notes and homeworks for math

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diff --git a/li/hw4.tex b/li/hw4.tex
new file mode 100644
index 0000000..558f9cb
--- /dev/null
+++ b/li/hw4.tex
@@ -0,0 +1,140 @@
+\def\bmatrix#1{\left[\matrix{#1}\right]}
+
+    {\noindent\bf Section 2.4}
+
+{\noindent\bf 6.}
+
+{\it (a)}
+
+It has a two-sided inverse if $r = m = n.$
+
+{\it (b)}
+
+It has infinitely many solutions if $r = m < n.$
+
+{\noindent\bf 12.}
+
+{\it (a)}
+
+Matrix $A$ is of rank 1 and equal to
+$$\bmatrix{1\cr 0\cr 2}\bmatrix{1&0&0&3}$$
+
+{\it (b)}
+
+Matrix $A$ is of rank 1 and equal to
+$$\bmatrix{2\cr 6}\bmatrix{1&-1}$$
+
+{\noindent\bf 18.}
+
+The row space has basis:
+$$\{\bmatrix{0\cr1\cr2\cr3\cr4}, \bmatrix{0\cr0\cr0\cr1\cr2}\},$$
+the null space has basis:
+$$\{\bmatrix{1\cr0\cr0\cr0\cr0}, \bmatrix{0\cr-2\cr1\cr0\cr0},
+\bmatrix{0\cr0\cr0\cr-2\cr0}\},$$
+the column space has basis:
+$$\{\bmatrix{1\cr1\cr0},\bmatrix{3\cr4\cr1}\},$$
+and the left null space has basis:
+$$\{\bmatrix{1\cr-1\cr1}\}.$$
+
+{\noindent\bf 32.}
+
+$A$ has column space of the xy-plane, and left null space of the z-axis.
+It also has row space of the yz-plane, and null space of the x-axis.
+
+$I+A$ has full rank, so its column space and row space are ${\bf R}^3,$
+and its null space and left null space are the zero vector.
+
+\iffalse % practice problems
+
+{\noindent\bf 2.}
+
+{\noindent\bf 3.}
+
+{\noindent\bf 8.}
+
+{\noindent\bf 9.}
+
+{\noindent\bf 10.}
+
+{\noindent\bf 16.}
+
+{\noindent\bf 17.}
+
+{\noindent\bf 21.}
+
+{\noindent\bf 25.}
+
+{\noindent\bf 27.}
+
+{\noindent\bf 35.}
+
+{\noindent\bf 37.}
+
+\fi
+
+    {\noindent\bf Section 2.6}
+
+{\noindent\bf 16.}
+
+$$\bmatrix{0&1&0&0\cr 0&0&1&0\cr 0&0&0&1\cr 1&0&0&0}$$
+If $A$ maps $(x_1, x_2, x_3, x_4)$ to $(x_2, x_3, x_4, x_1),$ $A^2$ maps
+$x$ to $(x_3, x_4, x_1, x_2),$ and $A^3$ takes $x$ to $(x_4, x_1, x_2,
+x_3),$ and $AA^3 = I = A^4,$ so $A^3 = A^{-1}$ by definition of the
+identity.
+
+{\noindent\bf 28.}
+
+{\it (a)}
+
+Range is $V^2,$ and kernel is $0.$
+
+{\it (b)}
+
+Range is $V^2,$ and kernel has basis $(0, 0, 1).$
+
+{\it (c)}
+
+Range is $0,$ and kernel is $V^2$
+
+{\it (d)}
+
+Range is the subspace with basis $(1, 1)$ and kernel has basis $(0, 1).$
+
+{\noindent\bf 36.}
+
+{\it (a)}
+
+$$\bmatrix{2&5\cr 1&3}$$
+
+{\it (b)}
+
+$$\bmatrix{3&-5\cr -1&2}$$
+
+{\it (c)}
+
+Because, by linearity, if $(2, 6) \mapsto (1, 0),$ $.5(2,6) = (1, 3)
+\mapsto (.5, 0).$
+
+{\noindent\bf 44.}
+
+This is equivalent to a 180$^\circ$ rotation.
+
+\iffalse % practice problems
+
+{\noindent\bf 6.}
+
+{\noindent\bf 7.}
+
+{\noindent\bf 8.}
+
+{\noindent\bf 9.}
+
+{\noindent\bf 17.}
+
+{\noindent\bf 40.}
+
+{\noindent\bf 45.}
+
+\fi
+
+\bye
diff --git a/zhilova/08_jensen b/zhilova/08_jensen
index 20a8158..efe8bdb 100644
--- a/zhilova/08_jensen
+++ b/zhilova/08_jensen
@@ -21,3 +21,40 @@ f is strictly convex <=> f' is strictly increasing on (a,b)
 (2) If f is twice differentiable on (a,b)
 f is convex <=> f'' \geq 0 on (a,b)
 f is strictly convex <=> f'' > 0 on (a,b)
+
+Transformations of an r.v.
+
+Where X is an r.v. with pdf f_X(x), cdf F_X(x).
+
+Y := g(X). f_Y(y) = ?
+
+(Case 1) g is differentiable and invertible on D_X (range of X).
+    f_Y(y) = f_X(g^{-1}(y)) * |d/dy g^{-1}(y)|
+    Also:
+        if monotonically increasing, F_Y(y) = F_X(g^{-1}(y))
+        if monotonically decreasing, F_Y(y) = 1 - F_X(g^{-1}(y))
+
+(Case 2) g is piecewise bijective.
+g is bijective on D_j where D_X = \cup_{j=1}^k D_j, with
+D_i \cap D_j = \empty if i =/= j. (i.e. D_1...D_k is a partition of D_X)
+
+Then apply (1) through a sum.
+    f_y(y) = \sum f_X(g_j^{-1}(y)) * |d/dy g_j^{-1}(y)| * Indicator(y in range of g_j)
+
+F_Y(y) = P(g(X) \leq y) = P(\sum_{j=1}^k g_j(X) \leq y)
+= \int_R f_X(x) * indicator(x : g(x) \leq y) dx
+= \sum_{j=1}^k \int_R f_X(x) * Indicator{x : g_j(x) \leq y} dx
+If g is monotonic increasing, the indicator is equivalent to x \leq
+g_j^{-1}(y)
+
+This gives rise to several other transformations. Ex: scale
+transformation ( g(x) = cx ), scale-position transformation ( g(x) =
+cx+d ).
+
+Def: Symmetric distribution is when f_X(x) = f_X(-x).
+
+If X is a symmetric distribution and E|X| < \infty, EX = 0.
+
+EX = \int_{-\infty}^\infty xf_X(x) dx
+   = \int_0^\infty xf_X(x) + (-x)f_X(x) dx
+   = 0, by symmetry and some rearrangement of the integral.
diff --git a/zhilova/hw1.tex b/zhilova/hw1.tex
new file mode 100644
index 0000000..d9f5d22
--- /dev/null
+++ b/zhilova/hw1.tex
@@ -0,0 +1,133 @@
+\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\Pr{\bb P}
+\def\E{\bb E}
+\newcount\qnum
+\def\q{\afterassignment\qq\qnum=}
+\def\qq{\qqq{\number\qnum}}
+\def\qqq#1{\bigskip\goodbreak\noindent{\bf#1)}\smallskip}
+\def\fr#1#2{{#1\over #2}}
+\def\var{\mathop{\rm var}\nolimits}
+
+\q1
+
+Let $A \setminus B := \{x \in A : x\not\in B\}.$
+$(A\setminus B) \cap B = \empty,$ by definition, so they are disjoint
+sets. Therefore, if $x \in A \land x \in B,$ $x\not\in A \setminus B$
+and $x\not\in B\setminus A,$ and also the two setminuses do not
+intersect with eachother, making them disjoint sets.
+$$\Pr(A\cup B) = \Pr((A\setminus B)\cup(A\cap B)\cup(B\setminus A)) =
+\Pr(A\setminus B) + \Pr(A\cap B) + \Pr(B\setminus A) + \Pr(A\cap B) -
+\Pr(A\cap B)$$$$ = \Pr((A\setminus B)\cup(A\cap B)) + \Pr((B\setminus
+A)\cup(A\cap B)) - \Pr(A\cap B) = \Pr(A) + \Pr(B) - \Pr(A\cap B).$$
+
+\q2
+
+$$\Pr(A\cap B) + \Pr(A\cap B^C) = \Pr(A\cap B) + \Pr(\{x\in A:x\not\in
+B\}) = \Pr((A\cap B) \cup (A\setminus B)) = \Pr(A),$$
+by disjointedness and set arithmetic.
+
+\q3
+
+$$\Pr(A_1 \cup (A_2^C \cap A_3^C)) = \Pr(A_1) + \Pr(A_2^C \cap A_3^C) -
+\Pr(A_1\cap A_2^C\cap A_3^C) = 1/6 + \Pr(A_2^C)\Pr(A_3^C) -
+\Pr(A_1)\Pr(A_2^C)\Pr(A_3^C)$$$$ = 1/6 + 25/36 - 25/216 = 161/216.$$
+
+\q4
+
+\noindent{\it (a)}
+
+Pairwise exclusive implies mutually exclusive:
+$$A_1\cap A_2\cap A_3 = A_1\cap (A_2\cap A_3) = A_1\cap\empty =
+\empty,$$
+and see {\it (b)} for a description of why these probabilities are
+therefore impossible.
+
+\noindent{\it (b)}
+
+No, if they were, all three sets would be disjoint, giving
+$$\Pr(A_1\cup A_2\cup A_3) = \Pr(A_1) + \Pr(A_2) + \Pr(A_3) = 1/2 + 1/4 +
+1/3 > 1,$$
+violating a property of the probability function ($0 \leq \Pr X \leq 1$)
+
+\q5
+
+\noindent{\it (a)}
+$$\E(Y) = \E(2 - 3X) = \E(2) - 3\E(X) = 2 - 3*2 = -4.$$
+
+\noindent{\it (b)}
+$$\var Y = \E(Y^2) - \E(Y)^2 = \E((2-3X)^2) - 16 = \E(4 - 12X + 9X^2) -
+16 = 4 - 12\E(X) + 9\E(X^2) - 16 = 18,$$
+by linearity of expectation.
+
+\q6
+
+$$P_X(k) = \left\{\vbox{\halign{$#$\hfil&\hskip3em $#$\hfil\cr{\lambda^k
+e^{-\lambda}\over k!}&k \geq 0\cr 0&{\rm otherwise}\cr}}\right..$$
+
+\noindent{\it (a)}
+
+$$\E(e^{tX}) = \sum_{x=0}^\infty e^{tx}P_X(x)
+= \sum_{x=0}^\infty {e^{tx}\lambda^xe^{-\lambda}\over x!}
+= e^{-\lambda}\sum_{x=0}^\infty {e^{(t+\ln \lambda)x}\over x!}
+= e^{-\lambda}\sum_{x=0}^\infty {(\lambda e^t)^x\over x!}
+= e^{-\lambda}e^{\lambda e^t},
+$$
+by definition of the $e^ax$ Taylor series.
+
+\noindent{\it (b)}
+
+The third-order moment $\E(X^3)$ will be the third derivative of the mgf
+$\E(e^{tX}).$ at $t=0,$ so we get
+$$e^{-\lambda}(\lambda e^t)^3 e^{\lambda e^t} = \lambda^3.$$
+
+\q7
+
+\noindent{\it (a)}
+
+The pmf is $1/16$ for $X = 0$ and $X=4,$ $4/16$ for $X=1$ and $X=3,$ and
+$6/16$ for $X=2,$ or in other terms, $P_X(k) = {{4\choose k}\over 2^4}.$
+
+\noindent{\it (b)}
+
+This can be immediately computed as $$\Pr(\hbox{$X$ is odd}) = \Pr(X = 1)
++ \Pr(X = 3) = 1/2,$$ by disjointedness of those events.
+
+\q8
+
+Chebyshev's inequality gives us
+$$\Pr(|X-\E X| < .1) \geq 1-{\sigma^2\over .1^2} \geq .95$$
+The maximum value of $\sigma = \var X$ is $.05*.1^2 = \sigma^2 \to
+\sigma = .1\sqrt{.05} \approx .022.$
+
+\q9
+
+{\it (a)}
+
+On $\{x \geq 0\} = \{y \geq 1\},$
+$$F_X(x) = 1 - e^{-\lambda x}.$$
+$$F_Y(y) = F_X(e^y) = 1 - e^{-\lambda e^y}.$$
+$$f_Y(y) = \lambda e^y e^{-\lambda e^y} = \lambda e^{y-\lambda e^y}.$$
+
+{\it (b)}
+
+$\lambda < 1$ gives convergence.
+
+\bye