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Let $r$ a random process defined by :

$$dr_t=\theta(t)dt + \sigma dW_t$$

$\theta$ is deterministic in $t$ and $W$ a brownian motion.

I don't know where my calculation below is going wrong :

Let $R=\int r(s)ds$

then : $$\frac{d}{dt}\mathop{\mathbb{E}}\left[ e^{-\int_t^Tr(s)ds}|\mathscr{F}_t \right] = \mathop{\mathbb{E}}\left[ \frac{d}{dt} e^{-(R_T - R_t)}|\mathscr{F}_t \right] = \mathop{\mathbb{E}}\left[ r(t) e^{-(R_T - R_t)}|\mathscr{F}_t \right] = r(t) \mathop{\mathbb{E}}\left[ e^{-\int_t^Tr(s)ds}|\mathscr{F}_t \right]$$

But regarding this question Bond dynamics in Ho Lee model my computation is not correct

Any help please?

JohnLord
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    What is your problem ? – Ezy Feb 14 '19 at 11:30
  • Regarding this question https://quant.stackexchange.com/questions/44002/bond-dynamics-in-ho-lee-model my computation is not correct – JohnLord Feb 14 '19 at 12:06
  • Your first equality is wrong. It is true that you can bring the derivative inside the expectation operator due to Lebesgue dominated convergence theorem (the function is bounded). However, you need to find the solution of the SDE for $r_t$ and then compute the mean of this Gaussian variable. – FunnyBuzer Feb 14 '19 at 12:37
  • @JohnLord then update your question to clarift. Also your notation is bad: do you wish to calculate partial derivative or total derivative ? Those are not the same. – Ezy Feb 14 '19 at 12:45
  • @FunnyBuzer why do you need the solution of $r_t$ SDE? after the derivative, the resulting $r_t$ is know given $F_t$ and we can simply take it out of the expectation. No? – JohnLord Feb 14 '19 at 13:42
  • If you express the partial derivative by definition, you will know where you were wrong. Note that you have $t$ in two places. Why do you ignore the other one? – Gordon Feb 14 '19 at 14:03
  • @JohnLord I suggest you revise the Leibniz integral rule: https://en.wikipedia.org/wiki/Leibniz_integral_rule – FunnyBuzer Feb 14 '19 at 14:04
  • @FunnyBuzer I didn't think of applying Leibniz formally in this example and I find it hard to do since the filtration is t-dependent. Can you show me how to do it taking into account the t-dependcy of the filtration? – JohnLord Feb 15 '19 at 08:45
  • Yes, that's why I suggest you solve the integral first, take the expectation (note that you'll have the a Gaussian integral in the exponent) and then differentiate. My answer below is a starting point. – FunnyBuzer Feb 15 '19 at 10:00

1 Answers1

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You can use the stochastic integration by parts to solve the integral inside the expectation: $$\int_t^Tr_sds=Tr_T-tr_t-\int_t^Tsdr_s=(T-t)r_t+\int_t^T(T-s)\underbrace{dr_s}_{:=\theta(s)ds+\sigma dW_s}$$ Solve this integral and then take the expectation and solve the derivative. For a complete answer see Ho and lee derivation for short rates model.

FunnyBuzer
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