Let $X_t = \sum_{i=1}^{N_t} Y_i$ and $N_t$ be a Poisson process with
intensity $\lambda >0$. Suppose $Y_i$ are i.i.d. (independent of
$N_t$) with normal distribution $N(m,\sigma^2)$. Compute $Var(X_t)$.
The MGF of $X_t$ is:
$\psi_{X_t}(u) = e^{\lambda t (\psi_Y(u) -1)}$
Then,
$\dfrac{d}{du}\psi_{X_t}(u) = \lambda t \dfrac{d}{du}\psi_{Y}(u) e^{\lambda t (\psi_Y(u) -1)}$
And,
$\dfrac{d^2}{du^2}\psi_{X_t}(u) = (\lambda t)^2 (\dfrac{d}{du}\psi_{Y}(u))^2 + \lambda t \dfrac{d^2}{du^2}\psi_{Y}(u) e^{\lambda t (\psi_Y(u) -1)}$
So,
$E(X_t^2) = \dfrac{d^2}{du^2}\psi_{X_t}(u) |_{u=0} = (\lambda t)^2 m^2 + \lambda t \sigma ^2 $
$=>Var(X_t) = (\lambda t)^2 m^2 + \lambda t \sigma ^2 – (\lambda t m)^2 = \lambda t \sigma ^2 $
But on wikipedia they have: $\lambda t m^2 + \lambda t \sigma ^2 $
Why is my answer not identical to wikipedia? They use a different method so I cannot track my error.
Best Answer
By the law of total variance, \begin{align} \mathrm{Var}(X(t)) &= \mathbb E[\mathrm{Var}(X(t))\mid N(t)] + \mathrm{Var}(\mathbb E[X(t)\mid N(t)])\\ &= \mathbb E[N(t)\mathrm{Var}(Y_1)] + \mathrm{Var}(N(t)\mathbb E[Y_1])\\ &= \sigma^2\mathbb E[N(t)] + m^2\mathrm{Var}(N(t))\\ &= \sigma^2\lambda t+ m^2\lambda t\\ &= (\sigma^2 + m^2)\lambda t. \end{align} The MGF of $X(t)$ is $$\varphi_{X(t)}(s)=e^{\lambda t\left(\varphi_{Y_1}(s)-1\right)} $$ where $$\varphi_{Y_1}(s) = e^{ms + \frac12\sigma^2s^2}.$$ We have \begin{align} \frac{\mathsf d}{\mathsf ds}\varphi_{Y_1}(s)&=\left(m+s \sigma^2\right) e^{ms+\frac12\sigma^2 s^2},\\ \frac{\mathsf d^2}{\mathsf d^2s}\varphi_{Y_1}(s)&=\sigma ^2 e^{ms+\frac12\sigma^2 s^2}+\left(m+s \sigma ^2\right)^2 e^{ms+\frac12\sigma^2 s^2}, \end{align} so \begin{align} \mathbb E[Y_1]&=\varphi_{Y_1}'(0)=m\\ \mathbb E[Y_1^2]&=\varphi_{Y_1}''(0)=m^2+\sigma^2. \\ \end{align} Hence \begin{align} \frac{\mathsf d}{\mathsf ds}\varphi_{X(t)}(s)&=\lambda t\varphi_{Y_1}'(s) e^{\lambda t\left(\varphi_{Y_1}(s)-1\right)} \\ \frac{\mathsf d^2}{\mathsf d^2s}\varphi_{X(t)}(s)&= \lambda t\left(\lambda t\varphi_{Y_1}'(s)^2 + \varphi_{Y_1}''(s)\right)e^{\lambda t\left(\varphi_{Y_1}(s)-1\right)} \end{align} so that \begin{align} \mathbb E[X(t)]&=\varphi_{X(t)}'(0)=\lambda tm \\ \mathbb E[X(t)^2]&=\varphi_{X(t)}''(0)= \lambda t(\lambda tm^2+m^2+\sigma^2). \\ \end{align} Therefore \begin{align} \operatorname{Var}(X(t)) &= E[X(t)^2] - E[X(t)]^2\\ &= \lambda t(\lambda tm^2+m^2+\sigma^2) - \lambda^2 t^2 m^2\\ &= \lambda t(\lambda tm^2+m^2+\sigma^2 - \lambda t m^2)\\ &= \lambda t(m^2+\sigma^2). \end{align}