# The efficient frontier in mean variance criterion

The efficient frontier is the portfolios with the minimum of variance ($$V$$) at a given mean ($$E$$) or a maximum of mean at a given variance，Why do the optimal portfolios in the effcient frontier, is the efficient frontier consistent with the maximum of expected utility?

For example, assume a logarithmic utility function U=log(1+R), through Taylor series expansion and keep the first three terms, we can get the expected utility $$EU=E-(E^2+V)/2$$, from this equation, take the partial derivative of $$E$$, it is not monotonical increase, it is the same with $$V$$, it seems the efficient frontier does not lead to the maximum of expected utility.

I understand if the utility function is expoential, the expected utility $$EU=E-\lambda V$$, which is exactly consistent with the efficient frontiers. I just do not know the other form of utility function, like the logarithmic form as above.

Any help would be appreciated.

• I'm not clear on your question. In the optimization framework, you can maximize the return, for a given level of variance or minimize the variance for a given level of return. But there's no claim that the resulting portfolios are identical. Maybe you could frame your question more clearly. – mark leeds Sep 19 at 13:49
• Very appreciate your reply! I understand the portfolios in the efficient frontier are not identical, and I have modified my question. I just want to know Why the efficient frontier is consistent with the maximum of expected utility for all the utility functions. – Aeeh Sep 19 at 17:50
• Now I understand your question more clearly. I don't know the answer but are you sure that it's consistent for all utility functions ? Maybe Markowitz assumed some ultility function. It's been too long since I read his paper and I'm not even sure that I ever did. Hopefully someone who is more up on this material can say something useful. – mark leeds Sep 19 at 18:52
• @Alecos: Is the efficient frontier dependent on one's utility function ? Seems like it would have to be because of risk and return relationship needing to go in same direction. I don't remember Markowitz assuming one but like I said, I don't know if I ever even read the original paper. Thanks. – mark leeds Sep 20 at 18:28

Using $$\mathbb E$$ for the expected value symbol, $$E_R$$ and $$V_R$$ for the mean and the avriance of returns $$R$$, for a utility function of the form

$$U(R) = \ln(1+R)$$

the second-order Taylor expansion gives (ignoring the remainder)

$$U(R) \approx \ln(1+E_R) + \frac{1}{1+E_R} (R- E_R) -\frac{1}{2(1+E_R)^2}(R-E_R)^2.$$

Then

$$\mathbb E[U(R)] \approx E_R + 0 - \frac{V_R}{2(1+E_R)^2}.$$

We then have

$$\frac{\partial \mathbb E[U(R)] }{\partial V_R} = -\frac{1}{2(1+E_R)^2} < 0$$

so decreasing the variance always increases expected utility, while

$$\frac{\partial \mathbb E[U(R)] }{\partial E_R} = 1 + \frac{V_R}{(1+E_R)^3} <0$$.

The reason why this derivative is always positive, is that even with negative returns, by construction "returns" cannot fall below $$-100\% = -1$$.
So increasing mean returns always increases expected utility.

I don't see where the problem is. What am I missing?

• Thanks for your reply. In your case, the Tylor series of $E[U(R)]$ at $E_R$. But I use the Tylor series at $0$, which means $U(R)=R-\frac{R^2}{2}$, and $E(U)=E-(E^2+V)/2$, so it is different with your result, which one is correct? – Aeeh Sep 20 at 15:50
• @Aeeh If you use the Taylor series at zero, then the derivative of expected utility with respect to the variance is again always negative, while the derivative with respect to the mean value is $1-E_R$, which will be positive as long as $E_R <1$ i.e. it does not exceed $100\%$. Since you are approximating from $E_R=0$, the fact that the derivative on the approximation stays positive so far away from the approximation's base, should be considered enough for all practical purposes. – Alecos Papadopoulos Sep 21 at 0:57
• Thanks! The Taylor series at zero, the $E_R$ should be very small, less than 1, it will have a good approximation. I always think the $E_R$ is a very large value, this is the key point. Actually, it depends on the scale/unit. – Aeeh Sep 21 at 16:48
• @Aeeh "Returns" are usually very small values, like $0.01$, or $0.15$. This is why the approximation $\ln(1+R) \sim R$ is acceptable. – Alecos Papadopoulos Sep 21 at 16:51
• I think the 'return' should in some kind of scale, for example, 100000 dollars is very large (lager than 1), but 0.01 million dollars (smaller than 1). So I think It depends on the scale. – Aeeh Sep 21 at 16:54