Probability of Attaining a Return Goal

Earlier we looked at calculating the probability of beating a fixed target. Now we will look at calculating the probability of beating a benchmark which is itself stochastic.

Let us consider two assets A and B with the following details:

MeanStandard DeviationCorrelation
A$$\mu_{A}=10\%$$$$\sigma_{A}=20\%$$$$\rho_{AB}=30\%$$
B$$\mu_{B}=12\%$$$$\sigma_{B}=26\%$$

We have a total of $10 million to invest. Our objective is to beat a benchmark.

Let us take the 50-50 portfolio, which has the following returns:

$$r_{1} = 0.5A + 0.5B$$

Suppose the benchmark has the following returns:

$$r_{2} = 0.4A + 0.6B$$

We need to find that probability that our portfolio will beat the benchmark index, i.e., $$P(r{1} > r{2})$$

This can be expressed as:

$$P(r{1} - r{2} > 0)$$

We can write this as:

$$P(0.5A + 0.5B - 0.4A - 0.6B > 0)$$

or

$$P(0.1A - 0.1B) > 0$$

0.1A - 0.1B is normally distributed.

Therefore, it's mean and standard deviation will be given as follows:

Mean, $$E(0.1A-0.1B) = 0.1\mu{A} - 0.1\mu{B} = 0.1(10\%)-0.1(12\%) = -0.2\%$$

Standard Deviation, $$\sigma(0.1A-0.1B) = \sqrt{0.1^{2}\sigma{A}^{2}+0.1^{2}\sigma{B}^{2}-2(0.1)(0.1)\sigma{A}\sigma{B}\rho}$$

$$= 2.76\%$$

We can write our probability as follows:

$$P(0.1A-0.1B>0) = P\left ( \frac{0.1A-0.1B+0.002}{0.0276} > \frac{0 + 0.002}{0.0276} \right )$$

$$= P(Z>0.0725)$$

where Z is the standard normal variable.

$$P(Z>0.0725) = 47.1\%$$, using 1-NORMSDIST(0.0725) in excel.

Therefore, the 50-50 portfolio has a 47.1% chance of beating the benchmark portfolio of 40-60.

This probability of beating the benchmar depends on the correlation between the assets. With high correlation, the probability will decrease and vice verse.