2 problems found
The random variable \(N\) takes positive integer values and has pgf (probability generating function) \(\G(t)\). The random variables \(X_i\), where \(i=1\), \(2\), \(3\), \(\ldots,\) are independently and identically distributed, each with pgf \({H}(t)\). The random variables \(X_i\) are also independent of \(N\). The random variable \(Y\) is defined by \[ Y= \sum_{i=1}^N X_i \;. \] Given that the pgf of \(Y\) is \(\G(H(t))\), show that \[ \E(Y) = \E(N)\E(X_i) \text{ and } \var(Y) = \var(N)\big(\E(X_i)\big)^2 + \E(N) \var(X_i) \,.\] A fair coin is tossed until a head occurs. The total number of tosses is \(N\). The coin is then tossed a further \(N\) times and the total number of heads in these \(N\) tosses is \(Y\). Find in this particular case the pgf of \(Y\), \(\E(Y)\), \(\var(Y)\) and \(\P(Y=r)\).
Solution: Recall that for a random variable \(Z\) with pgf \(F(t)\) we have \(F(1) = 1\), \(\E[Z] = F'(1)\) and \(\E[Z^2] = F''(1) +F'(1)\) so \begin{align*} && \E[Y] &= G'(H(1))H'(1) \\ &&&= G'(1)H'(1) \\ &&&= \E[N]\E[X_i] \\ \\ && \E[Y^2] &= G''(H(1))(H'(1))^2+G'(H(1))H''(1) + G'(H(1))H'(1) \\ &&&= G''(1)(H'(1))^2+G'(1)H''(1) + G'(1)H'(1) \\ &&&= (\E[N^2]-\E[N])(\E[X_i])^2 + \E[N](\E[X_i^2]-\E[X_i]) + \E[N]\E[X_i] \\ &&&= (\E[N^2]-\E[N])(\E[X_i])^2 + \E[N]\E[X_i^2] \\ && \var[Y] &= (\E[N^2]-\E[N])(\E[X_i])^2 + \E[N]\E[X_i^2] - (\E[N])^2(\E[X_i])^2\\ &&&= (\var[N]+(\E[N])^2-\E[N])(\E[X_i])^2 + \E[N](\var[X_i]+\E[X_i]^2) - (\E[N])^2(\E[X_i])^2\\ &&&= \var[N](\E[X_i])^2 + \E[N]\var[X_i] \end{align*} Notice that \(N \sim Geo(\tfrac12)\) and \(Y = \sum_{i=1}^N X_i\) where \(X_i\) are Bernoulli. We have that \(G(t) = \frac{\frac12}{1-\frac12z}\) and \(H(t) = \frac12+\frac12p\) so the pgf of \(Y\) is \(G(H(t) = \frac{\frac12}{1 - \frac14-\frac14p} = \frac{2}{3-p}\). \begin{align*} && \E[X_i] &= \frac12\\ && \var[X_i] &= \frac14 \\ && \E[N] &= 2 \\ && \var[N] &= 2 \\ \\ && \E[Y] &= 2 \cdot \frac12 = 1 \\ && \var[Y] &= 2 \cdot \frac14 + 2 \frac14 = 1 \\ && \mathbb{P}(Y=r) &= \tfrac23 \left ( \tfrac13 \right)^r \end{align*}
I choose a number from the integers \(1, 2, \ldots, (2n-1)\) and the outcome is the random variable \(N\). Calculate \( \E(N)\) and \(\E(N^2)\). I then repeat a certain experiment \(N\) times, the outcome of the \(i\)th experiment being the random variable \(X_i\) (\(1\le i \le N\)). For each \(i\), the random variable \(X_i\) has mean \(\mu\) and variance \(\sigma^2\), and \(X_i\) is independent of \(X_j\) for \(i\ne j\) and also independent of \(N\). The random variable \(Y\) is defined by \(Y= \sum\limits_{i=1}^NX_i\). Show that \(\E(Y)=n\mu\) and that \(\mathrm{Cov}(Y,N) = \frac13n(n-1)\mu\). Find \(\var(Y) \) in terms of \(n\), \(\sigma^2\) and \(\mu\).
Solution: \begin{align*} && \E[N] &= \sum_{i=1}^{2n-1} \frac{i}{2n-1} \\ &&&= \frac{2n(2n-1)}{2(2n-1)} = n\\ && \E[N^2] &= \sum_{i=1}^{2n-1} \frac{i^2}{2n-1} \\ &&&= \frac{(2n-1)(2n)(4n-1)}{6(2n-1)} \\ &&&= \frac{n(4n-1)}{3} \\ && \var[N] &= \frac{n(4n-1)}{3} - n^2 \\ &&&= \frac{n^2-n}{3} \end{align*} \begin{align*} && \E[Y] &= \E \left [ \E \left [ \sum_{i=1}^N X_i | N = k\right] \right]\\ &&&= \E \left[ N\mu \right] = n\mu \\ \\ && \mathrm{Cov}(Y,N) &= \mathbb{E}[XY] - \E[X]\E[Y] \\ &&&= \E \left [ \E \left [N \sum_{i=1}^N X_i | N = k\right] \right] - n^2 \mu \\ &&&= \E[N^2\mu] - n^2 \mu \\ &&&= \left ( \frac{n^2(4n-1)}{3} - n^2 \right) \mu \\ &&&= \frac{n^2-n}{3}\mu \\ \\ && \E[Y^2] &= \E \left [ \E \left [ \left ( \sum_{i=1}^N X_i \right) ^2\right ] \right] \\ &&&= \E \left [ \E \left [ \sum_{i=1}^N X_i ^2 + 2\sum_{i,j} X_iX_j\right ] \right] \\ &&&= \E \left [ \sum_{i=1}^N \left ( \E[X_i ^2] + 2\sum_{i,j} \E[X_i]\E[X_j]\right ) \right] \\ &&&= \E \left [ N(\sigma^2 + \mu^2) + (N^2-N)\mu^2\right] \\ &&&= n(\sigma^2+\mu^2) + \left ( \frac{n^2-n}{3}-n \right)\mu^2 \\ &&&= n\sigma^2 + \frac{n^2-n}{3} \mu^2 \\ \Rightarrow && \var[Y] &= n\sigma^2 + \frac{n^2-n}{3} \mu^2 - n^2\mu^2 \\ &&&= n\sigma^2 - \frac{2n^2+n}{3} \mu^2 \end{align*}