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Define stochastic processes $\displaystyle y_t := \exp\Big(-\int_0^t r_s \,ds \Big)$. The subscript denotes the time variable dependency. We apply the Ito's lemma recursively. In fact, the same procedure leads to the Ito version of the Taylor expansion to an arbitrary order.

Apply Ito's Lemma twice \begin{align} -(y_t-y_0) &=\int_0^tr_sy_sds \\ &=\int_0^tds\Big(r_0y_0+\int_0^s(y_udr_u+r_udy_u)\Big) \\ &=r_0t+\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u)+\int_0^tdB_u\,y_u\sigma_u(t-u) \tag1 \end{align} since $dr_tdy_t=-r_ty_tdr_tdt=0$. The last integral comes from exchanging the order of two integrations.

Apply Ito's lemma again \begin{align} d\big(y(\mu-r^2)\big) &= y\left[\Big(r^3-3\mu r+\mu\frac{\partial\mu}{\partial r}+\frac{\partial\mu}{\partial t}+\sigma^2\Big(\frac12\frac{\partial^2\mu}{\partial r^2}-1\Big)\Big)dt+y\sigma\Big(\frac{\partial\mu}{\partial r}-2r\Big)dB\right] \\ &=:\alpha\, dt+\beta\, dB_t. \end{align} Then interchange the order of integration, \begin{align} &2\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u) \\ = & y_0(\mu_0-r_0^2)t^2+\int_0^tds\ \alpha_s(t-s)^2 + \int_0^t dB\,\beta_s(t-s)^2. \end{align}

Substitute the above equation into Eq. (1), then take the expectation of Eq. (1). As $\mathbb E[\alpha]$ is bounded, $$\mathbb E[y_t] = 1-r_0t+\frac12(\mu_0-r_0^2)t^2+O(t^3)=\exp\Big(-r_0t+\Big(\frac12\mu_0-r_0^2\Big)t^2+O(t^3)\Big)$$$$\mathbb E[y_t] = 1-r_0t-\frac12(\mu_0-r_0^2)t^2+O(t^3)=\exp\Big(-r_0t-\frac12\mu_0t^2+O(t^3)\Big)$$ as $t\searrow 0$.

It remains to fill in details regarding the interchange of integration and expectation by Fubini's theorem, which should be justified using a priori estimates on the stochastic quantities using the strong linear growth assumptions.

Define stochastic processes $\displaystyle y_t := \exp\Big(-\int_0^t r_s \,ds \Big)$. The subscript denotes the time variable dependency. We apply the Ito's lemma recursively. In fact, the same procedure leads to the Ito version of the Taylor expansion to an arbitrary order.

Apply Ito's Lemma twice \begin{align} -(y_t-y_0) &=\int_0^tr_sy_sds \\ &=\int_0^tds\Big(r_0y_0+\int_0^s(y_udr_u+r_udy_u)\Big) \\ &=r_0t+\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u)+\int_0^tdB_u\,y_u\sigma_u(t-u) \tag1 \end{align} since $dr_tdy_t=-r_ty_tdr_tdt=0$. The last integral comes from exchanging the order of two integrations.

Apply Ito's lemma again \begin{align} d\big(y(\mu-r^2)\big) &= y\left[\Big(r^3-3\mu r+\mu\frac{\partial\mu}{\partial r}+\frac{\partial\mu}{\partial t}+\sigma^2\Big(\frac12\frac{\partial^2\mu}{\partial r^2}-1\Big)\Big)dt+y\sigma\Big(\frac{\partial\mu}{\partial r}-2r\Big)dB\right] \\ &=:\alpha\, dt+\beta\, dB_t. \end{align} Then interchange the order of integration, \begin{align} &2\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u) \\ = & y_0(\mu_0-r_0^2)t^2+\int_0^tds\ \alpha_s(t-s)^2 + \int_0^t dB\,\beta_s(t-s)^2. \end{align}

Substitute the above equation into Eq. (1), then take the expectation of Eq. (1). As $\mathbb E[\alpha]$ is bounded, $$\mathbb E[y_t] = 1-r_0t+\frac12(\mu_0-r_0^2)t^2+O(t^3)=\exp\Big(-r_0t+\Big(\frac12\mu_0-r_0^2\Big)t^2+O(t^3)\Big)$$ as $t\searrow 0$.

It remains to fill in details regarding the interchange of integration and expectation by Fubini's theorem, which should be justified using a priori estimates on the stochastic quantities using the strong linear growth assumptions.

Define stochastic processes $\displaystyle y_t := \exp\Big(-\int_0^t r_s \,ds \Big)$. The subscript denotes the time variable dependency. We apply the Ito's lemma recursively. In fact, the same procedure leads to the Ito version of the Taylor expansion to an arbitrary order.

Apply Ito's Lemma twice \begin{align} -(y_t-y_0) &=\int_0^tr_sy_sds \\ &=\int_0^tds\Big(r_0y_0+\int_0^s(y_udr_u+r_udy_u)\Big) \\ &=r_0t+\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u)+\int_0^tdB_u\,y_u\sigma_u(t-u) \tag1 \end{align} since $dr_tdy_t=-r_ty_tdr_tdt=0$. The last integral comes from exchanging the order of two integrations.

Apply Ito's lemma again \begin{align} d\big(y(\mu-r^2)\big) &= y\left[\Big(r^3-3\mu r+\mu\frac{\partial\mu}{\partial r}+\frac{\partial\mu}{\partial t}+\sigma^2\Big(\frac12\frac{\partial^2\mu}{\partial r^2}-1\Big)\Big)dt+y\sigma\Big(\frac{\partial\mu}{\partial r}-2r\Big)dB\right] \\ &=:\alpha\, dt+\beta\, dB_t. \end{align} Then interchange the order of integration, \begin{align} &2\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u) \\ = & y_0(\mu_0-r_0^2)t^2+\int_0^tds\ \alpha_s(t-s)^2 + \int_0^t dB\,\beta_s(t-s)^2. \end{align}

Substitute the above equation into Eq. (1), then take the expectation of Eq. (1). As $\mathbb E[\alpha]$ is bounded, $$\mathbb E[y_t] = 1-r_0t-\frac12(\mu_0-r_0^2)t^2+O(t^3)=\exp\Big(-r_0t-\frac12\mu_0t^2+O(t^3)\Big)$$ as $t\searrow 0$.

It remains to fill in details regarding the interchange of integration and expectation by Fubini's theorem, which should be justified using a priori estimates on the stochastic quantities using the strong linear growth assumptions.

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Hans
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Define stochastic processes $\displaystyle y_t := \exp\Big(-\int_0^t r_s \,ds \Big)$. The subscript denotes the time variable dependency. We apply the Ito's lemma recursively. In fact, the same procedure leads to the Ito version of the Taylor expansion to an arbitrary order.

Apply Ito's Lemma twice \begin{align} -(y_t-y_0) &=\int_0^tr_sy_sds \\ &=\int_0^tds\Big(r_0y_0+\int_0^s(y_udr_u+r_udy_u)\Big) \\ &=r_0t+\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u)+\int_0^tdB_u\,y_u\sigma_u(t-u) \end{align}\begin{align} -(y_t-y_0) &=\int_0^tr_sy_sds \\ &=\int_0^tds\Big(r_0y_0+\int_0^s(y_udr_u+r_udy_u)\Big) \\ &=r_0t+\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u)+\int_0^tdB_u\,y_u\sigma_u(t-u) \tag1 \end{align} since $dr_tdy_t=-r_ty_tdr_tdt=0$. The last integral comes from exchanging the order of two integrations.

Apply Ito's lemma again \begin{align} d\big(y(\mu-r^2)\big) &= y\left[\Big(r^3-3\mu r+\mu\frac{\partial\mu}{\partial r}+\frac{\partial\mu}{\partial t}+\sigma^2\Big(\frac12\frac{\partial^2\mu}{\partial r^2}-1\Big)\Big)dt+y\sigma\Big(\frac{\partial\mu}{\partial r}-2r\Big)dB\right] \\ &=:\alpha\, dt+\beta\, dB_t. \end{align} Then interchange the order of integration, \begin{align} &2\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u) \\ = & y_0(\mu_0-r_0^2)t^2+\int_0^tds\ \alpha_s(t-s)^2 + \int_0^t dB\,\beta_s(t-s)^2. \end{align}

Take expectation ofSubstitute the above expressionequation into Eq. (1), asthen take the expectation of Eq. (1). As $\mathbb E[\alpha]$ is bounded, $$\mathbb E[y_t] = 1-r_0t+\frac12(\mu_0-r_0^2)t^2+O(t^3)=\exp\Big(-r_0t+\Big(\frac12\mu_0-r_0^2\Big)t^2+O(t^3)\Big)$$ as $t\searrow 0$.

It remains to fill in details regarding the interchange of integration and expectation by Fubini's theorem, which should be justified using a priori estimates on the stochastic quantities using the strong linear growth assumptions.

Define stochastic processes $\displaystyle y_t := \exp\Big(-\int_0^t r_s \,ds \Big)$. The subscript denotes the time variable dependency. We apply the Ito's lemma recursively. In fact, the same procedure leads to the Ito version of the Taylor expansion to an arbitrary order.

Apply Ito's Lemma twice \begin{align} -(y_t-y_0) &=\int_0^tr_sy_sds \\ &=\int_0^tds\Big(r_0y_0+\int_0^s(y_udr_u+r_udy_u)\Big) \\ &=r_0t+\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u)+\int_0^tdB_u\,y_u\sigma_u(t-u) \end{align} since $dr_tdy_t=-r_ty_tdr_tdt=0$. The last integral comes from exchanging the order of two integrations.

Apply Ito's lemma again \begin{align} d\big(y(\mu-r^2)\big) &= y\left[\Big(r^3-3\mu r+\mu\frac{\partial\mu}{\partial r}+\frac{\partial\mu}{\partial t}+\sigma^2\Big(\frac12\frac{\partial^2\mu}{\partial r^2}-1\Big)\Big)dt+y\sigma\Big(\frac{\partial\mu}{\partial r}-2r\Big)dB\right] \\ &=:\alpha\, dt+\beta\, dB_t. \end{align} Then interchange the order of integration, \begin{align} &2\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u) \\ = & y_0(\mu_0-r_0^2)t^2+\int_0^tds\ \alpha_s(t-s)^2 + \int_0^t dB\,\beta_s(t-s)^2. \end{align}

Take expectation of the above expression, as $\mathbb E[\alpha]$ is bounded, $$\mathbb E[y_t] = 1-r_0t+\frac12(\mu_0-r_0^2)t^2+O(t^3)=\exp\Big(-r_0t+\Big(\frac12\mu_0-r_0^2\Big)t^2+O(t^3)\Big)$$ as $t\searrow 0$.

It remains to fill in details regarding the interchange of integration and expectation by Fubini's theorem, which should be justified using a priori estimates on the stochastic quantities using the strong linear growth assumptions.

Define stochastic processes $\displaystyle y_t := \exp\Big(-\int_0^t r_s \,ds \Big)$. The subscript denotes the time variable dependency. We apply the Ito's lemma recursively. In fact, the same procedure leads to the Ito version of the Taylor expansion to an arbitrary order.

Apply Ito's Lemma twice \begin{align} -(y_t-y_0) &=\int_0^tr_sy_sds \\ &=\int_0^tds\Big(r_0y_0+\int_0^s(y_udr_u+r_udy_u)\Big) \\ &=r_0t+\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u)+\int_0^tdB_u\,y_u\sigma_u(t-u) \tag1 \end{align} since $dr_tdy_t=-r_ty_tdr_tdt=0$. The last integral comes from exchanging the order of two integrations.

Apply Ito's lemma again \begin{align} d\big(y(\mu-r^2)\big) &= y\left[\Big(r^3-3\mu r+\mu\frac{\partial\mu}{\partial r}+\frac{\partial\mu}{\partial t}+\sigma^2\Big(\frac12\frac{\partial^2\mu}{\partial r^2}-1\Big)\Big)dt+y\sigma\Big(\frac{\partial\mu}{\partial r}-2r\Big)dB\right] \\ &=:\alpha\, dt+\beta\, dB_t. \end{align} Then interchange the order of integration, \begin{align} &2\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u) \\ = & y_0(\mu_0-r_0^2)t^2+\int_0^tds\ \alpha_s(t-s)^2 + \int_0^t dB\,\beta_s(t-s)^2. \end{align}

Substitute the above equation into Eq. (1), then take the expectation of Eq. (1). As $\mathbb E[\alpha]$ is bounded, $$\mathbb E[y_t] = 1-r_0t+\frac12(\mu_0-r_0^2)t^2+O(t^3)=\exp\Big(-r_0t+\Big(\frac12\mu_0-r_0^2\Big)t^2+O(t^3)\Big)$$ as $t\searrow 0$.

It remains to fill in details regarding the interchange of integration and expectation by Fubini's theorem, which should be justified using a priori estimates on the stochastic quantities using the strong linear growth assumptions.

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Hans
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Define stochastic processes $\displaystyle y_t := \exp\Big(-\int_0^t r_s \,ds \Big)$. The subscript denotes the time variable dependency. We apply the Ito's lemma recursively. In fact, the same procedure leads to the Ito version of the Taylor expansion to an arbitrary order.

Apply Ito's Lemma twice \begin{align} -(y_t-y_0) &=\int_0^tr_sy_sds \\ &=\int_0^tds\Big(r_0y_0+\int_0^s(y_udr_u+r_udy_u)\Big) \\ &=r_0t+\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u)+\int_0^tdB_u\,y_u\sigma_u(t-u) \end{align} since $dr_tdy_t=-r_ty_tdr_tdt=0$. The last integral comes from exchanging the order of two integrations.

Apply Ito's lemma again \begin{align} d\big(y(\mu-r^2)\big) &= y\left[\Big(r^3-3\mu r+\mu\frac{\partial\mu}{\partial r}+\frac{\partial\mu}{\partial t}+\sigma^2\Big(\frac12\frac{\partial^2\mu}{\partial r^2}-1\Big)\Big)dt+y\sigma\Big(\frac{\partial\mu}{\partial r}-2r\Big)dB\right] \\ &=:\alpha\, dt+\beta\, dB_t. \end{align} Then interchange the order of integration, \begin{align} &2\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u) \\ = & y_0(\mu_0-r_0^2)t^2+\int_0^tds\ \alpha_s(t-s)^2 + \int_0^t dB\,\beta_s(t-s)^2. \end{align}

Take expectation of the above expression, as $\mathbb E [ \alpha]$$\mathbb E[\alpha]$ is bounded, $$\mathbb E [y_t] = 1-r_0t+\frac12(\mu_0-r_0^2)t^2+O(t^3)=\exp\Big(-r_0t+\Big(\frac12\mu_0-r_0^2\Big)t^2+O(t^3)\Big)$$$$\mathbb E[y_t] = 1-r_0t+\frac12(\mu_0-r_0^2)t^2+O(t^3)=\exp\Big(-r_0t+\Big(\frac12\mu_0-r_0^2\Big)t^2+O(t^3)\Big)$$ as $t\searrow 0$.

It remains to fill in details regarding the interchange of integration and expectation by Fubini's theorem, which should be justified using a priori estimates on the stochastic quantities using the strong linear growth assumptions.

Define stochastic processes $\displaystyle y_t := \exp\Big(-\int_0^t r_s \,ds \Big)$. The subscript denotes the time variable dependency. We apply the Ito's lemma recursively. In fact, the same procedure leads to the Ito version of the Taylor expansion to an arbitrary order.

Apply Ito's Lemma twice \begin{align} -(y_t-y_0) &=\int_0^tr_sy_sds \\ &=\int_0^tds\Big(r_0y_0+\int_0^s(y_udr_u+r_udy_u)\Big) \\ &=r_0t+\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u)+\int_0^tdB_u\,y_u\sigma_u(t-u) \end{align} since $dr_tdy_t=-r_ty_tdr_tdt=0$. The last integral comes from exchanging the order of two integrations.

Apply Ito's lemma again \begin{align} d\big(y(\mu-r^2)\big) &= y\left[\Big(r^3-3\mu r+\mu\frac{\partial\mu}{\partial r}+\frac{\partial\mu}{\partial t}+\sigma^2\Big(\frac12\frac{\partial^2\mu}{\partial r^2}-1\Big)\Big)dt+y\sigma\Big(\frac{\partial\mu}{\partial r}-2r\Big)dB\right] \\ &=:\alpha\, dt+\beta\, dB_t. \end{align} Then interchange the order of integration, \begin{align} &2\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u) \\ = & y_0(\mu_0-r_0^2)t^2+\int_0^tds\ \alpha_s(t-s)^2 + \int_0^t dB\,\beta_s(t-s)^2. \end{align}

Take expectation of the above expression, as $\mathbb E [ \alpha]$ is bounded, $$\mathbb E [y_t] = 1-r_0t+\frac12(\mu_0-r_0^2)t^2+O(t^3)=\exp\Big(-r_0t+\Big(\frac12\mu_0-r_0^2\Big)t^2+O(t^3)\Big)$$ as $t\searrow 0$.

It remains to fill in details regarding the interchange of integration and expectation by Fubini's theorem, which should be justified using a priori estimates on the stochastic quantities using the strong linear growth assumptions.

Define stochastic processes $\displaystyle y_t := \exp\Big(-\int_0^t r_s \,ds \Big)$. The subscript denotes the time variable dependency. We apply the Ito's lemma recursively. In fact, the same procedure leads to the Ito version of the Taylor expansion to an arbitrary order.

Apply Ito's Lemma twice \begin{align} -(y_t-y_0) &=\int_0^tr_sy_sds \\ &=\int_0^tds\Big(r_0y_0+\int_0^s(y_udr_u+r_udy_u)\Big) \\ &=r_0t+\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u)+\int_0^tdB_u\,y_u\sigma_u(t-u) \end{align} since $dr_tdy_t=-r_ty_tdr_tdt=0$. The last integral comes from exchanging the order of two integrations.

Apply Ito's lemma again \begin{align} d\big(y(\mu-r^2)\big) &= y\left[\Big(r^3-3\mu r+\mu\frac{\partial\mu}{\partial r}+\frac{\partial\mu}{\partial t}+\sigma^2\Big(\frac12\frac{\partial^2\mu}{\partial r^2}-1\Big)\Big)dt+y\sigma\Big(\frac{\partial\mu}{\partial r}-2r\Big)dB\right] \\ &=:\alpha\, dt+\beta\, dB_t. \end{align} Then interchange the order of integration, \begin{align} &2\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u) \\ = & y_0(\mu_0-r_0^2)t^2+\int_0^tds\ \alpha_s(t-s)^2 + \int_0^t dB\,\beta_s(t-s)^2. \end{align}

Take expectation of the above expression, as $\mathbb E[\alpha]$ is bounded, $$\mathbb E[y_t] = 1-r_0t+\frac12(\mu_0-r_0^2)t^2+O(t^3)=\exp\Big(-r_0t+\Big(\frac12\mu_0-r_0^2\Big)t^2+O(t^3)\Big)$$ as $t\searrow 0$.

It remains to fill in details regarding the interchange of integration and expectation by Fubini's theorem, which should be justified using a priori estimates on the stochastic quantities using the strong linear growth assumptions.

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