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Let $N_t$ be a Poisson process with intensity $\lambda>0$ and $S_t$ follows a pure jump process $$dS_t=S_t(J_t-1)dN_t$$ where $J_t$ is the jump size variable if $N_t$ jumps at time $t$. Also, assume $J_t$ follows a lognormal distribution such that $\ln{ J_t}\sim N(\mu_j,\sigma_j^2)$ and $J_t$ is also independent of $N_t$ . How can we find $Var(S_T| S_t)$ ?

Thanks for any hint.

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    $\begingroup$ Have you tried applying ito's lemma (for discontinuous semi martingales) to $\ln(S_t)$ to find the solution of the SDE? $\endgroup$
    – Quantuple
    Nov 20, 2016 at 12:43
  • $\begingroup$ yes but I can't recieve the solution. please add your solution. $\endgroup$
    – math
    Nov 20, 2016 at 14:03
  • $\begingroup$ this document might help you better understand the nice answer provided by Behrouz Maleki: google.be/url?sa=t&source=web&rct=j&url=http://… $\endgroup$
    – Quantuple
    Nov 20, 2016 at 21:49

2 Answers 2

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Hint:

By application of Extended Ito's lemmma, we have $$d(\ln S_t)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\left(\frac{dS_t}{S_{t^-}}\right)^n\tag 1$$ Note $S_{t^-}$ denote the value of $S_t$ before a jump event. We know $d[N_t,N_t]=dN_t$, thus $$d(\ln S_t)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}(J_t-1)^n=\left(\ln{J_t} \right)dN_t\tag 2$$ By taking integrals on $[t,T]$, we have $$\int_{t}^{T}d(\ln S_u)=\int_{t}^{T}\left(\ln{J_u} \right)dN_u\tag 3$$ therefore $$\ln\left(\frac{S_T}{S_t}\right)=\sum_{n=1}^{N_{T-t}}\ln J_n\tag 4$$ in other words $$S_T=S_t\prod_{n=1}^{N_{T-t}}J_n\tag 5$$ Note $J_n \in (0, 2]$ is the jump size occurring at time instant $t_n$. Also $N_{T-t}=N_T-N_t$ is the total number of jumps in the time interval $(t,T]$. Since $\ln J_n\sim \mathcal{N}(\mu_J,\sigma^2_J)$ are independent and identically distribute, $\ln S_T=\ln S_t+\sum_{n=1}^{N_{T-t}}\ln J_n$ follows a normal distribution.Now apply equation $(5)$ $$\text{Var}\left(S_T\Big{|}S_t\right)=\text{Var}\left(S_t\prod_{n=1}^{N_{T-t}}J_n\Big{|}S_t\right)=S_t^2\,\text{Var}\left(\prod_{n=1}^{N_{T-t}}J_n\right)\\=S_t^2\,\text{Var}\left(\exp\left(\sum_{n=1}^{N_{T-t}}\ln J_n\right)\right)\\ $$ then $$\text{Var}\left(S_T\Big{|}S_t\right)=S_t^2\left(\underbrace{\mathbb{E}\left[\exp\left(\sum_{n=1}^{N_{T-t}}\ln J_n^2\right)\right]}_{I}-\underbrace{\mathbb{E}\left[\exp\left(\sum_{n=1}^{N_{T-t}}\ln J_n\right)\right]^2}_{J}\right)\tag 6$$ Note $$I=\mathbb{E}\left[\mathbb{E}\left[\exp\left(\sum_{n=1}^{N_{T-t}}\ln J_n^2\right)\right]\Big{|}N_{T-t}\right]=\exp\left(\lambda(T-t)(E[J_t^2]-1)\right)\\I=\exp\left(\lambda(T-t)e^{2\mu_J+2\sigma_J^2}-\lambda(T-t)\right)\tag 7$$ similarly $$J=\mathbb{E}\left[\mathbb{E}\left[\exp\left(\sum_{n=1}^{N_{T-t}}\ln J_n\right)\right]\Big{|}N_{T-t}\right]^2=\exp\left(2\lambda(T-t)e^{\mu_J+\frac 12\sigma_J^2}-2\lambda(T-t)\right)\tag 8$$ $(6)$, $(7)$ and $(8)$ $$\color{red}{\text{Var}\left(S_T\Big{|}S_t\right)=S_t^2\exp\left(\lambda(T-t)e^{2\mu_J+2\sigma_J^2}-\lambda(T-t)\right)\\\qquad\qquad\quad\quad -S_t^2\exp\left(2\lambda(T-t)e^{\mu_J+\frac 12\sigma_J^2}-2\lambda(T-t)\right)}$$

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  • $\begingroup$ Thanks but your solution is so hard. May you insert some details? $\endgroup$
    – math
    Nov 20, 2016 at 14:04
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    $\begingroup$ @math you are already lucky he took the time to develop all this given the fact you showed none of your effort in your question. $\endgroup$
    – SRKX
    Nov 20, 2016 at 14:58
  • $\begingroup$ @math I think my solution is clear and easy. $\endgroup$
    – user16651
    Nov 20, 2016 at 15:12
  • $\begingroup$ @SRKX I tried, but unfortunately I didn't solve my question.There is some complexities in Behrouz's solution $\endgroup$
    – math
    Nov 20, 2016 at 16:00
  • $\begingroup$ @math lots of clarifications were added, but if you're still struggling with something, be specific. $\endgroup$
    – SRKX
    Nov 20, 2016 at 16:05
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Alternatively, let $\{\tau_i\}_{i=1}^{\infty}$ be the jump time of the Poisson process $N$. Moreover, let \begin{align*} X_t = \int_0^t (J_s-1)dN_s. \end{align*} Here, we assume that the jump sizes $J(\tau_i)$, for $i=1, \ldots, \infty$, are independent identically distributed. Then, \begin{align*} X_t &= \sum_{0<s\le t}(J_s-1)1_{\Delta N(s) >0}. \end{align*} Note that \begin{align*} dS_t = S_{t-} dX_t, \end{align*} whose solution is the Doleans-Dade exponential, that is, \begin{align*} S_t &= S_0 \prod_{0 < s \le t} (1+ \Delta X_s)\\ &=S_0 \prod_{0 < s \le t} J_s 1_{\Delta N(s) >0}\\ &=S_0\prod_{i=1}^{N_t}J(\tau_i). \end{align*} For $T>t \ge 0$, \begin{align*} S_T &= S_t \prod_{i=N_t+1}^{N_T}J(\tau_i)\\ &= S_t \prod_{i=1}^{N_T-N_t}J(\tau_i) \quad \text{(in distribution)}\\ &= S_t \prod_{i=1}^{N_{T-t}}J(\tau_i) \quad \text{(in distribution)}. \end{align*} The remaining is the same as that of @Behrouz Maleki above.

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  • $\begingroup$ A nice approach +1 $\endgroup$
    – user16651
    Nov 21, 2016 at 14:30

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