2

enter image description here

In this problem i know that $X\sim B(m,n)$ and $(1-X)\sim B(n,m)$ After putting values in $Y_i$ i got this $\dfrac{x^2}{1-x^2}$ I have not idea what do further . I am confused.

Sri-Amirthan Theivendran
  • 31,116
Damandeep
  • 147

2 Answers2

1

Hint: Find the MGF of $Y_i$ (after you have the density) and then use it to compute the distribution of $U_n$ then $\alpha$ should be the standard deviation of the resulting distribution by the central limit theorem.

Edited: You should find this useful:

The following variance of the variable X divided by its mirror-image ($X/(1−X)$ results in the variance of the "inverted beta distribution" or beta prime distribution (also known as beta distribution of the second kind or Pearson's Type VI):

${\displaystyle \operatorname {var} \left[{\frac {1}{1-X}}\right]=\operatorname {E} \left[\left({\frac {1}{1-X}}-\operatorname {E} \left[{\frac {1}{1-X}}\right]\right)^{2}\right]=\operatorname {var} \left[{\frac {X}{1-X}}\right]=} $

${\displaystyle \operatorname {E} \left[\left({\frac {X}{1-X}}-\operatorname {E} \left[{\frac {X}{1-X}}\right]\right)^{2}\right]={\frac {\alpha (\alpha +\beta -1)}{(\beta -2)(\beta -1)^{2}}}{\text{ if }}\beta >2} $

Dr Richard Clare
  • 2,485
  • 12
  • 30
  • How do we find mgf of beta distribution? Never saw that in my books . – Damandeep Jan 19 '18 at 17:58
  • $ M_X(t) = {\displaystyle 1+\sum _{k=1}^{\infty }\left(\prod _{r=0}^{k-1}{\frac {\alpha +r}{\alpha +\beta +r}}\right){\frac {t^{k}}{k!}}}$ – Dr Richard Clare Jan 19 '18 at 18:01
  • $\frac{x^{\alpha-1}(1-x)^{\beta -1}} {B(\alpha,\beta)}!$ where ${\displaystyle \mathrm {B} (\alpha ,\beta )={\frac {\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}}}$ – Dr Richard Clare Jan 19 '18 at 18:03
  • My book never mentioned this mgf so i am pretty sure there is some other way we can do this problem. This question came in 2 marks. So there won't be any lengthy calculations. – Damandeep Jan 19 '18 at 18:10
  • Ok, since you do not cover that, I posted you another equivalent way without MFG. FInd what is the pdf of X/(1-X) then use the distribution of $U_n$ and compute it's variance and take the square root to find $\alpha$. – Dr Richard Clare Jan 19 '18 at 18:14
1

It suffices to find the standard deviation of $Y_{i}$. To this end first define a random variable $W\sim\text{Gamma}(p)$ ($p>0$) if $W$ has density $$ f_W(w)=\frac{1}{\Gamma(p)}w^{p-1}e^{-w}\quad (w>0). $$ It is easy to see that $EW^d=\frac{\Gamma(p+d)}{\Gamma(p)}$ by the definition of the gamma function. Now $X_i\stackrel{d}{=} X$ where $$ X=\frac{Z_{1}}{Z_{1}+Z_{2}};\quad Z_{1}\perp Z_{2};\quad Z_{1}\sim \text{Gamma}(6), Z_{2}\sim \text{Gamma}(4). $$ Then $$ Y_{i}\stackrel{d}{=} \frac{Z_{1}/(Z_{1}+Z_{2)}}{Z_{2}/(Z_{1}+Z_{2)}}=\frac{Z_{1}}{Z_{2}}. $$ Now $$ \text{Var}(Y_{i})=EZ_{1}^{2}Z_{2}^{-2}-(EZ_{1}Z_{2}^{-1})^2. $$ But then using the fact that $Z_{1}\perp Z_{2}$ we have that $$ EZ_{1}^{2}Z_{2}^{-2}=EZ_{1}^{2}\cdot EZ_{2}^{-2}=\frac{7(6)}{3(2)}=7 $$ and $$ EZ_{1}Z_{2}^{-1}=EZ_{1}\cdot EZ_{2}^{-1}=\frac{6}{3}=2. $$ Thus $$ \text{Var}(Y_{i})=7-2^2=3\implies\sigma(Y_{i})=\sqrt{3}. $$ The answer is (C).

Sri-Amirthan Theivendran
  • 31,116