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Jonathan Yu (jsyu)

21-217 HW5

\begin{enumerate}
  \item
    \begin{enumerate}
      \item
        First, let us compute
        \[P_X(3) = \sum_{i=1}^3 P_{X,Y}(3,i) = 1/9 + 1/18 + 1/9 = 5/18.\]
        Then we can compute
        \[P_{Y|X}(y|3) = P_{X,Y}(3,y) / P_X(3) = 18 P_{X,Y}(3,y) / 5.\]
        Then we have
        \begin{align*}
            P_{Y|X}(1|3) &= 18(1/9)/5\\
            P_{Y|X}(2|3) &= 18(1/18)/5\\
            P_{Y|X}(3|3) &= 18(1/9)/5.
        \end{align*}
      \item
        \begin{align*}
          E[X|Y = 2] &= \sum_{x=1}^3 x P_{X|Y}(x,2)\\
          &= 1(1/9) + 2(0) + 3(1/18) = 5/18\\
        \end{align*}
      \item
        We are given the joint PMF, so we can sum the rows and the columns
        \[\begin{array}{c|c|c|c|c}
           &1    &2   &3   &    \\
          1&1/9  &1/9 &0   &2/9 \\
          2&1/3  &0   &1/6 &1/2 \\
          3&1/9  &1/18&1/9 &5/18\\
           &5/9  &3/18&5/18&5/18\\
        \end{array}\]
        Note that $P_X(1)P_Y(1) = (2/9)(5/9) \neq 1/9 = P_{X,Y}(1,1)$.  Then
        $X$ and $Y$ are not independent.
    \end{enumerate}
  \item
    Let $X$ be the number of components that function tomorrow and 
    \[X_i = \left\{\begin{array}{rl}
      0&\text{ if component } i \text{ functions tomorrow}\\
      1&\text{ otherwise}
      \end{array}\right..\]
    \begin{enumerate}
      \item
        By linearity of expectations,
        \begin{align*}
          E[X | rain] &= E[\sum_{i=1}^n X_i | rain]\\
          &= \sum_{i=1}^n E[X_i | rain]\\
          &= \sum_{i=1}^n p_i.
        \end{align*}
      \item
        By linearity of expectations,
        \begin{align*}
          E[X] &= E[\sum_{i=1}^n X_i]\\
          &= \sum_{i=1}^n E[X_i]\\
          &= \sum_{i=1}^n (E[X_i | rain]\alpha + E[X_i | not\ rain](1-\alpha))\\
          &= \sum_{i=1}^n (\alpha p_i + (1-\alpha)q_i).
        \end{align*}
    \end{enumerate}
  \item
    Recall from lecture that 
    \begin{align*}
      P(X = k | X+Y = n) &= \frac{P(X=k,X+Y = n)}{P(X+Y = n)}\\
      &= \frac{P(X=k,Y = n-k)}{P(X+Y = n)}.
    \end{align*}
    First, we reduce the numerator to
    \begin{align*}
      P(X=k,Y = n-k) &= P(X=k)P(Y = n-k)&\text{independent}\\
      &= \left(e^{-\lambda} \frac{\lambda^k}{k!}\right)
        \left(e^{-\mu} \frac{\mu^{n-k}}{(n-k)!}\right)\\
      &= e^{-\lambda-\mu} \frac{\lambda^k \mu^{n-k}}{k!(n-k!)}\\
      &= e^{-\lambda-\mu} \lambda^k \mu^{n-k}\frac{1}{n!}{n \choose k}.
      \end{align*}
    Similarly, we can reduce the denominator to
    \begin{align*}
      P(X+Y = n) &= \sum_{k=0}^n P(X=k,Y = n-k)\\
      &= \sum_{k=0}^n P(X=k)P(Y = n-k)&\text{independent}\\
      &= e^{-\lambda-\mu}\frac{1}{n!}
        \sum_{k=0}^n \lambda^k \mu^{n-k}{n \choose k}.
      \end{align*}
    Then we have
    \begin{align*}
      P(X = k | X+Y = n) &= \frac{e^{-\lambda-\mu} \lambda^k
        \mu^{n-k}\frac{1}{n!}{n \choose k}}
        {e^{-\lambda-\mu}\frac{1}{n!}
        \sum_{k=0}^n \lambda^k \mu^{n-k}{n \choose k}}\\
      &=\frac{\lambda^k \mu^{n-k}{n \choose k}}{\sum_{k=0}^n \lambda^k
        \mu^{n-k}{n \choose k}}\\
      \end{align*}
    \item
      Let $X$ be the expected time it takes for the miner to escape.  Then
      \begin{align*}
        E[X] &= E[X | A]P(A) + E[X | B]P(B)\\
        &= 10(1/2) + E[X | B](1/2)\\
        &= 10(1/2) + (E[X] + 5)(1/2)\\
        E[X] &= 5 + E[X]/2 + 5/2\\
        E[X]/2 &= 5 + 5/2\\
        E[X] &= 10 + 5 = 15.
      \end{align*}
    \item
      We start by conditioning $E[X^3]$ on $X = 1$ to get
      \[E[X^3] = P(X = 1) E[X^3 | X = 1] + P(X > 1) E[X^3 | X > 1].\]
      Then by memoryless of $X$ and linearity of expectations,
      \[E[X^3 | X > 1] = E[(X + 1)^3] = E[X^3] + 3E[X^2] + 3E[X] + 1 = E[X^3] +
      3(2-p)/p^2 + 3/p + 1\]
      We now have
      \begin{align*}
        E[X^3] &= p(1) + (1-p)(E[X^3] + 3(2-p)/p^2 + 3/p + 1)\\
        E[X^3] &= p + E[X^3] + 3(2-p)/p^2 + 3/p + 1 - pE[X^3] - 3(2-p)/p - 3 - p\\
             0 &= 3(2-p)/p^2 + 3/p - pE[X^3] - 3(2-p)/p - 2\\
       pE[X^3] &= 3(2-p)/p^2 + 3/p - 3(2-p)/p - 2\\
        E[X^3] &= 3(2-p)/p^3 + 3/p^2 - 3(2-p)/p^2 - 2p\\
               &= \big(3(2-p) + 3p - 3(2-p)p - 2p^4\big) / p^3\\
               &= \big(6 - 3p + 3p - 6p - 3p^2 - 2p^4\big) / p^3\\
               &= \big(6 - 6p - 3p^2 - 2p^4\big) / p^3.
      \end{align*}
    \item
      \begin{enumerate}
        \item
          We can find the mean of the sample mean with
          \begin{align*}
            E[\bar{X}] &= \frac{1}{n} E[\sum_{i=1}^n X_i]\\
            &= \frac{1}{n} \sum_{i=1}^n E[X_i] &\text{by linearity of
            expectations}\\
            &= \frac{1}{n} \sum_{i=1}^n \lambda\\
            &= \frac{1}{n} n\lambda\\
            &= \lambda.
          \end{align*}
          The sample mean does not change when we change $n$. We can find the
          variance of the sample mean with
          \begin{align*}
            var(\bar{X})&= \frac{1}{n^2} var\left(\sum_{i=1}^n X_i\right)\\
            &= \frac{1}{n^2} \sum_{i=1}^n var\left(X_i\right)
                  &\text{independent}\\
            &= \frac{1}{n^2} \sum_{i=1}^n \lambda\\
            &= \frac{1}{n^2} n\lambda\\
            &= \frac{\lambda}{n}.
          \end{align*}
          For $n = 25$, we get $\frac{\lambda}{25}$ and for $n = 400$, we get
          $\frac{\lambda}{400}$.
        \item
          Note that $var(\bar{X}) = \frac{\lambda}{n}$.
          We get a better estimation of $\lambda$ with $n = 400$
          since it has a lower $var(\bar{X})$.
      \end{enumerate}
    \item
      \begin{enumerate}
        \item
          From the pdf we have
          \begin{align*}
            1 &= \int_0^1 1/2 dx + \int_1^2 1/3 dx + \int_2^3 c dx\\
            &= 1/2 + 1/3 + c\\
            1/6 &= c.
          \end{align*}
        \item
          From the pdf we have
          \begin{align*}
            P(1/2 < X < 5/2) &= \int_{1/2}^1 1/2 dx + \int_1^2 1/3 dx +
            \int_2^{5/2} 1/6 dx\\
            &= x/2 \Big|_{1/2}^1 +  x/3 \Big|_1^2 + x/6 \Big|_2^{5/2} \\
            &= 1/4 +  1/3 + 1/12 = 8/12 = 2/3.
          \end{align*}
      \end{enumerate}
    \item
      \begin{enumerate}
        \item
          \begin{align*}
            1 &= \int_{-1}^1 c(1-x^2) dx\\
            &= cx - \frac{x^3}{3} \Big|_{-1}^1 \\
            &= (c - \frac{1}{3}) - (-c - \frac{-1}{3})\\
            &= (c - \frac{1}{3}) + (c - \frac{1}{3})\\
            &= 2c - \frac{2}{3})\\
            \frac{5}{3})&= 2c\\
            \frac{5}{6})&= c.
          \end{align*}
        \item
          \begin{align*}
            P(-1/2 < X < 1/2) &= \int_{-1/2}^{1/2} 5/6(1-x^2) dx\\
            &= 5/6\Big\[x - \frac{x^3}{3}\Big\]_{-1/2}^{1/2} \\
            &= x/2 \Big|_{1/2}^1 +  x/3 \Big|_1^2 + x/6 \Big|_2^{5/2} \\
            &= 1/4 +  1/3 + 1/12 = 8/12 = 2/3.
          \end{align*}
      \end{enumerate}
\end{enumerate}
\end{document}