Find probability given moment generating function? So, I have a question that asks me to find the probability $P(Y<2)$ if $Y$ has a moment generating function $$M_Y(t) = (1-p+pe^t)^5$$
Is this a special distribution? Is there a trick I'm missing? Solving it algebraically/ with calculus gets really messy
 A: Suppose $$M_X(t) = 1-p+p\exp(t)=(1-p)\exp(0t)+p\exp(1t)$$
Hence $X$ is a Bernoulli distribution with success probability $p$.
It is raised to the power of $5$, $M_Y(t)=\prod_{i=1}^5M_{X_i}(t)=M_{\sum_{i=1}^5X_i}(t)$ means $5$ independent Bernoulli trials are sum up, Hence $Y$ is a binomial distirbution with $5$ trials and success probabiity $p$.
A: Method 1: Genarally analysing MGF
we have
$$M_Y(t)= (1-p+pe^t)^5$$
if we put 1-p=q or supposing p+q=1 we will get'
$$M_Y(t)=(q+pe^t)^5$$
we know that binomial random variable have MGF $$M_Y(t)=(q+pe^t)^n$$ after matching the corresponding terms with our give MGF we get n=5 hence
$$Y\sim Bin (5,p)$$
so
$$P(Y<2)=P(Y=0)+P(Y=1)$$
$$\Rightarrow P(Y<2)= \binom{5}{0}p^0(1-p)^5+\binom{5}{1}p^1(1-p)^4$$
$$\Rightarrow P(Y<2)=4p^5-15p^4+20p^3-10p^2+1$$
Now put your value of p (i.e generally called probability of success) and get your answer
Method 2 : By generating function method
we know that
$$ M_Y(log_e(t))= G_Y(t) $$
where $M_Y(\bullet)$ denotes Moment generating function of Y
and $G_Y(\bullet)$ represents generating function of Y, So we have to generally replace $t$ by $log_e(t)$ by doing that with the MGF you have given we will get
$$M_Y(log_e(t))=(1-p+pe^{log_et})^5$$
$$G_Y(t)=(1-p +pt)^5$$
$$G_Y(t)=p^5 t^5 - 5 p^5 t^4 + 10 p^5 t^3 - 10 p^5 t^2 + 5 p^5 t - p^5 + 5 p^4 t^4 - 20 p^4 t^3 + 30 p^4 t^2 - 20 p^4 t + 5 p^4 + 10 p^3 t^3 - 30 p^3 t^2 + 30 p^3 t - 10 p^3 + 10 p^2 t^2 - 20 p^2 t + 10 p^2 + 5 p t - 5 p + 1$$
as we know that
$$G_Y(t)=p_0+p_1t+p_2t^2\ldots\ldots\ldots\ldots$$
where
$$p_0 = P(Y=0)$$
$$p_1= P(Y=1)$$
$$p_2=P(Y=2)$$
$$\cdots$$
$$\cdots$$
as in our Generating function we can see that $$p_0=  - p^5+5 p^4 - 10 p^3 +10 p^2- 5 p + 1=P(Y=0)= constant \;term \;in\;G_Y(t)$$
$$p_1= 5 p^5- 20 p^4 +30 p^3- 20 p^2+5p =P(Y=1)= coefficient\; of\; t\; in\; G_Y(t)$$
so we have to find $$P(Y<2)=P(Y=0)+P(Y=1)$$
which is $$P(Y<2)=p_0+p_1$$
$$\Rightarrow P(Y<2)=(- p^5+5 p^4 - 10 p^3 +10 p^2- 5 p + 1)+(5 p^5- 20 p^4 +30 p^3- 20 p^2+5p)$$
$$\Rightarrow P(Y<2)=4p^5-15p^4+20p^3-10p^2+1$$
Here method 2 is lengthy but you should keep this method in mind because in some problems it is not lengthy and it is easy , here another thing that you should notice that t have not any negative powers it means  probability of negative  do not exist here but in some problems where  t have  any negative powers there probability for negative exists
