10

Here is a problem that came in a semester exam in our university few years back which I am struggling to solve.

If $X_1,X_2$ are independent $\beta$ random variables with densities $\beta(n_1,n_2)$ and $\beta(n_1+\dfrac{1}{2},n_2)$ respectively then show that $\sqrt{X_1X_2}$ follows $\beta(2n_1,2n_2)$.

I used the Jacobian method to obtain that the density of $Y=\sqrt{X_1X_2}$ is as follows: $$f_Y(y)=\dfrac{4y^{2n_1}}{B(n_1,n_2)B(n_1+\dfrac{1}{2},n_2)}\int_y^1\dfrac{1}{x^2}(1-x^2)^{n_2-1}(1-\dfrac{y^2}{x^2})^{n_2-1}dx$$

I am lost at this point actually. Now, in the main paper, I found a hint had been supplied. I tried to use the hint but could not obtain the desired expressions. The hint is verbatim as follows:

Hint: Derive a formula for the density of $Y=\sqrt{X_1X_2}$ in terms of the given densities of $X_1$ and $X_2$ and try to use a change of variable with $z=\dfrac{y^2}{x}$.

So at this point, I try to make use of this hint by considering this change of variable. Hence I get, $$f_Y(y)=\dfrac{4y^{2n_1}}{B(n_1,n_2)B(n_1+\dfrac{1}{2},n_2)}\int_{y^2}^y\dfrac{z^2}{y^4}(1-\dfrac{y^4}{z^2})^{n_2-1}(1-y^2.\dfrac{z^2}{y^4})^{n_2-1}\dfrac{y^2}{z^2}dz$$which after simplification turns out to be (writing $x$ for $z$)$$f_Y(y)=\dfrac{4y^{2n_1}}{B(n_1,n_2)B(n_1+\dfrac{1}{2},n_2)}\int_{y^2}^y\dfrac{1}{y^2}(1-\dfrac{y^4}{x^2})^{n_2-1}(1-\dfrac{x^2}{y^2})^{n_2-1}dx$$

I do not really know how to proceed. I am not even sure that I am interpreting the hint properly. Anyway, here goes the rest of the hint:

Observe that by using the change of variable $z=\dfrac{y^2}{x}$, the required density can be expressed in two ways to get by averaging $$f_Y(y)=constant.y^{2n_1-1}\int_{y^2}^1(1-\dfrac{y^2}{x})^{n_2-1}(1-x)^{n_2-1}(1+\dfrac{y}{x})\dfrac{1}{\sqrt{x}}dx$$Now divide the range of integration into $(y^2,y)$ and $(y,1)$ and write $(1-\dfrac{y^2}{x})(1-x)=(1-y)^2-(\dfrac{y}{\sqrt{x}}-\sqrt{x})^2$ and proceed with $u=\dfrac{y}{\sqrt{x}}-\sqrt{x}$.

Well, honestly, I cannot understand how one can use these hints: it seems I am getting nowhere. Help is appreciated. Thanks in advance.

Xi'an
  • 90,397
  • 9
  • 157
  • 575
Landon Carter
  • 1,295
  • 11
  • 21
  • I have seen a similar problem before which I had compiled some references to. See http://arxiv.org/pdf/1304.6671v1.pdf http://mathoverflow.net/questions/32782/square-root-of-betaa-b-distribution – Sid Mar 22 '15 at 18:55
  • @Sid Sorry but I could not find this problem in those references or anything similar. Could you kindly point out the places? Thanks!! – Landon Carter Mar 22 '15 at 19:00
  • Are you sure you applied the Jacobian method correctly? If I do it, I get: $$ f_Y(y) = \frac{2 y^{2n_1-1}}{B(n_1,n_2)B(n_1+0.5,n_2)} \int_{y^2}^1 \frac{1}{\sqrt{x}} [(1-\frac{y^2}{x})(1-x)]^{n_1-1} dx $$ I think you are also going to need the doubling formula $\Gamma(z)\Gamma(z+0.5)=2^{1-2z}\sqrt{\pi}\Gamma(2z)$, see http://en.wikipedia.org/wiki/Gamma_function – StijnDeVuyst Mar 22 '15 at 23:42
  • Apparently it seems that the formulae are the same. Maybe you have to use the change of variable $z=\sqrt{x}$ in your formula to obtain mine. I am talking about the Jacobian. – Landon Carter Mar 23 '15 at 03:13
  • I don't think they are the same. Doing the change of variable that you mention into my formula, I get something slightly simpler than what you have in the first integral of your OP. – StijnDeVuyst Mar 23 '15 at 08:04
  • Actually, you made a mistake in your expression. Inside the integral (in what you wrote) you actually have the power $n_2-1$ which you erroneously wrote $n_1-1$. But actually, I verified that the formulae turn out to be the same. – Landon Carter Mar 24 '15 at 01:34

1 Answers1

6

I would prove this in a different manner, using moment-generating functions. Or equivalently, by showing that the $q$th moment of $\sqrt{X_1X_2}$ is equal to the $q$th moment of a random variable $B$ with $\beta(2n_1,2n_2)$ distribution. If this is so for all $q=1,2,\ldots$, then by the strength of the moment problem, the exercise is proven.

For the last part, we obtain from http://en.wikipedia.org/wiki/Beta_distribution#Other_moments that the $q$th moment of $B$ is $$ \mathrm{E}[B^q] = \prod_{j=0}^{q-1} \frac{2n_1+j}{2n_1+2n_2+j} = \ldots = \frac{\Gamma(2n_1+q)\Gamma(2n_1+2n_2)}{\Gamma(2n_1)\Gamma(2n_1+2n_2+q)} $$ Now for the first part: $$ \mathrm{E}[(\sqrt{X_1X_2})^q] = \int\int (\sqrt{x_1x_2})^q f_{X_1}(x_1) f_{X_2}(x_2) dx_1dx_2 \\ = \int x^{q/2} f_{X_1}(x_1) d x_1 \cdot \int x_2^{q/2} f_{X_2}(x_2) d x_2\\ = \frac{1}{B(n_1,n_2)} \int x_1^{n_1+q/2-1}(1-x_1)^{n_2-1}dx_1 \cdot \frac{1}{B(n_1+\frac{1}{2},n_2)} \int x_2^{n_1+\frac{q+1}{2}-1}(1-x_2)^{n_2-1}dx_2\\ = \frac{B(n_1+\frac{q}{2},n_2)B(n_1+\frac{q+1}{2},n_2)}{B(n_1,n_2)B(n_1+\frac{1}{2},n_2)} $$ Now all that remains is to apply the definition $B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$ and then the doubling formula $\Gamma(\alpha)\Gamma(\alpha+\frac{1}{2})=2^{1-2\alpha}\sqrt{\pi}\Gamma(2\alpha)$. It then turns out that the first part and the second part are exactly the same.

StijnDeVuyst
  • 2,161
  • 11
  • 17
  • 2
    I do not think you can say that equality of moments implies equality of distribution. There are examples where this may not hold. – Landon Carter Mar 23 '15 at 03:01
  • I would be very interested to see such an example. If all the moments exist (as in the case above), I really think equality of all moments implies equality of distribution. – StijnDeVuyst Mar 23 '15 at 08:18
  • @yedaynara: To use the moment generating function, $q$ has to be any positive real number. But the equality still holds (with a wee more work). – Xi'an Mar 23 '15 at 12:23
  • 2
    StijnDeVuyst, sorry this is not an acceptable answer. I do have an example where the moments are equal but the distributions are not the same. The example is a bit complicated though. Regrettably I do not have the example with me now; it also came in one semester exam. But soon I will post the example in this thread if you are interested. Anyway I have worked the problem out myself. Thanks for your help. – Landon Carter Mar 23 '15 at 13:17
  • 3
    @yedaynara and Stijn: A (the?) classical example is due to Heyde: Consider the pdfs $f_b(x) = f_0(x)(1+b \sin(2\pi\log x))$ where $f_0$ is the pdf of the standard lognormal and $b \in [-1,1]$. **All** members of this family of distributions have the same moments (of all orders). Note that the standard lognormal is a member of this family and its moments have a nice closed form. – cardinal Mar 23 '15 at 13:47
  • 4
    There are, however, additional conditions (e.g., Carleman's) on the moments that will guarantee uniqueness of the distribution. This is known as the *Hamburger moment problem*. – cardinal Mar 23 '15 at 13:50
  • I knew a different example though. But this problem too came in another one of the semester exams :P – Landon Carter Mar 23 '15 at 13:53
  • 3
    Quote from http://web.williams.edu/Mathematics/sjmiller/public_html/book/papers/jcmp.pdf "...It is elementary linear algebra to verify that a positive measure with finite support is uniquely determined by its moments..." That settles the Carleman condition for M-determinacy for the Beta distributions in the OP. @cardinal and yedaynara are both correct that I was too quick to assume this. But apparently the finite support is what saves the day. – StijnDeVuyst Mar 23 '15 at 15:54
  • 2
    While I was commenting more on the general case in response to @yedaynara's lack of a quick example at hand; yes, Stijn, this is true for distributions of bounded support. It's also not hard to prove once one knows that probability measures are equal iff all bounded measurable functions have the same expected values under the measures in question. The moment condition implies all polynomials have identical expected values and Weierstrass says we can find a polynomial to approximate any (fixed) bounded continuous function. The result then follows by a $2\epsilon$ argument. – cardinal Mar 23 '15 at 16:54
  • I am sorry cardinal and StijnDeVuyst for my ignorance but actually I am a first year undergrad and have no idea about measure theory right now. But I will be studying measure theory a few semesters later though. Thank you for your valuable comments. – Landon Carter Mar 24 '15 at 01:32