Csdrhrt

$X sim mathcal{P}( lambda) $ and $Y sim mathcal{P}( mu)$ meaning that $X$ and $Y$ are Poisson distributions. What is the probability distribution law of $X + Y$. I know it is $X+Y sim mathcal{P}( lambda + mu)$ but I don't understand how to derive it.

This only holds if $X$ and $Y$ are independent, so we suppose this from now on. We have for $k ge 0$:
begin{align*}
P(X+ Y =k) &= sum_{i = 0}^k P(X+ Y = k, X = i)\
&= sum_{i=0}^k P(Y = k-i , X =i)\
&= sum_{i=0}^k P(Y = k-i)P(X=i)\
&= sum_{i=0}^k e^{-mu}frac{mu^{k-i}}{(k-i)!}e^{-lambda}frac{lambda^i}{i!}\
&= e^{-(mu + lambda)}frac 1{k!}sum_{i=0}^k frac{k!}{i!(k-i)!}mu^{k-i}lambda^i\
&= e^{-(mu + lambda)}frac 1{k!}sum_{i=0}^k binom kimu^{k-i}lambda^i\
&= frac{(mu + lambda)^k}{k!} cdot e^{-(mu + lambda)}
end{align*}
Hence, $X+ Y sim mathcal P(mu + lambda)$.

Another approach is to use characteristic functions. If $Xsim mathrm{po}(lambda)$, then the characteristic function of $X$ is (if this is unknown, just calculate it)
$$
varphi_X(t)=E[e^{itX}]=e^{lambda(e^{it}-1)},quad tinmathbb{R}.
$$
Now suppose that $X$ and $Y$ are independent Poisson distributed random variables with parameters $lambda$ and $mu$ respectively. Then due to the independence we have that
$$
varphi_{X+Y}(t)=varphi_X(t)varphi_Y(t)=e^{lambda(e^{it}-1)}e^{mu(e^{it}-1)}=e^{(mu+lambda)(e^{it}-1)},quad tinmathbb{R}.
$$
As the characteristic function completely determines the distribution, we conclude that $X+Ysimmathrm{po}(lambda+mu)$.

You can use Probability Generating Function(P.G.F). As poisson distribution is a discrete probability distribution, P.G.F. fits better in this case.For independent X and Y random variable which follows distribution Po($lambda$) and Po($mu$).
P.G.F of X is
begin{equation*}
begin{split}
P_X[t] = E[t^X]&= sum_{x=0}^{infty}t^xe^{-lambda}frac{lambda^x}{x!}\
&=sum_{x=0}^{infty}e^{-lambda}frac{(lambda t)^x}{x!}\
&=e^{-lambda}e^{lambda t}\
&=e^{-lambda (1-t)}\
end{split}
end{equation*}
P.G.F of Y is
begin{equation*}
begin{split}
P_Y[t] = E[t^Y]&= sum_{y=0}^{infty}t^ye^{-mu}frac{mu^y}{y!}\
&=sum_{y=0}^{infty}e^{-mu}frac{(mu t)^y}{y!}\
&=e^{-mu}e^{mu t}\
&=e^{-mu (1-t)}\
end{split}
end{equation*}

Now think about P.G.F of U = X+Y.
As X and Y are independent,
begin{equation*}
begin{split}
P_U(t)=P_{X+Y}(t)=P_X(t)P_Y(t)=E[t^{X+Y}]=E[t^X t^Y]&= E[t^X]E[t^Y]\
&= e^{-lambda (1-t)}e^{-mu (1-t)}\
&= e^{-(lambda+mu) (1-t)}\
end{split}
end{equation*}

Now this is the P.G.F of $Po(lambda + mu)$ distribution. Therefore,we can say U=X+Y follows Po($lambda+mu$)

In short, you can show this by using the fact that $$Pr(X+Y=k)=sum_{i=0}^kPr(X+Y=k, X=i).$$

If $X$ and $Y$ are independent, this is equal to
$$
Pr(X+Y=k)=sum_{i=0}^kPr(Y=k-i)Pr(X=i)
$$
which is
$$
begin{align}
Pr(X+Y=k)&=sum_{i=0}^kfrac{e^{-lambda_y}lambda_y^{k-i}}{(k-i)!}frac{e^{-lambda_x}lambda_x^i}{i!}\
&=e^{-lambda_y}e^{-lambda_x}sum_{i=0}^kfrac{lambda_y^{k-i}}{(k-i)!}frac{lambda_x^i}{i!}\
&=frac{e^{-(lambda_y+lambda_x)}}{k!}sum_{i=0}^kfrac{k!}{i!(k-i)!}lambda_y^{k-i}lambda_x^i\
&=frac{e^{-(lambda_y+lambda_x)}}{k!}sum_{i=0}^k{kchoose i}lambda_y^{k-i}lambda_x^i
end{align}
$$
The sum part is just
$$
sum_{i=0}^k{kchoose i}lambda_y^{k-i}lambda_x^i=(lambda_y+lambda_x)^k
$$
by the binomial theorem.
So the end result is
$$
begin{align}
Pr(X+Y=k)&=frac{e^{-(lambda_y+lambda_x)}}{k!}(lambda_y+lambda_x)^k
end{align}
$$
which is the pmf of $Po(lambda_y+lambda_x)$.

Using Moment Generating Function.

If $X sim mathcal{P}(lambda)$, $Y sim mathcal{P}(mu)$ and S=X+Y.

We know that MGF(Moment Generating Function) of $mathcal{P}(lambda)=e^{lambda(e^t-1)}$(See the end if you need proof)

MGF of S would be
$$begin{align}
M_S(t)&=E[e^{tS}]\&=E[e^{t(X+Y)}]\&=E[e^{tX}e^{tY}]\&=E[e^{tX}]E[e^{tY}]quad text{given }X,Ytext{ are independent}\&=e^{lambda(e^t-1)}e^{mu(e^t-1)}\&=e^{(lambda+mu)(e^t-1)}
end{align}$$

Thus S is a Poisson Distribution with parameter $lambda+mu$.

MGF of Poisson Distribution

If $X sim mathcal{P}(lambda)$, then by definition Probability Mass Function is

$$begin{align}
f_X(k)=frac{lambda^k}{k!}e^{-lambda},quad k in 0,1,2....
end{align}$$
It's MGF is
$$begin{align}
M_X(t)&=E[e^{tX}]\&=sum_{k=0}^{infty}frac{lambda^k}{k!}e^{-lambda}e^{tk}\&=e^{-lambda}sum_{k=0}^{infty}frac{lambda^ke^{tk}}{k!}\&=e^{-lambda}sum_{k=0}^{infty}frac{(lambda e^t)^k}{k!}\&=e^{-lambda}e^{lambda e^t}\&=e^{lambda e^t-lambda}\&=e^{lambda(e^t-1)}
end{align}$$