MGF and Variance of a Poisson process

Question

If $N(t)$ is a Poisson process with rate $\lambda$ and $X = 2N(1)+3N(4)$, find the variance and MGF of X.

$$\begin{align} Var(X) &= Var\big(2N(1)+3N(4)\big)\\ &= Var\big(2N(1)+3\big(N(4) - N(1)) + N(1) \big)\big)\\ &= Var\big(2N(1)+3\big(N(4) - N(1)\big) + 3N(1)\big)\\ &= Var\big(5N(1)+3\big(N(4) - N(1)\big)\big)\\ &= 25Var(N(1)) + 9Var(N(3)) + 0 \\ &= 52 \lambda \end{align}$$

And then for the MGF, since MGF of a Poisson process is $e^{\lambda t (e^{x} - 1)}$ where $t$ is the time and $x$ is the MGF parameter.

$$\begin{align} MGF(X) &= MGF\big(2N(1)+3N(4)\big)\\ &= MGF(5N(1)) \times MGF\big(3\big(N(4) - N(1)\big)\big)\\ &= 5MGF(N(1)) \times 3MGF\big(\big(N(4) - N(1)\big)\big)\\ &= 5MGF(N(1)) \times 3MGF(N(3))\\ &= 5e^{\lambda (e^{x} - 1)} \times 3e^{3\lambda (e^{x} - 1)}\\ &= 15e^{4\lambda (e^{x} - 1)} \end{align}$$

I think my variance is correct, that's why my MGF must be wrong since when I calculate the variance based on my MGF ($M''(0) - (M'(0))^2$), they don't match. Can anyone point out where I'm doing it wrong?

Thanks.

Answer 1

The variance is correct. The error in the MGF can be found if we go back to the definition of MGF:

$$M_X(z) = \operatorname{E}[e^{zX}]$$ for some random variable $X$.

So if $X = c N(t)$ for some scalar $c$ and Poisson counting process up to time $t$, then $$M_{cN(t)}(z) = \operatorname{E}[e^{(cz)N(t)}] = M_{N(t)}(cz);$$ that is to say, the MGF of the scaled counting process $cN(t)$ is equal to the MGF of the unscaled process $N(t)$ evaluated at $cz$. You cannot simply pull out the scaling constant as you have done.

Specifically then, $$M_{5N(1)}(z) = M_{N(1)}(5z) = e^{\lambda(e^{5z} - 1)},$$ and $$M_{3N(3)}(z) = M_{N(3)}(3z) = e^{(3\lambda)(e^{3z} - 1)}.$$

Answer 2

There's something else wrong with your MGF: it should be $1$ at $x=0$${}^\dagger$, not $15$. Note an arbitrary variable $Y$ satisfies $M_{aY}(x)=\Bbb Ee^{axY}=M_Y(ax)$, whereas you've assumed it's $aM_Y(x)$. The MGF should therefore be$$M(x)=\exp[\lambda(e^{5x}-1)]\exp[3\lambda(e^{3x}-1)]=\exp[\lambda(e^{5x}+3e^{3x}-4)].$$You'll find that gives a variance of $52\lambda$.

${}^\dagger$ Admittedly we normally call the MGF's parameter $k$ or $t$ rather than $x$, but no matter.