Inverse Transform Demonstration with Excel VBA

Given F(x) = 1-e^(-λ*x) as the cumulative distribution function, the inverse transform gives -1/λ*ln(U) as the function that has F(x) as its cdf, where U is the uniform distribution from [0, 1]. Here, the following VBA codes allow users to visualize this transformation in Microsoft Excel.

Upon the execution of this procedure, the user inputs a value for lambda. Then 10,000 simulations are run, initially generating a random number from [0, 1] and then inputting that random number into -1/λ*ln(U), and outputting the result in column A. At the end of the execution, column C contains the different x values from 0 to the maximum, in increments of 0.001. Column D reflects the cdf by counting entries of A that are smaller than the corresponding x value. Finally, column E calculates the true value of 1-e^(-λ*x). The idea is that the outputs in columns D and E are similar.

Sub InverseTransform()
'Demonstrates inverse transform of the cdf F(x) = 1-e^(-lambda*x)
Columns("A:E").Clear
Range("B1").Value = 1# * InputBox("Enter a positive value for lambda: ", "Input", 2)

'10,000 simulation trials to be performed
'Transform of F(x) gives -1/lambda*ln(U), where U is uniform distribution [0,1]
For i = 1 To 10000
Cells(i, 1) = -1 / Range("B1").Value * Log(Rnd)
Next i

'Determine the maximum for the range of numbers to work with
Range("B2").FormulaR1C1 = "=MAX(C[-1])"

'To determine the cumulative probability density, use 0.001 gradient from 0 to the maximum value as the counter
i = 1
While i / 1000 <= Range("B2").Value
Cells(i, 3).Value = i / 1000
'In column D, count entries in column A that are smaller than the counter, then divide by the number of trials
Cells(i, 4).FormulaR1C1 = "=COUNTIF(C[-3],""<""&RC[-1])/10000"
'In column E, calculate the true value of 1-e^(-lambda*x)
Cells(i, 5).FormulaR1C1 = "=(1-EXP(0-R1C2*RC[-2]))"
i = i + 1
Wend
Range("B2").Clear
End Sub

After the execution of this procedure, the user can perform further analysis. Graphing columns C through E does reveal that values in columns D and E are similar, as the points almost completely overlap. Error calculations from those two columns illustrate a similar result that the inverse transform method takes a function F(x), exponential function in this case, to produce a function whose cdf is F(x).