Question

. Let Yı, . . . , Ý, be a sample from N(0, σ*) distribution. Show that both Gi (Yi, . . . , X,; σ) = nHare pivots. j-1 72 and G (1) Recall the confidence interval based on Gi that we derived in class. (2) Let Z be N(0, 1) random variable. Find the expectation and variance of |Z. (3) If n is large, what is the approzimate distribution of (4) Use (3) to construct an approximate confidence interval for ơ2

Answer 1

$~Since~Y_i's~are~independently~N(0,\sigma^2)~then~\frac{Y_i^2}{\sigma^2}'s~are~also~independently~\chi^2_1~independently\\ hence~from~additive~property~of~\chi^2\\ G_1(Y_1,Y_2,...,Y_n)=\sum_{j=1}^n\frac{Y_j^2}{\sigma^2}\sim \chi^2_n~is~a~pivot.\\ We~know~\frac{Y}{\sigma}~is~N(0,1)~where~Y\sim N(0,\sigma^2).\\ Then~the~cdf~of~Z=\frac{|Y|}{\sigma}~is\\ F_Z(z)=P\left(Z\leq z \right )=P\left(|Y|\leq z\sigma \right )=\Phi(z)-\Phi(-z)=2\Phi(z)-1,~z>0\\ =0~otherwise\\ The~pdf~of~Z~is\\ f_Z(z)=\frac{\mathrm{d} F(z)}{\mathrm{d} z}=2\phi(z)=\sqrt{\frac{2}{\pi}}\exp\left(-\frac{z^2}{2} \right );~0<z<\infty\\ =0~otherwise\\ Hence~the~distribution~of~Z~is~free~from~\sigma^2,~hence~it~is~pivot~and~then\\ G(Y_1,Y_2,...,Y_n;\sigma)~is~also~pivot.\\ (1)~The~confidence~interval~for~\sigma~based~on~G_1~is\\ \left(\sqrt{\frac{\sum_{j=1}^nY_j^2}{\chi^2_{\alpha/2,n}}},~\sqrt{\frac{\sum_{j=1}^nY_j^2}{\chi^2_{1-\alpha/2,n}}} \right ).\\$$(2)~E(|Z|)=\sqrt{\frac{2}{\pi}}\int_0^\infty z\exp\left(-z^2/2 \right )dz=\sqrt{\frac{2}{\pi}}\int_0^\infty \exp\left(-z^2/2 \right )d(z^2/2)=\sqrt{\frac{2}{\pi}}\\ E(|Z|^2)=E(Z^2)=Var(Z)=1\\ Var(|Z|)=E(|Z|^2)-E^2(|Z|)=1-\frac{2}{\pi}=\frac{\pi-2}{\pi}.\\ (3)~Using~central~limit~theorem,\\ \frac{\sum_{j=1}^n\left(\frac{|Y_j|}{\sigma}-\sqrt{\frac{2}{\pi}} \right )}{\sqrt{\frac{n(\pi-2)}{\pi}}}\overset{L}{\rightarrow}N(0,1).\\ (4)~P\left(\sqrt{\frac{2}{\pi}}-Z_{\alpha/2}\sqrt{\frac{n(\pi-2)}{\pi}} \leq \sum_{j=1}^n\frac{|Y_j|}{\sigma}\leq \sqrt{\frac{2}{\pi}}&plus;Z_{\alpha/2}\sqrt{\frac{n(\pi-2)}{\pi}}\right )=1-\alpha\\ (1-\alpha)\times 100\%~C.I.~for~\sigma^2~is\\ \left( \left(\frac{\sum_{j=1}^n|Y_j|}{\sqrt{\frac{2}{\pi}}&plus;Z_{\alpha/2}\sqrt{\frac{n(\pi-2)}{\pi}}} \right )^2,~ \left(\frac{\sum_{j=1}^n|Y_j|}{\sqrt{\frac{2}{\pi}}-Z_{\alpha/2}\sqrt{\frac{n(\pi-2)}{\pi}}} \right )^2\right )$