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-1

$$\dfrac{dP_1}{dP}=\dfrac{f(B_1)}{f(B)}=\dfrac{\frac{1}{\sqrt{2\pi}\sigma_1} e^{ -\frac{(x-\mu_1)^2}{2\sigma_1^2} })}{\frac{1}{\sqrt{2\pi}\sigma} e^{ -\frac{(x-\mu)^2}{2\sigma^2} }}=\dfrac{\sigma_1}{\sigma}e^{\dfrac{(x-\mu)^2}{2\sigma^2} -\dfrac{(x-\mu_1)^2}{2\sigma_1^2}}=\dfrac{t}{t}e^{\dfrac{(x-0)^2}{2t^2} -\dfrac{(x-\mu t)^2}{2t^2}}=e^\dfrac{x^2+x^2-2x\mu ...


0

Note that $$\frac{dQ_{T_p}}{dQ}|_{T_0} = \frac{P(T_0, T_p)}{P(0, T_p)}\frac{A(0, T_0, T_n)}{A(T_0, T_0, T_n)}$$. Then $$E^{Q_{T_p}}\big(S(T_0, T_n)\big) = E^Q\bigg(S(T_0, T_n) \frac{P(T_0, T_p)}{P(0, T_p)}\frac{A(0, T_0, T_n)}{A(T_0, T_0, T_n)}\bigg) \\ = \frac{A(0, T_0, T_n)}{P(0, T_p)} E^Q\bigg(S(T_0, T_n) \frac{P(T_0, T_p)}{A(T_0, T_0, T_n)}\bigg).$$ That ...



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