# Using Normal Distribution to forecast active return

I wanted advice on how to go about forecasting active return via a standard normal distribution,

The asset is a security with annual volatility of 6%. The benchmark is a 5% annual return with 0% volatility (basically a straight line)

Without a benchmark, i would just use the annual volatility and state that over a year, there is a 15.8% chance of this security returning <-6%.

But how would I approach this with a benchmark that has 0% volatility.

I want to be able to state that: there is an 15.8% chance this security will underperform the benchmark by X%. What would 1 std be in active return terms? -1%?

## 1 Answer

If your benchmark has a volatility of zero and return 5% then for a benchmark investment of 1.0 after 1 year it is guaranteed to be worth 1.05, or $$X_b = 1.05$$

On the other hand your second investment needs to have an expected return, your calculations have assumed mean of 0% and vol of 6%, so the value at 1Y is given by $$X \sim \mathcal{N}(1.00, 0.06) \;,$$ which is why you have a 15.8% chance the investment is worth less than 0.94.

If you want to look at the outperformance, $Y$, then:

$$Y = X - X_b \sim \mathcal{N}(-0.05, 0.06)$$

The probability of $Y$ being less than zero, (i.e. underperforming) the benchmark is $$P(Y<0) = \Phi(0) = 79.8\%$$, and this makes much sense since the expectation of your investment is much less than the non volatile benchmark.

If you want there to be a 15.8% chance $X$ underperforms $X_b$ then you must assume that $X$ is drawn from $X \sim \mathcal{N}(1.05,0.06)$, i.e. the expected return of second investment is same as benchmark, so that $Y \sim \mathcal{N}(0,0.06)$.

• thank you. I was thinking that i would use -0.05 for the mean, but that seems unrealistic. I'll use the strategy's cagr - 5% for the mean – QFqs Jun 1 '18 at 13:46