X,Y are independent random variables with the following characteristic functions:
$ \phi_X(\theta) = \frac{1}{4}e^{i\theta}+\frac{3}{4}e^{i2\theta} \\ \phi_Y(\theta) = exp(e^{i\theta}-1)=e^{-1}\sum_{k=0}^\infty(\frac{e^{i\theta k}}{k!})$
Find: P(X+Y = 2)
Taking the product of the characteristic functions (since the characteristic function of the convolution/sum of X + Y is the product of their char functions), we get:
$\phi_{X+Y}(\theta)=\frac{1}{4}e^{-1}\sum_{K=0}^\infty(\frac{e^{i\theta (k+1)}}{k!})+\frac{3}{4}e^{-1}\sum_{K=0}^\infty(\frac{e^{i\theta (k+2)}}{k!}) \\$
Somehow, my lecturer goes from this line directly to: $\\$ $\therefore P(X+Y=2)=\frac{1}{4}e^{-1}\frac{1}{1!}+\frac{3}{4}e^{-1}\frac{1}{0!}=e^{-1} $
Could someone please explain how my lecturer has made that step above. I can't seem the work it out. I thought maybe he was using some property of the Poisson distribution, since Y is Poisson distributed with $\lambda=1$. But, I couldn't see how he did that when the (K+1) and (K+2) components were in the indices, for the characteristic function of the sum.
You can read off the distributions. For $X$, it is $1$ with probability $1/4$ and $2$ with probability $3/4$. Now $\Pr(X+Y=2)$ is an elementary calculation.