90 lines
2 KiB
Text
90 lines
2 KiB
Text
#import "@youwen/zen:0.1.0": *
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#import "@preview/cetz:0.3.2"
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#show: zen.with(
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title: "Homework 6",
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author: "Youwen Wu",
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date: "Winter 2025",
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)
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#show figure: it => {
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pad(y: 10pt, it)
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}
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#set enum(spacing: 2em)
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#let correction = content => {
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set text(fill: red)
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box(stroke: 1pt, inset: 5pt, content)
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}
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#let subproblems = content => {
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set enum(numbering: "a)")
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content
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}
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#rect[
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Initial score: $16/16$
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]
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#rect[
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#set text(fill: red)
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Revised score: $16/16$
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]
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1. $
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M_X (t) = EE[e^(X t)] = sum_(S_X) e^(x t) p_X (x) \
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= e^(-6t) 4 / 9 + e^(-2t) 1 / 9 + 2 / 9 (1+e^(3t))
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$
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2. We're looking for $EE[e^(X t)]$.
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$
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M_X (t) = integral_(-infinity) e^(x t) dot 1 / 2 e^(-|x|) dif x \
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1 / 2 integral_(-infinity)^0 e^(x t + x) dif x + 1 / 2 integral_0^infinity e^(-x(1-t)) dif x \
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= 1 / 2 [1 / (1+t) - 0] - 1 / 2 [0 - 1 / (1-t)] \
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= 1 / 2 [1 / (1+t) + 1 / (1-t)]
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$
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Note that the MGF is only defined for $t in (-1,1)$.
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3. #subproblems[
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1. Consider the MGF evaluated at 0
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$
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[(dif M_X (t)) / (dif t)]_(t=0) = [-4 / 3 e^(-4t) + 5 / 6 e^(5t)]_(t=0) = -1 / 2
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$
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For the variance we evaluate the second derivative instead.
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$
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[(dif^2 M_X (t)) / (dif t^2)]_(t=0) = [16 / 3 e^(-4t) + 25 / 6 e^(5t)]_(t=0) = 19 / 2
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$
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And then
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$
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"Var"(X) = 19 / 2 - (-1 / 2)^2 = 37 / 4
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$
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2. The PMF is $M_X (t) = sum _k e^(k t) p_X (k) = 1/2 + 1/3^(-4t) + 1/6 e^(5t)$
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Then $EE[X]$ and $EE[X^2]$ are
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$
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EE[X] = sum_k k dot p_X (k) = -4 / 3 + 5 / 6 = -1 / 2 \
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EE[X^2] = sum_k k^2 dot p_X (k) \
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= 16 / 3 + 25 / 6 = 19 / 2
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$
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So indeed our variance and mean match up.
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]
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4. The MGF is given by $X ~ "Pois"(3)$
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$
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M_X (t) = e^(3(e^t - 1))
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$
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So the answer is
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$
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P(X=4) = e^(-3) 3^4 / 4! = 0.16803
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$
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5. Let $Y = (X-1)^2$. The support of $Y$ is ${4,1,9}$. The PMF of $Y$ is
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$
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P(Y=4) = 1 / 7 \
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P(Y=1) = 2 / 7 \
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P(Y=9) = 4 / 7
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$
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6. $X ~ "Gamma"(2,1)$ and it has MGF
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$
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M_X (t) = 1 / ((1-t)^2)
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$
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