I denote by $W_0$ and $W_1$ the wealth of an investor at $t=0$ and $t=1$, respectively. Let $r_f$ be the risk free rate, $r$ the vector of returns of the risky assets in excess of the risk free rate, and $w$ the vector of weights of the risky assets. Here is the classical mean-variance optimization problem: $$\max_{w} E(W_1)-\frac{\gamma}{2}Var(W_1)$$ $$\textrm{st.}\hspace{0.5cm} W_1=W_0(1+r_f+w'r)$$

Injecting the constraint into the optimization problem, the first order condition is thus written as follows: $$\frac{1}{\gamma} E(r)=W_0Var(r)w$$

My point is that I would like to end up with the classical mean-variance first-order condition: $$\frac{1}{\gamma} E(r)=Var(r)w$$

But I still have this $W_0$ in the equation... Did I miss something? Could someone please help me? Thanks


1 Answer 1


I get the maximization problem $$ \max\limits_{w} \mathbb{E}\left[W_1\right] - \frac{\gamma}{2} Var(W_1) $$ $$ st. W_1 = W_0(1 + r_f + w^Tr)$$ So we have \begin{align*} L(w) &= \mathbb{E}\left[W_1\right] - \frac{\gamma}{2} Var(W_1)\\ & = \mathbb{E}\left[W_0(1 + r_f + w^Tr)\right] - \frac{\gamma}{2} Var(W_0(1 + r_f + w^Tr))\\ & = W_0 + W_0r_f + W_0w^T\mathbb{E}\left[r\right] - \frac{\gamma}{2}W_0^2 w^TwVar(r) \end{align*} Building the derivative w.r.t. $w$

\begin{align*} \partial L(w) / \partial w &= W_0 \mathbb{E}\left[r\right] - \frac{\gamma}{2}W_0^2 2wVar(r)\\ &= W_0 \mathbb{E}\left[r\right] - \gamma W_0^2 wVar(r) \overset{!}{=} 0 \\ \Leftrightarrow \frac{1}{\gamma}\mathbb{E}\left[r\right] & = W_0Var(r)w \end{align*} So, I get $ w = \frac{\mathbb{E}[r]}{\gamma W_0Var(r)}$. Set initial wealth $W_0 = 1$ and you get the desired result. The optimal investment strategy $w$ is $W_0$ dependent.

  • $\begingroup$ Thanks, that is basically the calculus I do (except that your matrix calculation is not appropriate: prefer $w'Var(r)w$) and the result I get. So I was wondering whether it was a classical assumption in research papers to set $W_0=1$. $\endgroup$
    – Dadoo
    Apr 4, 2019 at 21:26
  • $\begingroup$ Do you have a link to the exercise? $\endgroup$
    – SmurfAcco
    Apr 4, 2019 at 21:30
  • $\begingroup$ It is not an exercise, it is my work. $\endgroup$
    – Dadoo
    Apr 5, 2019 at 11:03
  • $\begingroup$ just provide another perspective. the unit between the first and second term is differen. so the penalty gamma should have a unit. so the disapper of Wo can be absorbed by gamma. if you try to maximam sharp ratio you can find what gamma will be $\endgroup$
    – XYQ
    Apr 5, 2019 at 22:41

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