I am trying to find the distriubtion of X when $M_X(t)=\frac 16 e^t+ \frac{2}6e^{2t}+\frac{3}6e^{3t}$ With some simple computations, I found that $Var(x)=5/9$, and $E(x)= 7/3$

However, since the given MGF does not match any common forms I found in a text, I was not able to match it to a random variable. I even tried to use the definition, $\frac 16 e^t+ \frac{2}6e^{2t}+\frac{3}6e^{3t}=E(e^{tX})$

Along with this MGF, I could not match $e^t/ (2-e^t)$ with a random variable X.

Having the answer to these would be nice, but I am more interested in the $process$ one would use to match a moment generating function to a random variable, aside from just looking at common forms of moment generating functions.

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    $\begingroup$ You have $E(X^n)=\frac{1}{6}\cdot 1^n + \frac{2}{6}2^n+\frac{3}{6}3^n.$ What does that look like? $\endgroup$ – Thomas Andrews Jan 19 '18 at 17:13
  • $\begingroup$ Terms with with linearly increasing coefficients and linearly increasing exponent bases summed, where the exponent bases are raised to a common power $n$. $\endgroup$ – wesssg Jan 19 '18 at 17:17

Read off the coefficients of the moment generating function. Note that $$ M_{X}(t)=E(e^{tX})=\sum_{x} e^{tx}P(X=x). $$ Hence $$ P(X=1)=1/6;\quad P(X=2)=2/6;\quad P(X=3)=3/6. $$

  • $\begingroup$ Yes, I understand that part of the problem. However, it didn't help me with identifying what random variable X was. Note that is how I found $E(x)$ and $Var(X)$ $\endgroup$ – wesssg Jan 19 '18 at 17:38
  • $\begingroup$ The question asks for the distribution of $X$. This is specified in the second line of my answer. $\endgroup$ – Foobaz John Jan 19 '18 at 17:52
  • $\begingroup$ As in, the classification of the random variable X. i.e. X is a $poisson$ random variable. $\endgroup$ – wesssg Jan 19 '18 at 18:23
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    $\begingroup$ @wesssg Not every distribution has its own name. Almost all don't. I don't think this one has a name, so the description of the distribution in this answer is about all one can say. $\endgroup$ – Daniel Fischer Jan 19 '18 at 18:58
  • $\begingroup$ @DanielFischer Thanks. I did not know that, because most of the time I see a moment generating function it is associated with a named distribution. $\endgroup$ – wesssg Jan 19 '18 at 21:31

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