Variance of an Exponentially Weighted Poisson Jump Sum

# Variance of Poisson integral


# Variance of Poisson integral












I am looking for the solution to the expression $$\text{Var}(\int_t^Te^{-a(T-s)}\ln(J)\text{d}N_s)$$ where $\ln(J)\sim N(\mu, \sigma^2)$ and $N$ has intensity $\lambda$. The term comes from a mean reverting spot process which has lognormal jumps. The log spot is obtained using Ito's formula and using the integrating factor method. The Jump part of this term has the for $$\int_t^Te^{-a(T-s)}\ln(J)\text{d}N_s.$$ I want to find the variance of this as it contributes to the volatility of the process.

I have tried to compute it using $$\mathbb{E}[\int_t^Te^{-a(T-s)}\ln(J)\text{d}N_s]=\lambda \mathbb{E}[\ln(J)]\int_t^Te^{-a(T-s)}\text{d}s$$ and $$\mathbb{E}[(\int_t^Te^{-a(T-s)}\ln(J)\text{d}N_s)^2]=\lambda \mathbb{E}[\ln(J)^2]\int_t^Te^{-2a(T-s)}\text{d}s,$$ however when combined to find the variance, it gives me negative values when $a$ is small. I have also tried computing it using probability methods i.e., simulating the jumps times using an exponential distribution and computing the random variable $$Y=\sum_{i=1}^{N_{T-t}}e^{-a(T-T_i)}\ln(J)_i$$ where $\ln(J)_i$ are draws from the above normal distribution then using the realisations of this, compute the variance. But it gives me a different answer to the above. Any help would be appreciated.

## Answer by Kurt G. (score 1)

https://quant.stackexchange.com/a/82092

Something is wrong with your second expectation (where you have the square in it). It also looks like you are silently assuming that $J$ and $N$ are independent.

Let's consider the martingale $$\tag1 M_t=N_t-\lambda t $$ whose quadratic variation is $$\tag2 [M]_t=N_t $$ as shown here.

By the Ito isometry for semimartingales we get \begin{align*}\tag3\require{ulem} &\mathbb E\left[\left(\textstyle\int_t^Te^{-a(T-s)}\,dM_s\right)^2\right]= \mathbb E\left[\textstyle\int_t^Te^{-2a(T-s)}\,d[M]_s\right]= \mathbb E\left[\textstyle\int_t^Te^{-2a(T-s)}\,dN_s\right]\\[2mm] &\stackrel{(*)}=\mathbb E\left[\textstyle\int_t^Te^{-2a(T-s)}\,\lambda\,ds\right]=\lambda e^{-2aT}\tfrac{e^{2aT}-e^{2at}}{2a}=\underline{\underline{\lambda\tfrac{1-e^{-2a(T-t)}}{2a}}}\,.\tag4 \end{align*} In (*) I have used that $$\tag5 \textstyle\int_0^te^{-2a(T-s)}dM_s=\int_0^te^{-2a(T-s)}d(N_s-\lambda s) $$ is a martingale.

By now we know that the variance of $$\tag6 \textstyle\int_t^Te^{-a(T-s)}\,dM_s $$ is given by the RHS of (4). But this is obviously also the variance of $$\tag7 Y:=\textstyle\int_t^Te^{-a(T-s)}\,dN_s $$ because (6) and (7) differ only by a deterministic term.

Correction.

The following is clear from the martingale property: $$\tag8 \mathbb E[Y]=\mathbb E\left[\textstyle\int_t^Te^{-a(T-s)}\,dN_s\right]= \textstyle\int_t^Te^{-a(T-s)}\,\lambda\,ds=\lambda e^{-aT}\frac{e^{aT}-e^{at}}{a} =\lambda \frac{1-e^{-a(T-t)}}{a}\,. $$ Therefore, by (4) and (8), \begin{align}\tag9 \mathbb E[Y^2]&=\operatorname{Var}[Y]+\mathbb E[Y]^2=\lambda\tfrac{1-e^{-2a(T-t)}}{2a}+\lambda^2\left(\tfrac{1-e^{-a(T-t)}}{a}\right)^2\,. \end{align} It is also clear that for two independent r.v. $X,Y$ we have $$\tag{10} \operatorname{Var}[XY]=\mathbb E[X^2]\mathbb E[Y^2]-\mathbb E[X]^2\mathbb E[Y]^2\,. $$ Applying this to $X=\ln J$ gives in total \begin{align} &\operatorname{Var}\left[\textstyle\int_t^Te^{-a(T-s)}\ln J\,dN_s\right]\\[2mm]&= \mathbb E[(\ln J)^2]\left\{ \lambda\tfrac{1-e^{-2a(T-t)}}{2a}+\lambda^2\left(\tfrac{1-e^{-a(T-t)}}{a}\right)^2\right\}\\[2mm]&~-\mathbb E[\ln J]^2\lambda^2\left(\tfrac{1-e^{-a(T-t)}}{a}\right)^2\\[2mm] &=\underline{\underline{\mathbb E[(\ln J)^2] \lambda\tfrac{1-e^{-2a(T-t)}}{2a}+\operatorname{Var}[\ln J]\lambda^2\left(\tfrac{1-e^{-a(T-t)}}{a}\right)^2}}\,.\tag{11} \end{align} Sanity check: For $a\to 0$ this becomes $$\tag{12} \mathbb E[(\ln J)^2]\lambda(T-t)+\operatorname{Var}[\ln J]\lambda^2(T-t)^2\,. $$ Setting $a=0$ from the beginning we get a variance of \begin{align} &\operatorname{Var}[\ln J(N_T-N_t)]\\[2mm] &\stackrel{(10)}=\mathbb E[(\ln J)^2]\mathbb E[(N_T-N_t)^2]-\mathbb E[\ln J]^2\mathbb E[N_T-N_t]^2\\[2mm] &=\mathbb E[(\ln J)^2]\Big\{\operatorname{Var}[N_T-N_t]+\mathbb E[N_T-N_t]^2\Big\}-\mathbb E[\ln J]^2\mathbb E[N_T-N_t]^2\\[2mm] &=\mathbb E[(\ln J)^2]\Big\{\lambda(T-t)+\lambda^2(T-t)^2\Big\} -\mathbb E[\ln J]^2\lambda^2(T-t)^2\tag{13} \end{align} which is seen to be equal to (12).