The situation as you wrote it has no random elements, are you sure you typed it correctly?
Hi everyone, I have a question regarding calculating the shortfall probability in the context of a retirement plan and would greatly appreciate any useful advice on how to solve it.
Imagine you have a retirement plan with a financial institution. At t=0, you deposit $10,000 with the instituion. At the start of every month, you will withdraw 5% of your account balance for own consumption. The remaining funds is carried over to the next period with an interest rate of 2%. On top of that, the first $6,000 of your account balance will receive an additional 1% as bonus. Since the amount you withdrew every month (x) is a percentage of the account balance brought forward from the previous month, it varies from month to month. Hence, you are interested to know what is the probability that the amount withdrawn will be less than your benchmark value of $400 (this is know as the shortfall probability i.e. P(x<$400)). Given these assumptions, what is the shortfall probability?
In my opinion, it is not possible to solve this question analytically and obtain a closed form solution because the evolution of the account balance depends on the interest rate earned and we cannot tell in advance what is the probability that it will earn "(2% + 1%) on the full amount" or "2% on full amount + 1% on the first $6,000". Since we cannot solve for the account balance and amount withdrawn analytically, we also cannot solve for its shortfall probability analytically. As such, I believe we have to rely on a Monte-carlo simulation to run say 10,000 independent paths and see what's the probability of the shortfall. Nevertheless, this is just my opinion and I may be mistaken. Therefore, any advice from anyone with ideas will be greatly appreciated.
Thank you!
Hi SpringFan 25, I will attempt to write it in a mathematical form. V_t is the value of your account at the beginning of period t. ω is your withdrawal rate (5% in our e.g.). R_t is the interest rate of 2% (which is not static, and we can just assume mean=2% with stdev=1% for simplicity).
You withdraw ω% of your account value at the start of every period. You will carry forward (1-ω)∙V_(t-1) from period t-1. This amount will earn R_t with an additional 1% on the first $6,000 only.
V_t=(1-ω)∙V_(t-1)∙(1+R_t ) +
(1-ω)∙V_(t-1)∙1% if (1-ω)∙V_(t-1)≤6,000
or
60,000∙1% if (1-ω)∙V_(t-1)>6,000
As you can see the value at the beginning of time t (V_t) depends on the value from the previous period (V_t-1) and this value from the previous period earns an additional interest which depends on whether the value is more than or less than $6,000. Therefore, we need to somehow know V has evolved previously to determine what V is in the current period, only then can we take ω∙V_t to determine the the amount withdrawn this period. Since we do not know ex-ante the probability of whether (1-ω)∙V_(t-1) is more than or less than $6,000, I believe here is where we need a Monte Carlo simulation to generate numerous independent paths to find different evolutions of V and finally to get the probability that P(x<$400) at each t. If you do not agree, perhaps for illustration purpose, can you can show how you would solve for V and the shortfall probability in period 10 analytically (i.e. without simulation)?
Thank you!
if the interest rate is random (which you did not say in post #1), then monte carlo simulation is appropriate.
Depending on the exact distribution of R_t there may be a closed form alternative, but it would be pretty hard to get at if it exists at all.