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diff --git a/zhilova/midterm2.tex b/zhilova/midterm2.tex new file mode 100644 index 0000000..8e74cf8 --- /dev/null +++ b/zhilova/midterm2.tex @@ -0,0 +1,146 @@ +\def\problem#1{\goodbreak\bigskip\noindent{\bf #1)}\smallskip\penalty500} +\newfam\rsfs +\newfam\bbold +\def\scr#1{{\fam\rsfs #1}} +\def\bb#1{{\fam\bbold #1}} +\let\oldcal\cal +\def\cal#1{{\oldcal #1}} +\font\rsfsten=rsfs10 +\font\rsfssev=rsfs7 +\font\rsfsfiv=rsfs5 +\textfont\rsfs=\rsfsten +\scriptfont\rsfs=\rsfssev +\scriptscriptfont\rsfs=\rsfsfiv +\font\bbten=msbm10 +\font\bbsev=msbm7 +\font\bbfiv=msbm5 +\textfont\bbold=\bbten +\scriptfont\bbold=\bbsev +\scriptscriptfont\bbold=\bbfiv + +\def\E{\bb E} +\def\P{\bb P} +\newcount\qnum +\def\fr#1#2{{#1\over #2}} +\def\var{\mathop{\rm var}\nolimits} +\def\cov{\mathop{\rm cov}\nolimits} +\def\dd#1#2{\fr{\partial #1}{\partial #2}} + +\problem{1} +\noindent{\bf (a)} +$$\E(Z) = \pmatrix{1\cr0}\qquad \var(Z) = \pmatrix{2&c\cr c&4}$$ + +\noindent{\bf (b)} +Covariance is a bilinear function, and $\var (kX) = \cov(kX,kX) = +k^2\var X.$ Also, $\cov(c, X) = 0,$ if $c$ is a constant, regardless of +$X.$ (because $\cov(c, X) = \E(cX) - \E(c)\E(X) = c\E(X) - c\E(X) = 0.$) +With linearity, this means $\cov(X+c, Y+b) = \cov(X, Y).$ +$$\rho(2X-1, 5-Y) = +{\cov(2X-1, 5-Y)\over\sqrt{\var(2X-1)}\sqrt{\var(5-Y)}} = +{\cov(2X, -Y)\over\sqrt{\var(2X)}\sqrt{\var(-Y)}} = $$$$ +{-2\cov(X, Y)\over2\sqrt{\var(X)}\sqrt{\var(Y)}} = +-{.5\over\sqrt{2}\sqrt{4}} = +-{\sqrt2\over8}.$$ + +\noindent{\bf (c)} + +$X$ and $Y$ are not necessarily independent from each other, although +independence would give a covariance of 0 ($p_XY(x,y) = p_X(x)p_Y(y) +\to \E(XY) = \E(X)\E(Y).$) + +Let $W \sim \cal N(0, 1),$ and $Z \sim 2\cal B(.5) - 1$ (i.e. it has a +.5 probability of being -1 or 1). +$X := \sqrt2 W + 1$ and $Y := 2ZW.$ +These are strictly dependent because $Y = \sqrt2Z(X-1),$ so $Y$ has +conditional distribution $\sqrt2(x-1)(2\cal B(.5) - 1),$ which is +clearly not equal to its normal distribution (which can be fairly easily +verified by symmetry of $W$). +However, they have covariance 0: +$$\E(XY) - \E X\E Y = \E((\sqrt2 W + 1)2ZW) - \E(\sqrt 2 W + 1)\E(2ZW) +$$$$ += \E(2\sqrt2 ZW^2) - \E(\sqrt 2 W)\E(2ZW) += 0 - 0\E(2ZW) = 0. +$$ + +% wikipedia says no + +\problem{2} +\noindent{\bf (a)} + +\def\idd#1#2{\dd{#1}{#2}^{-1}} +$Y_1 = 2X_2$ and $Y_2 = X_1 - X_2$ give us $X_2 = Y_1/2$ and +$X_1 = Y_2 + Y_1/2.$ This lets us compute Jacobian +$$J = \left|\matrix{\idd{x_1}{y_1} &\idd{x_1}{y_2}\cr + \idd{x_2}{y_1} &\idd{x_2}{y_2}}\right| + = \left|\matrix{2&1\cr2&0}\right| = -2.$$ +$$g(y_1, y_2) = \left\{\vcenter{\halign{\strut$#$,&\quad$#$\cr + |J|2e^{-(y_2+y_1/2)}e^{-y_1/2} & 0 < y_2+y_1/2 < y_1/2\cr + 0 & {\rm elsewhere}\cr + }}\right.$$ +$y_2+y_1/2 < y_1/2 \to y_2 < 0 \to -y_2 > 0.$ +And $0 < y_2 + y_1/2 \to y_1 > -2y_2 > 0.$ +$$ = \left\{\vcenter{\halign{\strut$#$,&\quad$#$\cr + 4e^{-y_2}e^{-y_1}&y_1 > -2y_2 > 0\cr + 0&{\rm elsewhere}\cr + }}\right.$$ + +\noindent{\bf (b)} + +$$g(y_1) = \int_{-\infty}^\infty g(y_1,y_2)dy_2 = +\bb I(y_1>0)\int_{-y_1/2}^0 4e^{-y_1}e^{-y_2} dy_2 = +\bb I(y_1>0)4e^{-y_1}(1-e^{y_1/2}).$$ +$$g(y_2) = \int_{-\infty}^\infty g(y_1,y_2)dy_1 = +\bb I(y_2<0)\int_{-2y_2}^\infty 4e^{-y_1}e^{-y_2} dy_1 = +\bb I(y_2<0)4e^{-y_2}(-e^{2y_2}) +.$$ + +\noindent{\bf (c)} + +They are independent iff $g(y_1,y_2) \bb I(y_1 > -2y_2 > 0)e^{-y_1-y_2} += g(y_1)g(y_2) = \bb I(y_1 > 0)\bb I(y_2 < 0) h(x),$ +where $h(x)$ is the strictly non-zero product of exponents that would +result, showing that they are dependent (if $y_1 = -y_2 = 1,$ the right +indicators are satisfied but not the left indicator, and since $h(x)$ is +non-zero, we see a contradiction.) + +\problem{3} +\noindent{\bf (a)} +We start determining the mgf from the pdf of $X,$ +$p_X(x) = +\fr1{\sqrt{2\pi}}e^{-\fr12x^2}.$ +$$\E(e^{tX}) = +\int_{-\infty}^\infty e^{tx}\fr1{\sqrt{2\pi}}e^{-\fr12x^2} dx = +\fr1{\sqrt{2\pi}} \int_{-\infty}^\infty e^{tx-\fr12x^2} dx = $$$$ +\fr1{\sqrt{2\pi}} \int_{-\infty}^\infty e^{tx-\fr12x^2-\fr12t^2 + \fr12t^2} dx = +e^{\fr12t^2} \int_{-\infty}^\infty \fr1{\sqrt{2\pi}}e^{-\fr12(x-t)^2} dx += e^{\fr12t^2}, +$$ +by the final integrand being a normal pdf and therefore integrating to +1. + +\noindent{\bf (b)} +$$M_Y(t) = \E(e^{t(aX+b)}) = \E(e^{bt}e^{atX}) = e^{bt}\E(e^{atX}) = +e^{bt}M_X(at) = e^{bt}e^{\fr12(at)^2}.$$ + +\noindent{\bf (c)} + +Theorem 1.9.2 states that two probability distribution functions are +alike if and only if their moment generating functions are equal in some +vicinity of zero. +The mgf of $Y$ corresponds to $\cal N(b, a^2),$ which has the following +mgf (by computation from its pdf definition): +$$\int_{-\infty}^\infty e^{tx}\fr1{a\sqrt{2\pi}}e^{-(x-b)^2\over +2a^2} dx = +\fr1{a\sqrt{2\pi}}\int_{-\infty}^\infty e^{2a^2tx - x^2 + 2bx - b^2\over +2a^2} dx =$$$$ +\fr1{a\sqrt{2\pi}}\int_{-\infty}^\infty e^{-(b+a^2t)^2 + +2(a^2t+b)x - x^2 - b^2 + (b+a^2t)^2\over 2a^2} dx = +e^{(b+a^2t)^2-b^2\over 2a^2}\int_{-\infty}^\infty +\fr1{a\sqrt{2\pi}}e^{-(b+a^2t+x)^2 \over 2a^2} dx = +e^{bt}e^{(at)^2\over2}, +$$ +proving $Y \sim \cal N(b, a^2),$ because this function is convergent +everywhere, and reusing the fact that the final integrand is a normal +pdf, so it must integrate to 1. + +\bye |