Easy mathematics ?

outlier · Post by **outlier** » Mon Jan 07, 2008 6:22 pm

Is [img]/User%20Files/4008/Latex-Equation-7392.gif[/img] when x small (x stochastic) ? Do anybody has a reference on this ?

pj · Post by pj » Mon Jan 07, 2008 6:29 pm

If x is constant yes.

Otherwise NO! .

Take a binomial for example.

Jaxx · Post by **Jaxx** » Mon Jan 07, 2008 6:44 pm

by jensen's inequality if f(x) is convex, then E(f(x)) >= f(E(x)) (with equality iff x is knst).

(edit : sorry cross posted)

outlier · Post by **outlier** » Mon Jan 07, 2008 6:54 pm

To pj: In the binomial case 0 or Y with Y small (I assumed x small is my post) than:

[img]/User%20Files/4008/Latex-Equation-7393.gif[/img]

so it holds in this case.

Is it always true?

outlier · Post by **outlier** » Mon Jan 07, 2008 6:57 pm

Thx Jaxx. I actually know that by jensen's inequality it is not equal, but I am wondering if the aproximation is ok when x small ?

doctorwes · Post by **doctorwes** » Mon Jan 07, 2008 7:04 pm

Well, when x is small, it is approximately constant.

pj · Post by pj » Mon Jan 07, 2008 7:26 pm

You mean ≈ as in

e^x≈1+x?

outlier · Post by **outlier** » Mon Jan 07, 2008 10:50 pm

When x constant first order aproximation on x gives [img]/User%20Files/4008/Latex-Equation-7395.gif[/img] as you said, when by Jensen inequality [img]/User%20Files/4008/Latex-Equation-7396.gif[/img]

meteor · Post by **meteor** » Mon Jan 07, 2008 11:30 pm

>>>but I am wondering if the aproximation is ok when x small ?

What do you exactly by x is small? X is a random variable?

What you can do is use the Delta method which basically tells you that:

if EX converges in proba to \theta; with in your case h(x)=exp(x) and assume var(X)=\sigma^2 then:

sqrt(n) E[h(x_n)-h(\theta)] converges in distribution to N(0, \sigma^2 h'(\theta))

meteor · Post by **meteor** » Mon Jan 07, 2008 11:42 pm

Or furthermore you can just taylor expand exp(x) and take the expectation of you expansion (but this assume that you know the moments of the distribution of X)