(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 $\frac{1}{10}.$
In general, the density for a uniformly distributed random variable X is given by
$f(x) = 1=n , \text{ where : n is a postive integer and } x = x_1, x_2, … , x_n$
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
$m(t) = \text{ (sum) } e^{tx} \frac{(1)}{(n)}$ 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.
Best Answer
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!
For this "general" case, there is really nothing further to do, since no genuine simplification is possible.
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 $$M_X(t)=\frac{e^{10t} -1}{10(e^t-1)}.$$