Question: Bob invests a certain sum of money in a scheme with a return of 22% p.a . After one year, he withdraws the entire amount (including the interest earned) and invests it in a new scheme with returns of 50% p.a (compounded annually) for the next two years. What is the compounded annual return on his initial investment over the 3 year period?
The answer to this problem is fairly simple if you assume initial investment to be say \$100 then calculate interest for 1st year at 22% then 2nd and 3rd year at 50% which would come out as \$274.5
Then return is \$174.5 over 3 years, using Compound Interest formula, you get rate of interest at around 40% for three years.
My question is can you skip all this lengthy process and use weighted averages to come up with the final answer?
$$
Average\ rate\ of\ Interest = \frac{1 * 22 + 2 * 50}{1 + 2} \approx 40.67\%
$$
The answer with this is off by 0.67%, it doesn't matter much. However, is using weighted averages a correct approach or am I getting the correct answer using a wrong approach?
Note: The goal of asking this question is to decide on a faster approach to this problem and not necessarily getting the final answer. If you have an approach faster than weighted averages (assuming it is correct), please feel free to post it as an answer.
Best Answer
The weighted averages a is not the correct approach. Here it is necessary to calculate the $\underline{\text{compounded}}$ annual return. After one period the you earn interest for the initial investment ($C_0$). Therefore after this period the initial investment increases to $C_1=(1+i_1)\cdot C_0$, where $i_k$ is the interest rate of the k-th period. And after n periods the growth of the investment $C_n=C_0\cdot \prod\limits_{k=1}^n (1+i_k)$. And the growth factor after $n$ periods is
$$g(n)=\frac{C_0\cdot \prod\limits_{k=1}^n (1+i_k)}{C_0}=\prod\limits_{k=1}^n (1+i_k)$$
To calculate the average growth rate we have to take the n-th root of $g(n)$ and then subtract 1. In other words, you obtain the compounded annual return. If you have a calculator it is not difficult to get the result. You just have to input $\left((1+i_1)\cdot (1+i_2)\cdot (1+i_3)\right)^{\frac13}$