2t}+\cdots + e^{mt}$ by noting that this is a geometric series with starting term $1$, common ratio $e^t$, and $m+1$ terms, and then using the geometric series formula. Have you learnt about geometric series? If not, it would be good to learn it first, since it comes up quite a lot. – Minus One-Twelfth Mar 21 '19 at 12:12
  • I'll look more deeply into that. It's just that I was given identities page where $\ \sum_{i=0}^n x^i = \frac{1-x^{n+1}}{1-x} $ . Basically I can just leave it as $\ the original series and summing it will be $\ 1 + e^t + ... $ ? – bm1125 Mar 21 '19 at 12:15
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    You should use that identity (don't just leave as $1+e^t+e^{2t}+\cdots+e^{mt}$). Note that $e^{ti} = \left(\color{blue}{e^t}\right)^i = \color{blue}{x}^i$, where $x=e^t$. So if you put $\color{blue}{x=e^t}$ (and replace $n$ with $m$) in that formula they gave you, you will get to the right answer. – Minus One-Twelfth Mar 21 '19 at 12:18