auto-update(nvim): 2025-03-02 23:45:29

documents/by-course/pstat-120a/hw6/main.typ

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+#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)
+  $

documents/by-course/pstat-120a/hw6/package.nix

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+{
+  pkgs,
+  typstPackagesCache,
+  typixLib,
+  cleanTypstSource,
+  flakeSelf,
+  ...
+}:
+let
+  src = cleanTypstSource ./.;
+  commonArgs = {
+    typstSource = "main.typ";
+
+    fontPaths = [
+      # Add paths to fonts here
+      # "${pkgs.roboto}/share/fonts/truetype"
+    ];
+
+    virtualPaths = [
+      # Add paths that must be locally accessible to typst here
+      # {
+      #   dest = "icons";
+      #   src = "${inputs.font-awesome}/svgs/regular";
+      # }
+    ];
+
+    XDG_CACHE_HOME = typstPackagesCache;
+    SOURCE_DATE_EPOCH = builtins.toString flakeSelf.lastModified;
+  };
+
+in
+typixLib.buildTypstProject (
+  commonArgs
+  // {
+    inherit src;
+  }
+)