#import "@youwen/zen:0.1.0": *
#import "@preview/cetz:0.3.2"

#show: zen.with(
  title: "Homework 6",
  author: "Youwen Wu",
  date: "Winter 2025",
)

#show figure: it => {
  pad(y: 10pt, it)
}
#set enum(spacing: 2em)

#let correction = content => {
  set text(fill: red)
  box(stroke: 1pt, inset: 5pt, content)
}

#let subproblems = content => {
  set enum(numbering: "a)")
  content
}

#rect[
  Initial score: $16/16$
]

#rect[
  #set text(fill: red)
  Revised score: $16/16$
]

1. $
    M_X (t) = EE[e^(X t)] = sum_(S_X) e^(x t) p_X (x) \
    = e^(-6t) 4 / 9 + e^(-2t) 1 / 9 + 2 / 9 (1+e^(3t))
  $

2. We're looking for $EE[e^(X t)]$.
  $
    M_X (t) = integral_(-infinity) e^(x t) dot 1 / 2 e^(-|x|) dif x \
    1 / 2 integral_(-infinity)^0 e^(x t + x) dif x + 1 / 2 integral_0^infinity e^(-x(1-t)) dif x \
    = 1 / 2 [1 / (1+t) - 0] - 1 / 2 [0 - 1 / (1-t)] \
    = 1 / 2 [1 / (1+t) + 1 / (1-t)]
  $
  Note that the MGF is only defined for $t in (-1,1)$.

3. #subproblems[
    1. Consider the MGF evaluated at 0
      $
        [(dif M_X (t)) / (dif t)]_(t=0) = [-4 / 3 e^(-4t) + 5 / 6 e^(5t)]_(t=0) = -1 / 2
      $
      For the variance we evaluate the second derivative instead.
      $
        [(dif^2 M_X (t)) / (dif t^2)]_(t=0) = [16 / 3 e^(-4t) + 25 / 6 e^(5t)]_(t=0) = 19 / 2
      $
      And then
      $
        "Var"(X) = 19 / 2 - (-1 / 2)^2 = 37 / 4
      $
    2. The PMF is $M_X (t) = sum _k e^(k t) p_X (k) = 1/2 + 1/3^(-4t) + 1/6 e^(5t)$

      Then $EE[X]$ and $EE[X^2]$ are
      $
        EE[X] = sum_k k dot p_X (k) = -4 / 3 + 5 / 6 = -1 / 2 \
        EE[X^2] = sum_k k^2 dot p_X (k) \
        = 16 / 3 + 25 / 6 = 19 / 2
      $
      So indeed our variance and mean match up.
  ]
4. The MGF is given by $X ~ "Pois"(3)$
  $
    M_X (t) = e^(3(e^t - 1))
  $
  So the answer is
  $
    P(X=4) = e^(-3) 3^4 / 4! = 0.16803
  $

5. Let $Y = (X-1)^2$. The support of $Y$ is ${4,1,9}$. The PMF of $Y$ is
  $
    P(Y=4) = 1 / 7 \
    P(Y=1) = 2 / 7 \
    P(Y=9) = 4 / 7
  $

6. $X ~ "Gamma"(2,1)$ and it has MGF
  $
    M_X (t) = 1 / ((1-t)^2)
  $