Moment Generating Function of an Unspecified Distribution This is the original question - I have inserted my own attempt at it, please critique as you wish.

Suppose we are interested in looking at the number of times a stock drops before it first increases in value. It is known that the stock value increases with probability $\theta$. Let $Y$ be the number of times the stock value drops before it first increases in value.


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*Find the MGF of Y, and hence E(Y).

*Now suppose that we are interested in 3 stocks. It is known that independently each stock's value will increase with probability $\theta$. Let $Y_i$ be the number of times the stock value drops before it first increases in value for stocks 1,2 and 3 respectively. Let $V=5Y_1-3Y_2+2Y_3$. Find the MGF of V.



I assume this can be modelled as a Poisson distribution. Hence it has pdf of $p(y)=\frac{\theta^x e^{-\theta}}{y!}$
$$E(e^{ty})=\sum\limits_{y=1}^\infty e^{ty}f(y)=e^{-\theta}e^{et\theta}\sum\limits_{y=1}^\infty \frac{(e^t\theta)^y e^{-(e^t\theta)}}{y!}$$
Which would result in MFG of $e^{\theta(e^t-1)}$. After differentiation and $t=0$, it is obtained that $E(Y)=\theta$.
Now as per the second part of the question, I'm unsure if I proceeded it correctly -
$$M_v(t)=E(e^{t(5Y_1-3Y_2+2Y_3)})=M_{Y_1}(5t)M_{Y_2}(-3t)M_{Y_3}(2t)$$
How do I further simplify it? Or did I approach it incorrectly? Thanks for all the help!
 A: From the definition of $Y$, i.e., "[l]et Y  be the number of times the stock value drops before it first increases in value," I assume this means that $Y$ follows a geometric distribution, with "failure" = dropping in the stock price and "success" = increase in the stock price. So the probability density function of $Y$ is $f_{Y}(y) = \theta(1-\theta)^{y}$. See if you understand how I derived the pdf from the problem statement. 
Then, the moment generating function of $Y$ is given by $M_{Y}(t) = \dfrac{\theta}{1-(1-\theta)e^{t}}$. A proof of this is as follows:
$M_{Y}(t) = E\left(e^{tY}\right) = \sum\limits_{y=0}^{\infty}e^{ty}*\theta(1-\theta)^{y}=\theta\sum\limits_{y=0}^{\infty}e^{ty}(1-\theta)^{y}=\theta\sum\limits_{y=0}^{\infty}[e^{t}(1-\theta)]^{y}$.
We assume that $|e^{t}(1-\theta)|<1$ in order for this series to converge. Since $0<\theta<1$, this is the same thing as saying that $|e^{t}| < \dfrac{1}{1-\theta}$, or $t<\text{ln} \left( \dfrac{1}{1-\theta}\right)$. The series is a geometric series which converges to $\dfrac{1}{1-(1-\theta)e^{t}}$, and multiplying by $\theta$, you get the desired result. So $\dfrac{\theta}{1-(1-\theta)e^{t}}$ is the moment generating function of $Y$. 
The expected value is given by $M^{'}_{Y}(t)|_{t=0}=\dfrac{1-\theta}{\theta}$, which I'll have you prove by yourself.
The way you did the second part looks correct. Just make sure you use the geometric distribution moment generating function for each random variable. 
