# Compounding vs Annualizing Returns in a Portfolio Optimization Context

This might be a rather basic question that might be closed... but I can't for the life of me understand why in many Google search results the annualization of daily returns is done like this:

r_yearly = (1+r_daily)^252 - 1

However, this is not the correct way to annualize returns in a portfolio optimization context. Meaning, if I were to annualize the daily returns and daily volatility of a stock to generate an annualized Sharpe ratio, it should be like:

r_yearly = r_daily * 252

vol_yearly = vol_daily * 252^0.5

Sharpe_yearly = r_yearly / vol_yearly

The reason why I am posting this is because I have wasted so much time searching about daily returns annualization and I have realized that there is a difference between annualizing and compounding. Most of the Google search results that stated annualization actually mean compounding. Can anyone confirm this and give some useful insights to understand/remember this well?

• One is for arithmetic returns, the other is for log returns. Oct 3 at 15:26
• Hi @amdopt, thanks. I suppose you are saying the 1st one (in my post) is for log and the second is arithmetic? I might have left out this question, but why is the annualization of volatility (shown above) done 'arithmetically'? Why is calculating Sharpe with the arithmetic returns correct but with the log returns wrong?
– Kai
Oct 3 at 15:45

Let's get two things straight:

1. Denote $$R$$ the average daily return of an asset. Then denote $$r \equiv \log(1+R)$$

2. Assume time period is 1 day. Then if working with levels the yearly return is: $$R_{year} = (1+R)^{252} - 1$$. If working in logs: $$r_{year} = 252*r$$.

Now for daily returns: $$\log(1+R) \approx R$$ which implies: $$252*R \approx 252*r$$ and therefore it does not matter much.

• Thanks @phdstudent, should little r equal to big R when both are in their annualized forms? Because when I tried both r's with their annualization methods, they gave different values.
– Kai
Oct 3 at 18:19
• Also, for your 2nd point, I assume you mean R_year = (1+R)^252 - 1?
– Kai
Oct 3 at 18:21
• For the second point you are correct. If you do the annualization follwoing point 2, then $1+R_A = (1+R_m)^{252}$ Take logs on both sides to get: $r_a = 252 log(1+Rm) = 252 * r_m$ Oct 3 at 18:42