MGF of Poisson Z=X+2Y

Question

If $X\sim P(2)$ and $Y \sim P(3)$ using the moment generating function, what kind of distribution has random variable $Z=X+2Y$.

So far as I know :

$$ M_X(t)=e^{-\lambda(1-e^t)}=e^{-2(1-e^t)} $$ $$ M_Y(t)=e^{-\lambda(1-e^t)}=e^{-3(1-e^t)} $$ $$ M_{2Y}(t)=e^{-\lambda(1-e^t)}=e^{-3(1-e^{2t})} $$ $$ M_{X+Y}(t)=e^{-5(1-e^t)} $$

Since

$$ X+Y \sim P(\lambda_1+\lambda_2) $$

so I am guessing that $$ Z=X+2Y \sim P(\lambda_1+2\lambda_2)$$ but I can find it from the equation

$$M_{X+2Y}(t)=e^{-2(1-e^t)} e^{-3(1-e^{2t})}=e^{(-2(1-e^t)-3(1-e^{2t})}$$

I don't know how to rearrange this equation to get the distribution. How to get rid of "$2t$".

Answer 1

No, your $Z=X+2Y$ will not have a poisson distribution (you should expect it to have to small point probabilities at odd integers, and too high at even, compared to Poisson).

You didn't really say that $X$ and $Y$ are independent, but I will assume so much. EDIT: I found that this distribution is known as the Hermite distribution.

For this problem it will be convenient to use pgf's (probability generating function). The pgf for a Poisson variable with parameter $\lambda$ is $\DeclareMathOperator{\E}{\mathbb{E}}G(z)=\E z^X=e^{\lambda (z-1)}$. Then we calculate the pgf of $Z$ as $$ G_Z(z)= \E e^{X+2Y} = \E z^X \cdot \E e^{2Y}=G_X(z)G_Y(z^2) $$ and inserting the parameter values 2 for $X$, 3 for $Y$ we get $$ G_Z(z)=\exp\left\{ 2(z-1)+3(z^2-1) \right\} $$ which certainly is not the pgf of any Poisson random variable.

We can find the probability mass function by expanding the pgf in a power series. I will use maple:

G := z -> exp( 2*(z-1) + 3*(z^2-1) )
                              /             2\
                 G := z -> exp\2 z - 5 + 3 z /
S1 := series(G(z), z=0, 100) :
S2 :=  sort( convert(S1,polynom) ) :
ps := map(i -> coeff(S2,z,i), [seq(i,i=0..99)]) :
add(ps)
1230201581592231835090021705882732508724506091485915653564875278\

  48000526534413904564309823833442956185744700756387658829897786\

  002504758907395407/8289033054959573892412837535227749840302277\

  58541379236843775436718019022859048977460196496524216398837958\

  212205265551360000000000000000000000 exp(-5)
evalf(%)
                          1.000000000
evalf(ps)
[                                                             
[0.006737946999, 0.01347589400, 0.03368973500, 0.04941161132, 

  0.07524040818, 0.08939009688, 0.1050371071, 0.1066306851, 

  0.1054355016, 0.09451723482, 0.08216474795, 0.06649390044, 

  0.05216469070, 0.03871482954, 0.02788698595, 0.01920419661, 

  0.01285814431, 0.008290674603, 0.005207234170, 0.003166237682, 

  0.001878794019, 0.001083572101, 0.0006109049235, 

  0.0003357931503, 0.0001807089934, 0.00009504707551, 

  0.00004901338891, 0.00002475219374, 0.00001227088289, 

  0.000005967411318, 0.000002852003999, 0.000001338983094, 

                -7                -7                -7  
  6.184371932 10  , 2.809325136 10  , 1.256614172 10  , 

                -8                -8                -8  
  5.534051189 10  , 2.401804243 10  , 1.027240962 10  , 

                -9                -9                -10  
  4.332975626 10  , 1.802574590 10  , 7.400750733 10   , 

                -10                -10                -11  
  2.998926265 10   , 1.200056117 10   , 4.742713913 10   , 

                -11                -12                -12  
  1.852018065 10   , 7.146737691 10   , 2.726403463 10   , 

                -12                -13                -13  
  1.028366661 10   , 3.836490437 10   , 1.415815929 10   , 

                -14                -14                -15  
  5.170114897 10   , 1.868415403 10   , 6.684138498 10   , 

                -15                -16                -16  
  2.367418890 10   , 8.303642362 10   , 2.884589421 10   , 

                -17                -17                -17  
  9.926970182 10   , 3.384724660 10   , 1.143642594 10   , 

                -18                -18                -19  
  3.829768330 10   , 1.271301538 10   , 4.183805419 10   , 

                -19                -20                -20  
  1.365253276 10   , 4.417990328 10   , 1.417987144 10   , 

                -21                -21                -22  
  4.514448654 10   , 1.425880696 10   , 4.468425869 10   , 

                -22                -23                -23  
  1.389554316 10   , 4.288357079 10   , 1.313571045 10   , 

                -24                -24                -25  
  3.993983741 10   , 1.205586530 10   , 3.613024042 10   , 

                -25                -26                -27  
  1.075151891 10   , 3.177126405 10   , 9.324127135 10   , 

                -27                -28                -28  
  2.717868033 10   , 7.869294731 10   , 2.263426221 10   , 

                -29                -29                -30  
  6.467827602 10   , 1.836311462 10   , 5.180437626 10   , 

                -30                -31                -31  
  1.452283891 10   , 4.046094469 10   , 1.120343793 10   , 

                -32                -33                -33  
  3.083401674 10   , 8.435336890 10   , 2.294031526 10   , 

                -34                -34                -35  
  6.202256673 10   , 1.667182276 10   , 4.455813691 10   , 

                -35                -36                -37  
  1.184158304 10   , 3.129376210 10   , 8.224281987 10   , 

                -37                -38                -38  
  2.149590912 10   , 5.588007681 10   , 1.444860516 10   , 

                -39                -40]
  3.716098685 10   , 9.507457408 10   ]     

Answer 2

Taking the characteristic functions of the underlying random variables you have:

$$\begin{equation} \begin{aligned} \varphi_X(t) &= \exp \Big( 2 (e^{it}-1) \Big), \\[6pt] \varphi_{2Y}(t) &= \exp \Big( 3 (e^{2it}-1) \Big). \\[6pt] \end{aligned} \end{equation}$$

Assuming that $X$ and $Y$ are independent, you then have:

$$\varphi_Z(t) = \varphi_{X}(t) \cdot \varphi_{2Y}(t) = \exp \Big( 3 e^{2it} +2 e^{it} -5 \Big).$$

The distribution of $Z$ can be obtained by inverting the characteristic function via standard methods for inverse-Fourier transformation. Using a well-known inversion formula we have:

$$\begin{equation} \begin{aligned} p_Z(z) &= \lim_{T \rightarrow \infty} \frac{1}{2T} \int \limits_{-T}^{+T} e^{-itz} \varphi_Z(t) dt \\[6pt] &= \lim_{T \rightarrow \infty} \frac{1}{2T} \int \limits_{-T}^{+T} \exp \Big( 3 e^{2it} +2 e^{it} - itz -5 \Big) dt \\[6pt] \end{aligned} \end{equation}$$

This integral is hard to solve, so it is easier to proceed in this case by taking a standard convolution. Doing this yields the closed form solution:

$$\begin{equation} \begin{aligned} p_Z(z) &= \sum_{y=0}^{\lfloor z/2 \rfloor} p_Y(y) \cdot p_X(z-2y) \\[6pt] &= \exp(-5) \sum_{y=0}^{\lfloor z/2 \rfloor} \frac{3^y 2^{z-2y}}{y! (z-2y)!} \\[6pt] &= 2^z \exp(-5) \sum_{y=0}^{\lfloor z/2 \rfloor} \frac{(\tfrac{3}{4})^y}{y! (z-2y)!}. \\[6pt] \end{aligned} \end{equation}$$

You can create a function for this PMF in R as follows:

#Create probability mass function for Z
PMF <- function(z) { YY <- floor(z/2);
                     Y <- 0:YY;
                     for (i in 0:YY) { Y[i+1] <- dpois(i, 3)*dpois(z-2*i, 2); }
                     sum(Y); }

#Create data frame of PMF values from Z = 0,...,25
ZZ   <- 25;
Z    <- 0:ZZ;
Prob <- rep(0,length(ZZ));
for (z in 0:ZZ) { Prob[z+1] <- PMF(z) }
DATA <- data.frame(z = Z, Prob =  Prob)

#Plot the PMF
library(ggplot2);

FIGURE <- ggplot(data = DATA, aes(x = z, y = Prob)) +
          geom_bar(stat = 'identity', fill = 'red') +
          theme(plot.title = element_text(hjust = 0.5, face = 'bold')) + 
          ggtitle('Probability Mass Function') +
          xlab('z') + ylab('Probability');
FIGURE;