# What is the meaning of the following mathematical equations? [closed]

Let's say that we have a discrete probability distribution, where

$$x_i$$ represents each of the possible outcomes (discrete set of possible outcomes), and

$$L$$ represents the expected value we want to achieve, by manipulating the original probability distribution (keeping the same values of x_i, but adjusting the corresponding probabilities.

To do so, I am asked to use an optimization method, while implementing the two following restrictions,

$$\Delta P(x_i) = \alpha /(x_i - L)$$

$$\Delta \sum_{i=-n}^{n} P(x_i) = 0$$

How should I interpret the above restrictions, so I can implement them computationally?

• What's the source for the question? Sep 25, 2022 at 20:31
• Pure academic optimization problem. I believe my question is self explanatory. No problem if you don't know the answer :) Sep 25, 2022 at 20:38
• The second constraint is just that the total probability is conserved / does not change under optimization, which makes sense. And the first is some kind of obscure condition on the allowed deviation for each probability. Since it's a condition and not an inequality I frankly do not see the optimization.
– user34971
Sep 25, 2022 at 20:55
• In that case, wouldn't the second constraint mean the sum of all the probabilities would still be equal to 1? Because that is also stated in the exercise, which leads me to think it does not imply the same constraint (sum Pi = 1), but rather something different. Sep 25, 2022 at 20:56
• Yes the sum of the new probabilities is 1, just like the sum of the old probabilities. That's what I wrote.
– user34971
Sep 25, 2022 at 20:58

First condition says how much you change the probability mass as a function of alpha. Second one that probability must sum up to one. The two conditions imply that $$\sum_i (x_i-L)^{-1}=0$$ for thr original problem, and that the mean $$L \neq x_j$$ for all $$j$$. If this is originally fulfilled, you are free to chose any real $$\alpha$$, so not much of an optimization. However, you should also add that each probability is in $$[0,1]$$ before and after, which will bound your possible alpha.