Mean Preserving Discretisation of a Truncated Continuous Random Variable

Question

As suggested by the title, I am trying to determine how to generate a discretised version of a continuous random variable, that is truncated to the right by some value, say $M$. I would like the resultant discretised random variable to have the same mean as the continuous version.

If I ignore truncation for the moment, I have found the following result that I am not able to verify:

If $Y$ is the discretised version of some continuous r.v. $X$, for bandwidth $h$, if we let

$f_{j} = P(Y = jh)$

Then:

$f_{0} = 1- \frac{E[X\hspace{0.1cm}\wedge\hspace{0.1cm}h]}{h}$

$f_{j} = \frac{2E[X\hspace{0.1cm}\wedge\hspace{0.1cm}jh] - E[X\hspace{0.1cm}\wedge\hspace{0.1cm}(j-1)h] - E[X\hspace{0.1cm}\wedge\hspace{0.1cm}(j+1)h]}{h}$

And $Y$ will have the same mean as $X$.

I am struggling to explain these formulae intuitively (although I can sort of see that we are finding the average of $X$ at increments along the doman, where the increments are determined by the value of $h$).

I am certainly struggling to derive a mathematical and rigorous proof.

Can anyone help on both parts?

Perhaps if I grasp this then I will be able to determine the formulae for the truncated version.

Thanks for any help.

Answer 1

I suppose you're looking for a completely general method. But many important continuous distributions are approximated by discrete distributions that are already programmed into major software packages. For example, $\mathsf{Norm}(\mu=200,\sigma=10) \approx \mathsf{Binom}(n=400,1/2).$ For other normal distributions, you could re-scale the binomial. Using R:

For most purposes, the discrepancy between the discrete and continuous distributions is smaller than the margin of simulation error.

> x = rnorm(10^6, 200, 10);  mean(x);  sd(x)
[1] 200.0146
[1] 9.999379
> y = rbinom(10^6, 400, .5); mean(y);  sd(y)
[1] 199.9938
[1] 10.00215