MGF and Variance of a Poisson process

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.

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 $$M_{3N(3)}(z) = M_{N(3)}(3z) = e^{(3\lambda)(e^{3z} - 1)}.$$

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.