Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

My goal is to work out on the following equation:

$$Mean(P)(1-Mean (P))\left (h + (1 - 2 h) (1 - \frac {Var (P)} {Mean (P) (1 - Mean (P))}) (1 - Mean (P)) + \frac {Var (P)} {Mean (P) (1 - Mean (P))} (1 - h) \right)$$

where $P$ is a sample, $Mean(P)$ is the Mean value of the sample $P$: $Mean(P) = 1/n \sum_{i=1}^n p_i$. $Var(P)$ is the Variance of the random variable $P$.

How can I Simplify this kind of equation with Mathematica?

I thought I could eventually define a sequence of $p_i$ of undefined length and work with it. Is something similar feasible?

Here is the Mathematica code:

Mean[P](1-Mean[P])( h + (1-2h)(1-(Variance[P])/(Mean[P](1-Mean[P])))*
  (1-Mean[P]) + (Variance[P])/(Mean[P](1-Mean[P])) (1-h))
share|improve this question
    
@Öskå Thanks for your comment. As I don't really know how to treat this kind of calculation I don't really know what code to write. But I added edited my question. Hope that makes sense. I realize that when running Simplify[..] on my expression I get a result! Maybe that was all I wanted. But if I run Simplify[Mean[X^2]-Mean[X]^2] Mathematica does not answer Var[X]. Thanks for your help –  Remi.b May 17 at 11:46
    
I am confused by several aspects of the question. First, you say $Exp[P]$ is the expected value of $P$, but you then define it as the sample mean (which is not the same thing). Second, your sample mean is defined as the sum from 0 to $n$ (i.e. $n+1$ values), but you divide by $n$, not $n+1$. Doesn't make sense. Third, you then define $Var[P]$ as the population variance, but do you actually want it to be that, or the sample variance? And if the latter, which definition of sample variance: $1/n$ or $1/(n-1)$ ? –  wolfies May 17 at 11:47
    
@wolfies Thanks your comment helps. Yes, I actually talk about sample mean and variance not about statistical moments, you're correct. And moreover I made the mistake in defining it! I edited the question to fix this mistake. –  Remi.b May 17 at 11:50
    
@Remi.b If you define P = Symbol["p" <> ToString@#] & /@ Range@10, what result would you expect? –  Öskå May 17 at 12:21
    
@Remi.b If my answer suits your needs please accept it. –  olliepower May 19 at 14:17

1 Answer 1

Using FullSimplify we obtain

fun0[P_,h_] := Mean[P]*(1 - Mean[P])*(h + (1 - 2 h) (1 - (Variance[P])/(Mean[P] (1 - Mean[P])))*
  (1 - Mean[P]) + (Variance[P])/(Mean[P](1 - Mean[P])) (1 - h))
fun[P_,h_] := FullSimplify@fun0[P]
fun[P,h]
-(-1 + Mean[P]) Mean[P] (1 - h + (-1 + 2 h) Mean[P]) + 
  (h + Mean[P] - 2 h Mean[P]) Variance[P]

Declare variables to see if the simplified equation is the same

P = {1, 2, 3};
h = 1;
fun0[P,h]
-5

as does

fun[P,h]
-5
share|improve this answer
    
Thanks for the edit Oska –  olliepower May 17 at 22:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.