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I am trying to understand stochastic calculus and got stuck calculating the following. I need the distribution of a zero bond under the black model, so I am deriving the variance using the second moment of $\int_{t}^{t+h} r_s ds$

Given the differential equation for Rt
$dr_t = \mu dt + \sigma dWt^Q$ with $W$ standard Brownian motion

I want to calculate

$\mathbb{E}_t^Q\big[\big(\int_{t}^{t+h} r_s ds\big)^2\big]$

I have gotten this far:

$=\mathbb{E}_t^Q\big(\int_{t}^{t+h} r_s ds\int_{t}^{t+h}r_udu\big)$
$=\int_{t}^{t+h}\int_{t}^{t+h} \big(\mathbb{E}_t^Q r_s r_u\big) dsdu$

Now I know that $r_s$ and $r_u$ are normally distributed, but where to go from here?

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Could you maybe use $Var[\int_t^{t+h}r_sds]=E[(\int_t^{t+h}r_sds)^2]-(E[\int_t^{t+h}r_sds])^2 $. Then if $r_s$ is Normal, then the integral itself follows Normal distribution with $\mu=\int_t^{t+h}E[r(s)]ds$ and $\sigma^2=\int_t^{t+h} \int_t^{t+h}E[r_s r_t]dtds$

Then you yo must solve the SDE and plug everything in

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  • $\begingroup$ I am actually looking for the steps that solve the integral $\int_{t}^{t+h}\int_{t}^{t+h} \big(\mathbb{E}_t^Q r_s r_u\big) dsdu$= ... into $1/3 \sigma^2 h^3 + (E[\int_t^{t+h}r_sds])^2$ Sadly your answer does not help me yet. $\endgroup$ – Neb Feb 21 '16 at 13:02
  • $\begingroup$ can you tell me what $E[r_t]$ is? $\endgroup$ – Trelokoritso Feb 21 '16 at 13:13
  • $\begingroup$ $r_t$ is known at time t, so the conditional expection is just $ r_t$ $\endgroup$ – Neb Feb 21 '16 at 13:36

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