Question

1. [26 pts Let Uı, , Un be independent, identically distributed Unifomn random variables with (continu- ous) support on (0, b), where b> 0 is a parameter. (a) Define the random variable Y :--Σί 1 log(U,), where log is the natural logarithm function. De- termine the probability density function (pdf) p(y; b) ofY by explicitly computing it (b) Based on the pdf you found in part (a) above, determine the third moment of Y, i.e., EY] (c) Suppose now that you are given two independent observations yı and y2 of Y. Determine the maximum likelihood estimate of b based on the observations yi and y

Answer 1

$(a)~Suppose~U\sim Uniform(0,b)~then~transform~U\rightarrow Z\\ where,~z=-2\log (u/b)\Rightarrow \frac{\mathrm{d} z}{\mathrm{d} u}=-2\times \frac{b}{u}\times \frac{1}{b}~or,~|Jacobian|=\frac{u}{2}=\frac{be^{-z/2}}{2}\\ The~pdf~of~U~is\\ f_U(u)=\frac{1}{2}e^{-\frac{z}{2}},~z>0\\ =0~otherwise\\ Hence~Z\sim \chi^2_2\\ Since~U_i's~are~independent~then~from~additive~property~of~\chi^2\\ V=-2\sum_{i=1}^n\log\left(\frac{U_i}{b} \right )\sim \chi^2_{2n}\\ The~pdf~of~V~is\\ f_V(v)=\frac{1}{\Gamma(n)2^n}e^{-\frac{v}{2}}v^{n-1},~v>0\\ =0~otherwise\\ Transform~V\rightarrow Y~where\\ v=2y+2n\log b\Rightarrow \frac{\mathrm{d} v}{\mathrm{d} y}=2,~|Jacobian|=2\\ The~pdf~of~Y~is\\ f_Y(y)=\frac{1}{\Gamma(n)}e^{-y-n\log b}(y+n\log b)^{n-1},~y>0\\ =0~otherwise\\ (b)~E((Y+n\log b)^r)=\frac{1}{\Gamma(n)}\int_0^\infty \frac{1}{\Gamma(n)}e^{-y-n\log b}(y+n\log b)^{r+n-1}dy=\frac{\Gamma(n+r)}{\Gamma(n)}$

$E((Y+n\log b)^3)=\frac{\Gamma(n+3)}{\Gamma(n)}=(n+2)(n+1)n\\ or,~E(Y^3)+3n\log bE(Y^2)+3(n\log b)^2E(Y)+(n\log b)^3=n(n+1)(n+2).....(1)\\ E((Y+n\log b)^2)=\frac{\Gamma(n+2)}{\Gamma(n)}=(n+1)n\\ E(Y^2)+2n\log bE(Y)+n^2(\log b)^2=n(n+1).........(2)\\ E((Y+n\log b))=\frac{\Gamma(n+1)}{\Gamma(n)}=n\\ E(Y)=n(1-\log b)...............(3)\\ From~(2)~and~(3)\Rightarrow E(Y^2)=n(n+1)-2n^2\log b(1-\log b)-n^2(\log b)^2\\ =n(n+1)+n^2(\log b)^2-2n^2\log b=n^2(1+(\log b)^2)+n-2n^2\log b.......(4)\\ Then~E(Y^3)=n(n+1)(n+2)-3n\log b\left(n^2(1+(\log b)^2)+n-2n^2\log b \right )\\ -3(n\log b)^2n(1-\log b)-(n\log b)^3\\ (c)~The~Likelihood~function~of~b=L(b)\\ =\frac{1}{(\Gamma(n))^2}e^{-y_1-y_2-2n\log b}(y_1+n\log b)^{n-1}(y_2+n\log b)^{n-1}\\ \log(L(b))=-2\log(\Gamma(n))-(y_1+y_2+2n\log b)+(n-1)\left(\log (y_1+n\log b)+\log(y_2+n\log b) \right )\\$

$\frac{\mathrm{d} \log L(b)}{\mathrm{d} b}=-\frac{2n}{b}+\frac{(n-1)}{b}\left(\frac{1}{y_1+n\log b}+\frac{1}{y_2+n\log b} \right )=0.....(5)\\ Solving~(5)~using~numerical~iteration~we~get~MLE~of~b.$