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I have a strategy in Samuelson model with zero safe rate defined as $$Z_t^{\Pi}=\frac{X_t^{\Pi}}{X_t^{\rho}} \quad \quad (1)$$ where $$\frac{dX_t^{\Pi}}{X_t^{\Pi}} = \mu \pi dt + \sigma \pi \ dW_t \quad \quad (2)$$ $$ \frac{dX_t^{\rho}}{X_t^{\rho}} = \mu \rho dt + \sigma \rho \ dW_t \quad \quad (3)$$

what gives the following dynamic $$\frac{dZ_t^{\Pi}}{Z_t^{\Pi}} = (\mu -\sigma^2 \rho )(\pi - \rho) dt + \sigma (\pi - \rho)\ dW_t \quad \quad (4)$$

To prove that $\rho=\frac{\mu}{\sigma^2} $ is the optimal strategy for the $$\max_{\Pi} E [ logX_T^{\Pi}] \quad \quad (5)$$

Using (1), logarithmic property, Jensen's inequality and supermartingale property I can derive the below inequality $$ E \big{[} log(X_t^{\Pi}) \big{]} \leq E \big{[} log(X_t^{\rho}) \big{]} \quad \quad (6)$$

The question I have is how the inequality (5) implies that the optimal strategy is $\rho=\frac{\mu}{\sigma^2} $?

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  • $\begingroup$ should $\max_{\Pi} E [ logX_T^{\Pi}]$ in (5) be $\max_{\pi} E [ logZ_T^{\Pi}]$ $\endgroup$ – Gordon May 16 '16 at 13:03
  • $\begingroup$ no, it is $\max_{\Pi} E [ logX_T^{\Pi}]$ in the text. The solution manual suggests $$ E [ logX_T^{\Pi}] - E [ logX_T^{\rho}] = E [ logZ_T^{\Pi}] \leq log E [ Z_T^{\Pi}] \leq 0$$ $\endgroup$ – Michal May 16 '16 at 13:11
  • $\begingroup$ Then why do you need $Z$ and $X^{\rho}$? The dynamics for $X$ does not involve $\rho$. $\endgroup$ – Gordon May 16 '16 at 13:14
  • $\begingroup$ not sure, if it was $Z_T^{\Pi}$ then the drift of the $\frac{dZ_t^{\Pi}}{Z_t^{\Pi}} $ is equal zero when $\rho=\frac{\mu}{\sigma^2}$, but how does this imply the optimal strategy? $\endgroup$ – Michal May 16 '16 at 13:19
  • $\begingroup$ Note also that, though $\pi$ is a parameter, while $\Pi$ is not. That is the reason I assume that the problem is $\max_{\pi}E[\ln Z_T^{\Pi}]$. Please double check your questions. Or should it be $\max_{\rho}E[\ln Z_T^{\Pi}]$ $\endgroup$ – Gordon May 16 '16 at 13:23
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I assume that the problem is $$\max_{\pi} E\left(\ln Z_T^{\Pi} \right).$$ Note that $\ln Z_t^{\Pi} = \ln X_t^{\Pi} -\ln X_t^{\rho}$. Moreover, \begin{align*} d\ln Z_t^{\Pi} &= d\ln X_t^{\Pi} -d\ln X_t^{\rho}\\ &=\Big[\big(\mu \pi - \frac{1}{2}\sigma^2 \pi^2\big) - \big(\mu \rho- \frac{1}{2}\sigma^2 \rho^2\big) \Big]dt + \sigma(\pi-\rho)dW_t. \end{align*} Then \begin{align*} E\left(\ln Z_T^{\Pi} \right) = \Big[\big(\mu \pi - \frac{1}{2}\sigma^2 \pi^2\big) - \big(\mu \rho- \frac{1}{2}\sigma^2 \rho^2\big) \Big]T. \end{align*} Consequently, \begin{align*} \max_{\pi}E\left(\ln Z_T^{\Pi} \right) &= \max_{\pi}\Big[\big(\mu \pi - \frac{1}{2}\sigma^2 \pi^2\big) - \big(\mu \rho- \frac{1}{2}\sigma^2 \rho^2\big) \Big]T\\ &=\max_{\pi}\Big[-\frac{1}{2}\sigma^2\big( \pi - \frac{\mu}{\sigma^2} \big)^2 + \frac{1}{2}\mu^2- \big(\mu \rho- \frac{1}{2}\sigma^2 \rho^2\big) \Big]T, \end{align*} which is a maximization problem for a quadratic function of $\pi$. It is then clear that the maximum is achieved at $\pi = \frac{\mu}{\sigma^2}$.

EDIT

Consider the problem $\max_{\pi}E\left(\ln X_T^{\Pi} \right)$. Note that \begin{align*} d\ln X_t^{\Pi} = \big(\mu \pi - \frac{1}{2}\sigma^2 \pi^2\big)dt + \sigma \pi dW_t. \end{align*} Then, \begin{align*} \max_{\pi}E\left(\ln X_T^{\Pi} \right) &= \max_{\pi}\big(\mu \pi - \frac{1}{2}\sigma^2 \pi^2\big)T\\ &=\max_{\pi}\Big[-\frac{1}{2}\sigma^2\big( \pi - \frac{\mu}{\sigma^2} \big)^2 + \frac{1}{2}\mu^2 \Big]T, \end{align*} which is again a maximization problem for a quadratic function of $\pi$, and the maximum is achieved at $\pi = \frac{\mu}{\sigma^2}$.

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  • $\begingroup$ yes that's clear. how about if it indeed was $\max_{\Pi} E [ logX_T^{\Pi}]$ ? would you use the Jensen's inequality? the (6) itself is not showing anything yet in this case, is it? $\endgroup$ – Michal May 16 '16 at 14:43
  • $\begingroup$ @Michal: See updates above. However, I do not know how do you get (6). I am not using the Jensen's inequality. $\endgroup$ – Gordon May 16 '16 at 15:01
  • $\begingroup$ $$E [ logZ_T^{\Pi}] \leq log E [ Z_T^{\Pi}]$$ ,$$E [ logX_T^{\Pi}] - E [ logX_T^{\rho}] = E [ logZ_T^{\Pi}] \leq log E [ Z_T^{\Pi}] $$ $\endgroup$ – Michal May 16 '16 at 15:11
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    $\begingroup$ @Michal: Jensen's inequality is not needed for the question that you have asked. $\endgroup$ – Gordon May 16 '16 at 15:30

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