We derive a deterministic function to model a systematic withdrawal plan. Table 1 outlines the variables used in this function.

Procedure for Month 1:

Step 1: We multiply the initial amount in the fund α by $\left(1+r_g/100\right)$ to calculate the appreciated amount at Month 1’s end:

$\alpha \left(1+\frac{r_g}{100}\right)$

Step 2: We multiply the initial monthly withdrawal amount β by $\left(1+r_i/100\right)$ to calculate the amount to be withdrawn at Month 1’s end:

$\beta \left(1+\frac{r_i}{100}\right)$

Step 3: We subtract the amount to be withdrawn at Month 1’s end from the appreciated amount at Month 1’s end:

$f\left(1\right)=\alpha \left(1+\frac{r_g}{100}\right)-\beta \left(1+\frac{r_i}{100}\right)$ (1)

Step 4: We set the amount at Month 1’s end to the amount at the start of Month 2.

Procedure for Month 2:

Step 1: We multiply the amount at the start of Month 2 by $\left(1+r_g/100\right)$ to calculate the appreciated amount at Month 2’s end:

$\left(\alpha \left(1+\frac{r_g}{100}\right)-\beta \left(1+\frac{r_i}{100}\right)\right)\left(1+\frac{r_g}{100}\right)$

Step 2: We multiply the amount to be withdrawn at Month 1’s end by $\left(1+r_i/100\right)$ to calculate the amount to be withdrawn at Month 2’s end:

$\beta {\left(1+\frac{r_i}{100}\right)}^2$

Step 3: We subtract the amount to be withdrawn at Month 2’s end from the appreciated amount at Month 2’s end, which gives

$\begin{array}{c}f\left(2\right)=\left(\alpha \left(1+\frac{r_g}{100}\right)-\beta \left(1+\frac{r_i}{100}\right)\right)\left(1+\frac{r_g}{100}\right)-\beta {\left(1+\frac{r_i}{100}\right)}^2\\ =\alpha {\left(1+\frac{r_g}{100}\right)}^2-\left(1+\frac{r_g}{100}\right)\beta \left(1+\frac{r_i}{100}\right)-\beta {\left(1+\frac{r_i}{100}\right)}^2\end{array}$ (2)

Step 4: We set the amount at Month 2’s end to the amount at the start of Month 3. At this stage, two simplifications avoid excessive complexity in the model equation.

Simplification 1: A variable m_g is set to the value $\left(1+\frac{r_g}{100}\right)$.

Simplification 2: A variable m_i is set to the value $\left(1+\frac{r_i}{100}\right)$.

Using the simplifications, the expression for the fund’s value at Month 1’s end is:

$f\left(1\right)=\alpha m_g-\beta m_i$ (3)

Using the simplifications, the expression for the fund’s value at Month 2’s end is:

$f\left(2\right)=\alpha m_g^2-m_g\beta m_i-\beta m_i^2$ (4)

The procedure used for Months 1 and 2 is repeated for Months 3, 4, 5, and so on until the fund reaches zero.

Table 2 illustrates the expression for the fund’s value at every month’s end after withdrawal. The first five months have been included in the table for representation purposes.

We factor out β from every term except the first term.

For example, for Month 3, we have

$\begin{array}{c}f\left(3\right)=\alpha m_g^3-m_g^2\beta m_i-m_g\beta m_i^2-\beta m_i^3\\ =\alpha m_g^3-\beta \left(m_g^2{m}_i+m_g{m}_i^2+m_i^3\right)\end{array}$ (5)

We highlight the exponents of m_g and m_i in every term. Table 3 displays the amount in the fund at the month’s end with highlighted exponents.

For every month, the expression can be written as a subtraction of two distinct expressions: Expression 1 and Expression 2 ( Figure 2 ).

For every month, Table 4 illustrates Expression 1 and the exponent of the variable m_g in Expression 1.

Let the variable x represent the month number. For example, the value of x for Month 3 is 3.

Then, Expression 1 can be modeled by the function:

$f\left(x\right)=\alpha m_g^x$ (6)

For every month, it is evident that the expression that the variable β multiplies is a series ( Figure 3 ).

Step 1: We isolate the series from the variable β.

Step 2: We isolate every term of the series.

Step 3: We isolate the exponent of the variable m_g and the exponent of the variable m_i from every term of the series ( Table 5 ).

Step 4: By observing Table 6 , we deduce that the month number is equal to the number of terms in the series corresponding to that month. Then, the variable x (defined in Section 2.5) represents both the month number and the number of terms in the series corresponding to that month.

Step 5: Let j denote the exponent of the variable m_g, and let k denote the exponent of the variable m_i

Step 6: Therefore, by observing the trends in the values of j and k in Table 7 , we formulate a generalized expression for representing the series for any arbitrary month x:

$\underset{k=1}{\overset{x}{{\displaystyle \sum }}}{m}_g^j{m}_i^k$

Step 7: By observing the rightmost column of Table 6 , we can deduce that the sum of j and k for any arbitrary month x is equal to the month number x.

Then, we have:

$j+k=x$

$j=x-k$ (7)

We substitute j with x – k in the generalized expression for the series:

$\underset{k=1}{\overset{x}{{\displaystyle \sum }}}{m}_g^{x-k}{m}_i^k$

Step 8: We combine the generalized expression for the series with the variable β, which was isolated in Step 1, to obtain the function modelling Expression 2:

$f\left(x\right)=\beta \left(\underset{k=1}{\overset{x}{{\displaystyle \sum }}}{m}_g^{x-k}{m}_i^k\right)$ (8)

As shown in Figure 2 , the main expression is formed by subtracting Expression 2 from Expression 1.

Therefore, the function modelling the main expression:

$f\left(x\right)=a{m}_g^x-\beta \left(\underset{k=1}{\overset{x}{{\displaystyle \sum }}}{m}_g^{x-k}{m}_i^k\right)$ (9)

We replace our simplifications m_g and m_i with their original expansions to obtain:

$f\left(x\right)=\alpha {\left(1+\frac{r_g}{100}\right)}^x-\beta \left(\underset{k=1}{\overset{x}{{\displaystyle \sum }}}{\left(1+\frac{r_g}{100}\right)}^{x-k}{\left(1+\frac{r_i}{100}\right)}^k\right)$ (10)

The function above is a function of the month number x, and f(x) represents the amount in the fund at the end of x months. The function models the fund of a systematic withdrawal plan, for which the variables α, β, r_g, and r_i can be inputted.

The function that models the fund assumes that r_g is constant. However, r_g, which denotes the monthly growth rate of the S&P 500, is bound to vary. Therefore, we must account for the uncertainty in r_g in the model. A dataset of 168 r_g values was retrieved [4] . Each r_g value corresponds to its respective month from January 2010 to January 2024. To check if the randomness in r_g values follows a predictable pattern, exploratory data analysis is conducted by generating a histogram of r_g values ( Figure 4 ).

Observing the shape of the histogram, we can intuitively hypothesize that the variable r_g is likely to be normally distributed. To validate our hypothesis, we shall check the normality of r_g using a one-sample Kolmogorov–Smirnov test [5] .

Hypotheses for the one-sample Kolmogorov–Smirnov test on r_g:

H: r_g is a random variable drawn from a normal distribution with μ equal to the mean of r_g and σ equal to the standard deviation of r_g.

H_A: r_g is not a random variable drawn from a normal distribution with μ equal to the mean of r_g and σ equal to the standard deviation of r_g.

The Kolmogorov–Smirnov test outputs a KS Statistic of 0.0764 and a p-value of 0.2582. Since the p-value of 0.2582 is more than the alpha value of 0.05, the null hypothesis is accepted. This comparison proves that r_g is random variable normally distributed with μ as 0.6933 (% per month) and σ as 4.0266 (% per month).

Since we know that r_g is a normally distributed random variable, we can run stochastic simulations for the function that models the fund. Each time we run a stochastic simulation; we note the number of months it takes for the function to fall below zero. In other words, we note how long the fund of a systematic withdrawal plan lasts. Let t be a new variable, where t refers to the number of months it takes for the fund to fall below zero. Since the function that models the fund depends on r_g, the uncertainty in r_g carries over to t. By running sufficient stochastic simulations, we can generate a histogram of t values to check if the randomness in t values follows a predictable pattern. We set realistic example values for α as $500000 and β as $5000 for the stochastic simulations. r_g is sampled from a normal distribution with parameters (0.6933, 4.0266), and r_i is 1.0021( Figure 5 ).

Pseudocode for plotting a histogram of t values by running stochastic simulations:

loop i from 1 to 10000

f(x) = α

loop x while f(x) > 0

r_g = np.random.normal(mu, sigma)

f(x) = (f(x)*(1 + (r_g/100))) – (β*(1 + (r_i/100))^x)

x++

end loop

t_values[i] = x

end loop

plot histogram of t_values

Figure 6 suggests that outliers exist when t equals 10,000 months. These outliers are halted at 10,000 months due to the technical limitations of the R language. However, these outliers signify extreme cases when t approaches infinity. These cases happen when stochastic variations result in unrealistic consistently high r_g values, which allow the fund to last infinitely. However, these cases are significantly rare, with a 0.58% (58 in 10,000 simulations) probability. Therefore, we remove the outliers to improve robustness, and we plot a new histogram.

The shape of the histogram of t values in Figure 7 suggests that the variable t is likely to be log-normally distributed. To validate our hypothesis, we shall log-transform [6] the t values and check the normality of the log-transformed t values using a one-sample Kolmogorov–Smirnov test.

The Kolmogorov–Smirnov test outputs a KS Statistic of 0.0488 and a p-value of 0.1318. Since the p-value is more than the alpha value of 0.05, the null hypothesis is accepted ( Table 8 ). This comparison proves that t is a random variable drawn from a log-normal distribution with μ_t equal to 4.8466 and σ_t equal to 0.2721. Therefore, a continuous probability distribution function for t values, based on the log-normal distribution [7] is given by:

$P\left(x\right)=\frac{1}{2}\left(1+\text{erf}\left(\frac{\ln \left(x\right)-4.8466}{0.2721\sqrt{2}}\right)\right)$ (11)

Equation (11) applies specifically to the example given in this study. However, the log-normal nature of the distribution of t values holds for all values of α and β, as the shape of the distribution depends on r_g, and r_g follows a fixed normal distribution.

a. t values are rounded to the nearest integer.

Table 9 displays the percentiles of t values according to the log-normal distribution given by Equation (11). The deterministic model from Section 2, which does not account for randomness, predicts a t value of 137 months for α = 500,000, β = 5000, r_g = 0.6933, and r_i = 1.0021. r_g was set to 0.6933, which is the mean and best estimate of r_g, as the model requires a constant value. This prediction is validated by our stochastic analysis, as 137 is close to 127, which is the median of the log-normal distribution of t values. However, the deterministic model and the median are not a reasonable estimate of the fund’s longevity; they approximately provide a 50% safety margin, as 50% of the time, we will get a t value of less than 127 months, solely due to random chance. We must have a 90% safety margin, chosen in several equity funds, to have a high level of confidence in the fund’s longevity.

Our goal is to develop an algorithm to output α, the initial fund requirement, with a 90% safety margin, based on inputs of β and t_Target Let t_Target be a new variable, which represents the number of months we require our fund to last.

Step 1: Given inputs β and t_Target, we use the deterministic model from Section 2 to find the 50% safety margin value for α.

When we set Equation (10) from Section 2 to equal zero, the x in the equation represents the number of months the fund of the systematic withdrawal plan lasts.

$\alpha {\left(1+\frac{r_g}{100}\right)}^x-\beta \left(\underset{k=1}{\overset{x}{{\displaystyle \sum }}}{\left(1+\frac{r_g}{100}\right)}^{x-k}{\left(1+\frac{r_i}{100}\right)}^k\right)=0$ (12)

We rearrange Equation (13) to obtain:

$\alpha {\left(1+\frac{r_g}{100}\right)}^x=\beta \left(\underset{k=1}{\overset{x}{{\displaystyle \sum }}}{\left(1+\frac{r_g}{100}\right)}^{x-k}{\left(1+\frac{r_i}{100}\right)}^k\right)$

$\alpha =\frac{\beta \left({{\displaystyle \sum }}_{k=1}^x{\left(1+\frac{r_g}{100}\right)}^{x-k}{\left(1+\frac{r_i}{100}\right)}^k\right)}{{\left(1+\frac{r_g}{100}\right)}^x}$

$\alpha =\frac{\beta \left({{\displaystyle \sum }}_{k=1}^{t_{target}}{\left(1+\frac{r_g}{100}\right)}^{t_{target}-k}{\left(1+\frac{r_i}{100}\right)}^k\right)}{{\left(1+\frac{r_g}{100}\right)}^{t_{target}}}$ (13)

We plug the values of β, r_g = 0.6933, r_i = 1.0021, and t_Target into Equation (13) to calculate the 50% safety value of α.

Step 2: We run stochastic simulations to find the log-normal distribution of t values for the 50% safety value of α using the pseudocode from Section 3.3.

Step 3: We identify the 90^th percentile t value from the log-normal distribution of t values, and we plug it into Equation (13) to find the 90% safety value of α.

For example, we set the same example value from Section 3.3 of β as $5000, and we set the target t value as 127 months from Section 3.5. Running the stochastic algorithm from Figure 8 on these parameters returns α as $629,065. Compared to the lower initial α value of $500,000 from Section 3.3, which had a 50% safety margin, the higher α value of $629,065 suggested by our algorithm gives a 90% safety margin.