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For the calculus of a financial derivatives, I need to compute the next expectation:

$$\mathbb{E}\left((\sum_{i=1}^{N_T} (J_i-k))_+\mid J_1+\cdots + J_{N_t}=x \right)$$

where $$(X_t-k)_+= \begin{cases} X_t-k & \text{if } X_t \geq k \\[6pt] 0 & \text{if } X_t < k \end{cases} $$

where $N_t$ is a Poisson Process, so $\sum_{i=1}^{N_T} J_i$ a compound Poisson process and $t <T$ fixed (non random). Furthermore $J_i$ are independent and identically distributed random variables and also independet of $N_t$.

Note: I think it would be a good starting point would be the next:

$$\mathbb{E}\left((\sum_{i=1}^{N_T} J_i-k))_+\mid J_1+\cdots + J_{N_t} = x \right) = \mathbb{E}\left(x+(\sum_{i=N_t}^{N_T} J_i-k)_+ \right) = \mathbb{E}\left(\mathbb{E} \left(x+\sum_{i=N_t}^{N_T} J_i-k\mid N_T-N_t=j\right)_+\right) =\mathbb{P}(N_T-N_t=j)(\mathbb{E}(\left(x+\sum_{i=1}^j J_i-k\right)_+ = \sum^{\infty}_{n=0} \mathbb E \left[ \left( x + \sum^{n}_{i=1}J_i -k\right)^+ \right] \frac{e^{-\lambda \tau}(\lambda \tau)^n}{n!} $$ Where, $$\mathbb E \left[ \left( x + \sum^{n}_{i=1}J_i -k\right)^+ \right] =\mathbb{E} \left[ \left(\sum^{n}_{i=1}J_i -(k-x)\right) 1_{S_n \geq k-x} \right]= \mathbb{E} (\sum^{n}_{i=1}J_i -(k-x))\mathbb{E}(1_{S_n \geq k-x})=(\mathbb{E} (\sum^{n}_{i=1}J_i)-k+x)\mathbb{P}(S_n \geq k-x)=(n\mathbb{E}(J_1)-k+x)(\mathbb{P}(S_n \geq k-x))$$

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    $\begingroup$ Are you not given the distribution of the jumps? $\endgroup$ – Slug Pue Jun 22 '15 at 19:16
  • $\begingroup$ No, I have to do for any distribution. However, if I obtain for example for the exponential jump distribution, it would be fine. $\endgroup$ – Edin_91 Jun 22 '15 at 19:22
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    $\begingroup$ and btw, should it actually be $$ E \left( (\sum^{N_T}_{i=1} J_i - k)_+ \mid J_1 + \cdots J_{N_t} = x \right) $$ instead of $$ E \left( \sum^{N_T}_{i=1} (J_i - k) \mid J_1 + \cdots J_{N_t} = x \right)_+ $$ ? Because as it is now, you are subtracting the strike $k$ from each jump $\endgroup$ – Slug Pue Jun 22 '15 at 19:33
  • $\begingroup$ Yes, you are completely ritght :) $\endgroup$ – Edin_91 Jun 22 '15 at 19:38
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    $\begingroup$ Are $t$ and $T$ supposed to be fixed, i.e. non-random? ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 22 '15 at 19:43
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Denote $J_1 + \cdots J_{N_t}$ by $S_{N_t}$. $$ \mathbb E \left[ \left( \sum^{N_T}_{i=1}J_i -k\right)^+ \mid S_{N_t}=x \right] = \mathbb E \left[ \left( \sum^{N_t}_{i=1}J_i + \sum^{N_T}_{i=N_t+1}J_i -k\right)^+ \mid S_{N_t}=x \right]\\ =\mathbb E \left[ \left( x + \sum^{N_T}_{i=N_t+1}J_i -k\right)^+ \mid S_{N_t}=x \right]\\ =\mathbb E \left[ \left( x + \sum^{N_{T-t}}_{i=1}J_i -k\right)^+ \right] $$ where the last line follows by independence of the jumps and independent/stationary increments of the Poisson process. Now condition on the no. of jumps in the interval $[0, T-t]$: $$ = \mathbb E \left[ \mathbb E \left[ \left( x + \sum^{N_{T-t}}_{i=1}J_i -k\right)^+ \mid N_{T-t} = n \right]\right]\\ = \sum^{\infty}_{n=0} \mathbb E \left[ \left( x + \sum^{n}_{i=1}J_i -k\right)^+ \right]\mathbb P(N_{T-t} = n) \\ = \sum^{\infty}_{n=0} \mathbb E \left[ \left( x + \sum^{n}_{i=1}J_i -k\right)^+ \right] \frac{e^{-\lambda \tau}(\lambda \tau)^n}{n!} $$ where $\tau = T-t$ and $\lambda$ is the intensity of the jumps . So now it remains to caluclate the expectation, which will depend on the distribution of the jumps: $$ \mathbb E \left[ \left( x + \sum^{n}_{i=1}J_i -k\right)^+ \right] = \mathbb E \left[ \left(\sum^{n}_{i=1}J_i -(k-x)\right) 1_{S_n \geq k-x} \right]\\ =\mathbb E \left[ \left(\sum^{n}_{i=1}J_i\right) 1_{S_n \geq k-x} \right] - (k-x)\mathbb P \left( S_n \geq k-x \right) $$ If for example $J_i \sim \mathcal N (\mu, \sigma^2)$ then $S_n \sim \mathcal N (n\mu, n\sigma^2)$ so in particlar $$ \mathbb E \left[ \left(\sum^{n}_{i=1}J_i\right) 1_{S_n \geq k-x} \right] = \int_{k-x}^\infty u \frac{1}{\sqrt{2 \pi n}\sigma} \exp \left( \frac{(u-n \mu)^2}{2n\sigma^2} \right) du $$ As you can see, inserting this expectation into the infinite series and then evaluate the series will not exactly be trivial.

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  • $\begingroup$ Thanks a lot Slungpue! It is what I have written, obviosly better explicated and fixing my error (x is summing and not multiplying). But how can I continue in that point? $\endgroup$ – Edin_91 Jun 22 '15 at 20:33
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    $\begingroup$ @Edin_91 you will have to find the distribution of the random variable $Y_n = \sum^n J_j$ and calculate $E[(Y_n-(k-x))^+]$ and then plug this into the infinite series. But if $J_i$ is for example Normally distributed you're not gonna get a closed-form expression (as far as I know). So what to do next highly depends on the distribution of the jumps. May I ask where this problem is from? Is it an exercise from a book or something else? $\endgroup$ – Slug Pue Jun 22 '15 at 21:40
  • $\begingroup$ Is it okay the part that I have completed in my answer? it can´t be developed more? $\endgroup$ – Edin_91 Jun 23 '15 at 13:02
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    $\begingroup$ @Edin_91 I added some lines for demonstration. Your derivation beyond what I wrote is wrong, unfortunately. The expectation of the product in not a product of the expectations, as the random variables are not independent. $\endgroup$ – Slug Pue Jun 23 '15 at 13:30
  • $\begingroup$ @ Slungpue It wouldn´t be, $$\mathbb E \left[ \left(\sum^{n}_{i=1}J_i\right) 1_{S_n \geq k-x} \right] = \int_{k-x}^\infty u \frac{1}{\sigma \sqrt{2 \pi n}} \exp \left( \frac{(x-n \mu)^2}{2n\sigma^2} \right) du$$ And, $$\mathbb E \left[ \left( x + \sum^{n}_{i=1}J_i -k\right)^+ \right] = \mathbb E \left[ \left(\sum^{n}_{i=1}J_i -(k-x)\right) 1_{S_n \geq k-x} \right]\\ =\mathbb E \left[ \left(\sum^{n}_{i=1}J_i\right) 1_{S_n \geq k-x} \right] - (k-x)\mathbb P \left( S_n \geq k-x \right)$$ $\endgroup$ – Edin_91 Jun 23 '15 at 13:51

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