Deriving Vasicek Bond Prices by Reordering Stochastic Integrals

# Cumulative Integration with regard to Vasicek Model's Bond Price and its Forward Price


# Cumulative Integration with regard to Vasicek Model's Bond Price and its Forward Price












### (My Question)

Please show me how to compute the following expectation with its computation process. Besides, $B_t$ is S.B.M.

$$E\left[ \exp \left( - \int^T_t \int^u_0 \sigma e^{-b(u-s)} d B_s du \right) \right]$$

(Thank you for your help in advance.)

### (Cross-link)

I have posted the same question on https://math.stackexchange.com/questions/3343590/cumulative-integration-with-regard-to-vasicek-models-bond-price-and-its-forward

### (Original Question)

Solve $P(t, T)$ with the following model

$$dr_t=-br_t dt + \sigma dB_t$$

### (My consideration)

- Fist,

$$r_u=e^{-bu} r_0 + \int^u_0 \sigma e^{-b(u-s)} dB_s$$

- Second,

\begin{eqnarray} P(t, T) &=& E \left[ \exp \left( - \int^T_t r_u du \right) \middle| \mathcal{F}_t \right] \\ &=& E\left[ \exp \left( - \int^T_t \left(e^{-bu} r_0 + \int^u_0 \sigma e^{-b(u-s)} dB_s \right) du \right) \middle| \mathcal{F}_t \right] \\ &=& E\left[ \exp \left( - \int^T_t e^{-bu} r_0 du - \int^T_t \int^u_0 \sigma e^{-b(u-s)} dB_s du \right) \middle| \mathcal{F}_t \right] \\ &=& \frac{r_0}{b} (e^{-bT}-e^{-bt}) E\left[ \exp \left(- \int^T_t \int^u_0 \sigma e^{-b(u-s)} dB_s du \right) \middle| \mathcal{F}_t \right] \end{eqnarray}

- Third, I assume to use the following formula, but I cannot have any idea to replace the integration order.

$$E\left[ \exp \left( \int^T_t f(s) dB_s \right) \middle| \mathcal{F}_t \right] = \exp \left( \frac{1}{2} \int^T_t f(s)^2 ds \right) $$

### Thank you for your help in advance.

## Answer by koji (score 1, accepted)

https://quant.stackexchange.com/a/47523

### (My answer)

### the Vasicek Bond Price and its Forward Price

- Recall the result of Exercise 5.2.(1) or Exercise 4.5.(10). \begin{eqnarray} P(t, T) &=& E \left[ \exp \left( - \int^T_t r_u du \right) \middle| \mathcal{F}_t \right] \\ &=& E \left[ \exp \left( - \int^T_t \left( e^{-bu} r_0 + \sigma \int^u_0 e^{-b(u-s)} dB_s \right) du\right) \middle| \mathcal{F}_t \right] \\ &=& E \left[ \exp \left( - r_0 \int^T_t e^{-bu} du - \int^T_t\int^u_0 \sigma e^{-b(u-s)} dB_s du \right) \middle| \mathcal{F}_t \right] \\ &=& E \left[ \exp \left( - r_0 \left[ -\frac{1}{b} e^{-bu} \right]^T_t - \int^T_t\int^u_0 \sigma e^{-b(u-s)} dB_s du \right) \middle| \mathcal{F}_t \right] \\ %&=& E \left[ \exp \left( \frac{r_0}{b} (e^{-bT} -e^{-bt} ) - \int^T_t\int^u_0 \sigma e^{-b(u-s)} dB_s du \right) \middle| \mathcal{F}_t \right] \\ &=& \exp \left( \frac{r_0}{b} (e^{-bT} -e^{-bt} ) \right) \nonumber \\ && \qquad E \left[ \exp \left( - \int^T_t\int^u_0 \sigma e^{-b(u-s)} dB_s du \right) \middle| \mathcal{F}_t \right] \end{eqnarray}

- Here, one computes the exponential part inside of expectation using the following formula (Cumulative Integration Formula Replacing $du$ and $dB_s$. (Developed on Sep 5, 2019 by Kojima)) \begin{eqnarray} \int^t_0 \int^u_0 dB_s \ du &=& \int^t_0 \int^u_s du \ dB_s \\ \int^T_t \int^u_0 dB_s \ du &=& \int^T_0 \int^u_s du \ dB_s - \int^t_0 \int^u_s du \ dB_s \end{eqnarray}

- Therefore, \begin{eqnarray} && - \int^T_t\int^u_0 \sigma e^{-b(u-s)} dB_s \ du \nonumber \\ && \qquad = - \int^T_0 \int^u_0 \sigma e^{-b(u-s)} dB_s \ du + \int^t_0 \int^u_0 \sigma e^{-b(u-s)} dB_s \ du \\ && \qquad = - \int^T_0 \int^T_s \sigma e^{-b(u-s)} du \ dB_s + \int^t_0 \int^t_s \sigma e^{-b(u-s)} du \ dB_s \\ && \qquad = - \int^T_0 \left[ -\frac{ \sigma }{b} e^{-b(u-s)} \right]^T_s dB_s + \int^t_0 \left[ -\frac{ \sigma }{b} e^{-b(u-s)} \right]^t_s dB_s \\ && \qquad = \int^T_0 \frac{ \sigma }{b} \left( e^{-b(T-s)} -1 \right) dB_s - \int^t_0 \frac{ \sigma }{b} \left( e^{-b(t-s)} -1 \right) dB_s \\ && \qquad = \int^T_t \frac{ \sigma }{b} \left( e^{-b(T-s)} -1 \right) dB_s \nonumber \\ && \qquad \qquad + \int^t_0 \frac{ \sigma }{b} \left( e^{-b(T-s)} -1 \right) dB_s - \int^t_0 \frac{ \sigma }{b} \left( e^{-b(t-s)} -1 \right) dB_s \end{eqnarray}

- Substitute the above results into the exponential part. \begin{eqnarray} P(t, T) &=& \exp \left( \frac{r_0}{b} (e^{-bT} -e^{-bt} ) \right) \nonumber \\ && \qquad E \left[ \exp \left( - \int^T_t\int^u_0 \sigma e^{-b(u-s)} dB_s du \right) \middle| \mathcal{F}_t \right] \\ &=& \exp \left( \frac{r_0}{b} (e^{-bT} -e^{-bt} ) \right) \nonumber \\ && \ \cdot \exp \left( \int^t_0 \frac{ \sigma }{b} \left( e^{-b(T-s)} -1 \right) dB_s - \int^t_0 \frac{ \sigma }{b} \left( e^{-b(t-s)} -1 \right) dB_s \right) \\ && \ E \left[ \exp \left( \int^T_t \frac{ \sigma }{b} \left( e^{-b(T-s)} -1 \right) dB_s \right) \middle| \mathcal{F}_t \right] \end{eqnarray}

- Here, use the following formula. \begin{eqnarray} E\left[ \exp \left( \int^T_t f(s) dB_s \right) \middle| \mathcal{F}_t \right] = \exp \left( \frac{1}{2} \int^T_t f(s)^2 ds \right) \end{eqnarray}

- Therefore, \begin{eqnarray} P(t, T) &=& \exp \left( \frac{r_0}{b} (e^{-bT} -e^{-bt} ) \right) \nonumber \\ && \ \cdot \exp \left( \int^t_0 \frac{ \sigma }{b} \left( e^{-b(T-s)} -1 \right) dB_s - \int^t_0 \frac{ \sigma }{b} \left( e^{-b(t-s)} -1 \right) dB_s \right) \\ && \ E \left[ \exp \left( \int^T_t \frac{ \sigma }{b} \left( e^{-b(T-s)} -1 \right) dB_s \right) \middle| \mathcal{F}_t \right] \\ &=& \exp \left( \frac{r_0}{b} (e^{-bT} -e^{-bt} ) \right) \nonumber \\ && \ \cdot \exp \left( \int^t_0 \frac{ \sigma }{b} \left( e^{-b(T-s)} -1 \right) dB_s - \int^t_0 \frac{ \sigma }{b} \left( e^{-b(t-s)} -1 \right) dB_s \right) \\ && \ \cdot \exp \left( \frac{1}{2} \int^T_t \frac{ \sigma^2 }{b^2} \left( e^{-b(T-s)} -1 \right)^2 ds \right) \end{eqnarray}

- One computes the exponential part with letting it $\mathscr{P}(t, T, r)$. \begin{eqnarray} \mathscr{P}(t, T, r) &=& \frac{r_0}{b} (e^{-bT} -e^{-bt} ) \nonumber \\ && \qquad + \int^t_0 \frac{ \sigma }{b} \left( e^{-b(T-s)} -1 \right) dB_s - \int^t_0 \frac{ \sigma }{b} \left( e^{-b(t-s)} -1 \right) dB_s \nonumber \\ && \qquad + \frac{1}{2} \int^T_t \frac{ \sigma^2 }{b^2} \left( e^{-b(T-s)} -1 \right)^2 ds \\ &=& \frac{r_0}{b} (e^{-bT} -e^{-bt} ) \nonumber \\ && \qquad + \int^t_0 \frac{ \sigma }{b} e^{-b(T-s)} dB_s - \int^t_0 \frac{ \sigma }{b} e^{-b(t-s)} dB_s \nonumber \\ && \qquad + \frac{ \sigma^2 }{2b^2} \int^T_t \left( e^{-2b(T-s)} -2e^{-b(T-s)} + 1 \right) ds \\ &=& \frac{r_0}{b} (e^{-bT} -e^{-bt} ) \nonumber \\ && \qquad + \frac{1}{b} e^{-bT} e^{bt}\int^t_0 \sigma e^{-b(t-s)} dB_s - \frac{1}{b} \int^t_0 \sigma e^{-b(t-s)} dB_s \nonumber \\ && \qquad + \frac{ \sigma^2 }{2b^2} \left[ \frac{1}{2b} e^{-2b(T-s)} - \frac{2}{b} e^{-b(T-s)} + s \right]^T_t \end{eqnarray}

- Here, use the result. \begin{eqnarray} \sigma \int^t_0 e^{-b(t-s)} dB_s &=& r_t - e^{-bt} r_0 \\ \mathscr{P}(t, T, r) &=& \frac{r_0}{b} (e^{-bT} -e^{-bt} ) \nonumber \\ && \qquad + \frac{1}{b} e^{-bT} e^{bt}\int^t_0 \sigma e^{-b(t-s)} dB_s - \frac{1}{b} \int^t_0 \sigma e^{-b(t-s)} dB_s \nonumber \\ && \qquad + \frac{ \sigma^2 }{2b^2} \left[ \frac{1}{2b} e^{-2b(T-s)} - \frac{2}{b} e^{-b(T-s)} + s \right]^T_t \\ &=& \frac{r_0}{b} (e^{-bT} -e^{-bt} ) \nonumber \\ && \qquad + \frac{1}{b} e^{-b(T-t)} (r_t - e^{-bt} r_0 )- \frac{1}{b} (r_t - e^{-bt} r_0 ) \nonumber \\ && \qquad + \frac{ \sigma^2 }{2b^2} \frac{1}{2b} ( 1- e^{-2b(T-t)} ) \nonumber \\ && \qquad - \frac{ \sigma^2 }{2b^2} \frac{2}{b} ( 1- e^{-b(T-t)} ) + \frac{ \sigma^2 }{2b^2} (T-t) \end{eqnarray}

- Substitute the above results into the exponential part. Hence, one reaches the following result for $P(t, T)$. \begin{eqnarray} P(t, T) &=& \exp \left( \frac{r_0}{b} (e^{-bT} -e^{-bt} ) \right) \nonumber \\ && \qquad \cdot \exp \left( \frac{1}{b} e^{-b(T-t)} (r_t - e^{-bt} r_0 )- \frac{1}{b} (r_t - e^{-bt} r_0 ) \right) \nonumber \\ && \qquad \cdot \exp \left(\frac{ \sigma^2 }{4b^3} ( 1- e^{-2b(T-t)} ) \nonumber \right) \\ && \qquad \cdot \exp \left( -\frac{ \sigma^2 }{b^3} ( 1- e^{-b(T-t)} ) + \frac{ \sigma^2 }{2b^2} (T-t) \right) \end{eqnarray}

- Recall the definition of the forward rate $f(t, T, S)$. \begin{eqnarray} f(t, T, S) &\equiv& - \frac{ \log P(t, S) - \log P(t, T)}{S-T} \\ f(t, T_1, T_2) &\equiv& - \frac{ \log P(t, T_2) - \log P(t, T_1)}{T_2-T_1} \end{eqnarray}

- Here, one reaches $-\log P(t, T_2)$. \begin{eqnarray} -\log P(t, T_2) &=& - \frac{r_0}{b} e^{-bT_2} + \frac{r_0}{b} e^{-bt} -\frac{1}{b} e^{-b(T_2-t)} r_t + \frac{1}{b} e^{-b(T_2-t)} e^{-bt} r_0 \nonumber \\ && \qquad + \frac{1}{b} r_t - \frac{1}{b} e^{-bt} r_0 - \frac{\sigma^2}{4b^3} + \frac{\sigma^2}{4b^3} e^{-2b(T_2 -t)} \nonumber \\ && \qquad + \frac{\sigma^2}{b^3} -\frac{\sigma^2}{b^3}e^{-b(T_2 -t)} - \frac{\sigma^2}{2b^2} T_2 + \frac{\sigma^2}{2b^2} t \\ &=& - \frac{r_0}{b} e^{-bT_2} + \frac{r_0}{b} e^{-bt} -\frac{1}{b} e^{-b(T_2-t)} r_t + \frac{1}{b} e^{-bT_2} r_0 \nonumber \\ && \qquad + \frac{1}{b} r_t - \frac{1}{b} e^{-bt} r_0 - \frac{\sigma^2}{4b^3} + \frac{\sigma^2}{4b^3} e^{-2b(T_2 -t)} \nonumber \\ && \qquad + \frac{\sigma^2}{b^3} -\frac{\sigma^2}{b^3}e^{-b(T_2 -t)} - \frac{\sigma^2}{2b^2} T_2 + \frac{\sigma^2}{2b^2} t \end{eqnarray}

- Moreover, one reaches $\log P(t, T_1)$. \begin{eqnarray} \log P(t, T_1) &=& \frac{r_0}{b} e^{-bT_1} - \frac{r_0}{b} e^{-bt} + \frac{1}{b} e^{-b(T_1-t)} r_t - \frac{1}{b} e^{-bT_1} r_0 \nonumber \\ && \qquad - \frac{1}{b} r_t + \frac{1}{b} e^{-bt} r_0 + \frac{\sigma^2}{4b^3} - \frac{\sigma^2}{4b^3} e^{-2b(T_1 -t)} \nonumber \\ && \qquad - \frac{\sigma^2}{b^3} + \frac{\sigma^2}{b^3}e^{-b(T_1 -t)} + \frac{\sigma^2}{2b^2} T_1 - \frac{\sigma^2}{2b^2} t \end{eqnarray}

- Therefore, one reaches the following equation. \begin{eqnarray} -\log P(t, T_2) + \log P(t, T_1) &=& - \frac{r_t}{b} ( e^{-b(T_2-t)} - e^{-b(T_1-t)} ) \nonumber \\ && \qquad + \frac{\sigma^2}{4b^3} ( e^{-2b(T_2-t)} - e^{-2b(T_1-t)} ) \nonumber \\ && \qquad - \frac{\sigma^2}{b^3} ( e^{-b(T_2-t)} - e^{-b(T_1-t)} ) \nonumber \\ && \qquad - \frac{\sigma^2}{2b^2} (T_2 -T_1) \\ &=& - \frac{\sigma^2}{2b^2} (T_2 -T_1) \nonumber \\ && \qquad - \left( \frac{r_t}{b} + \frac{\sigma^2}{b^3} \right) ( e^{-b(T_2-t)} - e^{-b(T_1-t)} ) \nonumber \\ && \qquad + \frac{\sigma^2}{4b^3} ( e^{-2b(T_2-t)} - e^{-2b(T_1-t)} ) \end{eqnarray}

- Hence, substitute the above result into the definition of the forward rate $f(t, T_1, T_2)$. \begin{eqnarray} f(t, T_1, T_2) &\equiv& - \frac{ \log P(t, T_2) - \log P(t, T_1)}{T_2-T_1} \\ &=& - \frac{\sigma^2}{2b^2} - \frac{1}{T_2 -T_1 } \left( \frac{r_t}{b} + \frac{\sigma^2}{b^3} \right) ( e^{-b(T_2-t)} - e^{-b(T_1-t)} ) \nonumber \\ && \qquad + \frac{1}{T_2 -T_1 } \frac{\sigma^2}{4b^3} ( e^{-2b(T_2-t)} - e^{-2b(T_1-t)} ) \end{eqnarray}

$\square$

## Answer by koji (score 1)

https://quant.stackexchange.com/a/47524

### (Cumulative Integration Formula Replacing $du$ and $dB_s$ )

- I have developed formulas to solve this by myself!

\begin{eqnarray} \int^t_0 \int^u_0 dB_s \ du &=& \int^t_0 \int^u_s du \ dB_s \\ \int^T_t \int^u_0 dB_s \ du &=& \int^T_0 \int^u_s du \ dB_s - \int^t_0 \int^u_s du \ dB_s \end{eqnarray}

- Therefore, we can use the following formula as I assume.

$$E\left[ \exp \left( \int^T_t f(s) dB_s \right) \middle| \mathcal{F}_t \right] = \exp \left( \frac{1}{2} \int^T_t f(s)^2 ds \right) $$

- I solved the Vasicek Bond Price computation using the formulas.