Question

Suppose that X1, X2, ..., Xn is an iid sample, each with probability p of being distributed as uniform over (-1/2,1/2) and with probability 1 - p of being distributed as uniform over (a) Find the cumulative distribution function (cdf) and the probability density function (pdf) of X1 (b) Find the maximum likelihood estimator (MLE) of p. c) Find another estimator of p using the method of moments (MOM)

Answer 1

$The~pdf~of~X_1~is\\ f_{X_1}(x_1)=p\times \frac{1}{\frac{1}{2}&plus;\frac{1}{2}}=p~when~-1/2<x_1\leq 0\\ =p\times \frac{1}{\frac{1}{2}&plus;\frac{1}{2}}&plus;(1-p)\times \frac{1}{1-0}=1~if~0<x_1<1/2\\ =(1-p)\times \frac{1}{1-0}=(1-p)~if~1/2\leq x_1<1\\ =0~otherwise\\ The~cdf~of~X_1~is\\ F_{X_1}(x_1)=P(X_1\leq x_1)=0~if~x_1\leq -1/2\\ =p\int_{-1/2}^{x_1}dx_1=p(x_1&plus;1/2)~if~-1/2<x_1<0\\ =p\int_{-1/2}^{0}dx_1&plus;p\int_{0}^{x_1}dx_1&plus;(1-p)\int_{0}^{x_1}dx_1=\frac{p}{2}&plus;x_1~if~0<x_1<1/2\\ =p\int_{-1/2}^{0}dx_1&plus;p\int_{0}^{1/2}dx_1&plus;(1-p)\int_{0}^{1/2}dx_1&plus;(1-p)\int_{1/2}^{x_1}dx_1\\ =p&plus;(1-p)x_1,~1/2\leq x_1<1\\ =1~if~x_1\geq 1\\$

$(b)~Write~f_{X_i}(x_i)=\sigma (u_i)p^{y_i}(1-p)^{z_i}\\ where,~\sigma (u_i)=1~if~-1/2<x_i<1\\ =0~otherwise\\ ~y_i=1~if~x_i\in(-1/2,~0]\\ =0~otherwise\\ z_i=1~if~x_i\in [1/2,1)\\ =0~otherwise\\ L(p|x_1,x_2,...,x_n)=\prod_{i=1}^nf_{X_i}(x_i)=\prod_{i=1}^n\sigma(u_i)p^{y_i}(1-p)^{z_i}\\ =\left(\prod_{i=1}^n\sigma(u_i) \right )\times p^{\sum_{i=1}^ny_i}(1-p)^{\sum_{i=1}^nz_i}\\ \log L(p|x_1,x_2,...,x_n)=l(p)\\ =\log\left(\prod_{i=1}^n\sigma(u_i) \right )&plus;(\sum_{i=1}^ny_i)\log p&plus;(\sum_{i=1}^nz_i)\log (1-p)\\ \frac{\mathrm{d} l(p)}{\mathrm{d} p}=\frac{\sum_{i=1}^ny_i}{p}-\frac{\sum_{i=1}^nz_i}{1-p}=0\Rightarrow \hat{p}=\frac{\sum_{i=1}^ny_i}{\sum_{i=1}^ny_i&plus;\sum_{i=1}^nz_i}\\ \frac{\mathrm{d}^2 l(p)}{\mathrm{d} p^2}=-\frac{\sum_{i=1}^ny_i}{p^2}-\frac{\sum_{i=1}^nz_i}{(1-p)^2}<0~for~p=\hat{p}.\\ Hence~MLE~of~p=\frac{\sum_{i=1}^ny_i}{\sum_{i=1}^ny_i&plus;\sum_{i=1}^nz_i}$

$(c)~E(X_1)=p\int_{-1/2}^0xdx&plus;\int_{0}^{1/2}xdx&plus;(1-p)\int_{1/2}^1xdx\\ =\frac{p}{8}&plus;\frac{1}{8}&plus;\frac{3(1-p)}{8}=\frac{2-p}{4}~or,~E(2-4X_1)=p\\ Now,~E(2-4\bar{X})=p~where,~\bar{X}=\frac{1}{n}\sum_{i=1}^nX_i.\\\\ Hence~MOM~estimator~of~p=2-4\bar{X}.$