# Moment Generating Function for a discrete random distribution

(Discrete uniform distribution) A discrete random variable is said to be uniformly distributed if it assumes a nite number of values with each value occurring with the same probability. If we consider the generation of a single random digits, then Y , the number generated, is uniformly distributed with each possible digit, 0, 1, 2, ... , 9 occurring with probability 1/10. In general, the density for a uniformly distributed random variable X is given by

f(x) = 1=n , where : n is a postive integer and x = x1, x2, ... , xn

Find the moment generating function for the discrete uniform random variable X.

• I don't know how to approach this with what I have from class... all I can come up with is

m(t) = (sum) e^tx (1/n) which I know is complete rubbish. I am grasping so little of this so any assistance in what a moment generating function is and the concepts needed for this question would be greatly appreciated.

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Actually, what you wrote is fairly close to what needs to be done. Your random variable $X$ takes on the values $x_1,x_2,\dots,x_n$, each with probability $\dfrac{1}{n}$.
The moment generating function $M_X(t)$ of $X$ is, by definition, $E(e^{tX})$. By the usual formula for expectation, $$M_X(t)= E(e^{tX})=\sum_{k=1}^n \frac{1}{n}e^{tx_k}.$$ That's all there is to it!
In the special case where the $x_i$ are separated by constant amounts, we can do some simplification. Let's look at your random digits example. In that case, $$M_X(t)=\frac{1}{10}\left(e^{0}+e^t+e^{2t}+\cdots +e^{9t}\right).$$ We have here a geometric series with common ratio $e^t$. By the usual formula for the sum of a finite geometric series, we get $$M_X(t)=\frac{e^{10t} -1}{10(e^t-1)}.$$
It doesn't matter. We can get to your version by multiplying top and bottom by $-1$. And it is $e^{10t}$, not $e^{10}$. – André Nicolas Oct 26 '12 at 7:05