Question about a property of pdfs

[Obie]

I've encountered this question in some theoretical work I've been doing. Suppose you have two pdfs fH(x) and fL(x) with the same support (bounded or infinite... you can assume whatever you want about it).

Suppose that the expected value of a random variable with pdf fL(x) is greater than the expected value of a random variable with pdf fH(x).

Is it true that the integral over the common support INT{ (fH(x)^2)/fL(x) } dx is larger than one? Is it true if the CDF FH(x)<=FL(X)?

I've checked family of distributions x^a, exponential, Normal, binomial and the claim is true in these classes, but I don't know if its true in general...

 

[uart]

I've encountered this question in some theoretical work I've been doing. Suppose you have two pdfs fH(x) and fL(x) with the same support (bounded or infinite... you can assume whatever you want about it).

Suppose that the expected value of a random variable with pdf fL(x) is greater than the expected value of a random variable with pdf fH(x).

Is it true if the CDF FH(x)<=FL(X)?

You mean for all x right? In that case the answer is definitely no. I had no problems finding simple counter examples for this. (Can post one later if you wish, they're very easy to find).

Is it true that the integral over the common support INT{ (fH(x)^2)/fL(x) } dx is larger than one?

I pretty sure that the answer is yes, but be careful as this may not mean exactly what you're expecting it means. Note however that I also think that INT{ (fL(x)^2)/fH(x) } dx is larger than one as well! In fact I believe that for any given density function f1, that the density function f2 which minimizes the integral of f2(x)^2/f1(x) dx is f2 = f1. That means that for any two density functions that are not identical then your integral will be greater than unity, regardless of which one you put on the numerator and which one on the denominator.

 

[uart]

Is it true if the CDF FH(x)<=FL(X)?

Hang on, are you actually asking if that inequality is true for the CDF evaluated at the mean or are you asking if it's true for all x as I assumed above?

Did you really mean to ask if the following is true.

F_H(\overline{x}_H) \leq F_L(\overline{x}_L)

Is that what you're asking?

 

[uart]

Suppose that the expected value of a random variable with pdf fL(x) is greater than the expected value of a random variable with pdf fH(x).

Also did you mean that the other way around? As in "the expected value of a random variable with pdf fL(x) is less than the expected value of a random variable with pdf fH(x)".

 

[Obie]

Hi Uart.

Thanks for taking the time to reply. I apologize for the confusing phrasing of the question. Please allow me to rephrase the question in a more precse way.

Consider two Random Variables X1, X2, with the same support, for which E[X1]<E[X2]. Let the CDF of X1 be F1 and the cdf of X2 be F2, with pdfs f1, f2. Notice that if, F2(x)<=F1(x) for all x in the support, this is a sufficient, but not necessary condition for E[X1]<E[X2]. This is the property called first order stochastic dominance.

Question 1)

does E[X1]<E[X2] imply that the integral over the support of (f2(x)^2)/f1(x) is greater than 1?

It seems your answer is that the claim is true regrdless of the relationship between the expected values. This would be great. I would appreciate any further explanation. Did you set the problem up as a constrained optimization for one of the pdfs?

Question 2)
If the answer to question1) is no, then would the addition of the property of first order dominance make it true. That is, is it tru that if F2(x)<=F1(x) for all x in the support, (implies E[X1]<E[X2]) that the intgral of the previous ratio is larger than one.

Thanks so much for your help. Much appreciated.

 

[uart]

Ok, here's a brief "outline proof" of that first conjecture.

Let f2 be the equal to f1 plus a perturbating function n(x),

f_2(x) = f_1(x) + n(x)

Integrating both sides shows that the integral of the perturbing function, n(x), must be zero ( since \int f_1 = 1 and \int f_2 = 1 ).

Substituting f2 = (f1 + n) into your integral gives,

\int \frac{f_2^2}{f_1} = \int f_2 + 2 \int n + \int \frac{n^2}{f_1}

= 1 + 0 + integral of non-negative function.

Therefore \int \frac{f_2^2}{f_1} \geq 1

 

[uart]

BTW. Note that for finite functions that n must be non-zero at more than a finite number of points in order to make the two means differ. It follows therefore that the last part of the above integral (the part I referred to as "integral of nonnegative function") is strictly positive. In this case we can make the stronger statement that if the two means are different then,

\int \frac{f_2^2}{f_1} &gt; 1

 

[Obie]

Thamks so much. I guess I thought it was going to be more complicated than it actually is!