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If a random variable has a moment generation function $$M(t) = a + (1-a) \, e^t$$ with $ 0 < a < 1$

a) Determine the Distribution of $X$

b) Show that the moment of r-class of $X$ is equal to $E[X^r] = 1-a$ , for $1,2,\ldots$


So this is my best best approach till now; We know that $$M_x(t) = \mathbb{E} \left[e^{tx} \right] = \int_{ -\infty}^{+\infty} e^{t x} f(x) dx $$

Also I know that $M_x(t=0)=1$.

So, if I could find $f(x)$, I could easily then find $F(x)$ (distribution).

The problem is that I am doing cyrcles all over my papers .

Should I find $\text{Var}(X)$? thats easy... and then try from this?

Can't think of anything else.

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This is not a "do my homework for me" site. Show us what you've tried and where are you stuck. And dont' copy the exercise statement, as if given orders. Read math.stackexchange.com/questions/how-to-ask meta.math.stackexchange.com/questions/1803/… –  leonbloy Jun 2 '12 at 21:33
@ leonbloy never said that this is a "do my homework" site ... –  Liam N Jun 3 '12 at 0:45
Back on the problem , it seems that the Var(X) equals to zero ... also, all the derivatives of Mx(t) are all equal . –  Liam N Jun 3 '12 at 1:02
I edited the math, check if all is ok. –  leonbloy Jun 3 '12 at 1:03
"Var(X) equals to zero " Why? –  leonbloy Jun 3 '12 at 1:14

2 Answers 2

HINT Think of discrete random variables.

Move your mouse over the gray area below for another hint.

HINT Bernoulli random variable.

Move your mouse over the gray area below for the complete answer.

$M(t) = a + (1-a) \exp(t)$ corresponds to a bernoulli random variable $X$ where $$X = \begin{cases} 0 & \text{ with probability }a\\ 1 & \text{ with probability }1-a \end{cases}$$ Hence, $$\mathbb{E}(X^r) = 0^r \times a + 1^r \times (1-a) = (1-a)$$

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You formula for the MGF is ok; now, in the particular case when the variable has a discrete probability function (its density is a sum of Dirac deltas), say $p_k = P(X=k) $, the integral turns into a sum:

$$M_x(t) = \sum_{k} p_k e^{t k} $$

Then, by inspection , you can guess your density: it's the sum of just two exponentials (one with $k=0$)

$$M_x(t) = p_0 e^{0t} + p_1 e^{1 t} = p_0 + p_1 e^t = a + (1-a) \, e^t $$

so it's a Bernoulli with $p=1-a$ Can you go on from here?

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oh my god i see.... thanks for your answers.. Bernoulli .. voila! and its discrete of course... i thought it was continouus.. due to the stupid a, i guess =/ Need sleep asap. –  Liam N Jun 3 '12 at 1:15

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