I am given a random variable X with all moments equal: $ EX^n = c, \ n \ge1 $
I needed to determine the distribution of X, so I did the following:
given that $ M_X(t) = 1 + \sum_{n=1}^{\infty} \frac {t^n}{n!} \cdot EX^n, \ |t| < h$ and since $EX^n = c$ in this case, we can say
$M_X(t) = 1 + \sum_{n=1}^{\infty} \frac{t^n}{n!} \cdot c = 1 + \sum_{n=1}^{\infty}\frac{t^n}{n!} \cdot p $, thus the distribution of X is Be(c) = Bernoulli(c)
Is this correct or did I make incorrect assumptions?
Please help! Thank you!
The last equation looks strange... where did that $p$ come from? If you are just showing that our given $M_X(t)$ happens to be (by inspection) equal to the $M_X(t)$ of a Bernoulli of parameter $p$... well, you are right, but it doesn't look as a deduction but as a fortunate find.
The natural way, for me, would be to recognize that the sum is just an exponential, so
$$M_X(t) = 1 + \sum_{n=1}^{\infty} \frac{t^n}{n!} c= 1 + c (e^t-1)=(1-c)+ce^t$$
Again, you can inspect a table of MCF and find that you have a Bernoulli.
But you don't need that, you can work out the inversion yourself. Knowing that for a discrete random variable $$M_X(t) = E(e^{tX})=\sum_x P(X=x) e^{xt}$$
you should see that by writing
$$M_X(t) = (1-c) e^{0 \cdot t}+ce^{1\cdot t}$$
the original $P(X)$ turns to be, indeed, a Bernoulli with parameter $c$.