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Worked solutions. From the course's own solution sets, licensed CC BY-NC-SA 4.0.

Solutions — The Bernoulli Process

Department of Electrical Engineering & Computer Science

6.041/6.431: Probabilistic Systems Analysis

(Fall 2010)

Recitation 13 Solutions

October 21, 2010

  1. (a) We begin by writing the definition for \(\mathbf{E}[Z \mid X, Y]\) $$ \mathbf{E}[Z \mid X = x, Y = y] = \sum_z z p_{Z \mid X, Y}(z \mid x, y) $$ Since \(\mathbf{E}[Z \mid X, Y]\) is a function of the random variables \(X\) and \(Y\), and is equal to \(\mathbf{E}[Z \mid X = x, Y = y]\) whenever \(X = x\) and \(Y = y\), which happens with probability \(p_{X, Y}(x, y)\), using the expected value rule, we have $$ \begin{aligned} \mathbf{E}[\mathbf{E}[Z \mid X, Y]] &= \sum_x \sum_y \mathbf{E}[Z \mid X = x, Y = y] p_{X, Y}(x, y) \ &= \sum_x \sum_y \sum_z z p_{Z \mid X, Y}(z \mid x, y) p_{X, Y}(x, y) \ &= \sum_x \sum_y \sum_z z p_{X, Y, Z}(x, y, z) \ &= \mathbf{E}[Z] \end{aligned} $$

(b) We start with the definition for \(\mathbf{E}[Z \mid X, Y]\) which is a function of the random variables \(X\) and \(Y\), and is equal to \(\mathbf{E}[Z \mid X = x, Y = y]\) whenever \(X = x\) and \(Y = y\), so $$ \mathbf{E}[Z \mid X = x, Y = y] = \sum_z z p_{Z \mid X, Y}(z \mid x, y) $$ Proceeding as above, but conditioning on the event \(X = x\), we have $$ \begin{aligned} \mathbf{E}[\mathbf{E}[Z \mid X, Y = y] \mid X = x] &= \sum_y \mathbf{E}[Z \mid X = x, Y = y] p_{Y \mid X}(y \mid x) \ &= \sum_y \sum_z z p_{Z \mid X, Y}(z \mid x, y) p_{Y \mid X}(y \mid x) \ &= \sum_y \sum_z z p_{Y, Z \mid X}(y, z \mid x) \ &= \mathbf{E}[Z \mid X = x] \end{aligned} $$ Since this is true for all possible values of \(x\), we have \(\mathbf{E}[\mathbf{E}[Z \mid Y, X] \mid X] = \mathbf{E}[Z \mid X]\).

(c) We take expectations of both sides of the formula in part (b) to obtain $$ \mathbf{E}[\mathbf{E}[Z \mid X]] = \mathbf{E}[\mathbf{E}[\mathbf{E}[Z \mid X, Y] \mid X]]. $$ By the law of iterated expectations, the left-hand side above is \(\mathbf{E}[Z]\), which establishes the desired result.

  1. Let \(Y\) be the length of the piece after we break for the first time. Let \(X\) be the length after we break for the second time.

(a) The law of iterated expectations states: $$ \mathbf{E}[X] = \mathbf{E}[\mathbf{E}[X \mid Y]] $$ We have \(\mathbf{E}[X \mid Y] = \frac{Y}{2}\) and \(\mathbf{E}[Y] = \frac{\ell}{2}\). So then: $$ \mathbf{E}[X] = \mathbf{E}[\mathbf{E}[X \mid Y]] = \mathbf{E}[Y/2] = \frac{1}{2} \mathbf{E}[Y] = \frac{1}{2} \frac{\ell}{2} = \frac{\ell}{4} $$

(b) We use the Law of Total Variance to find \(\text{var}(X)\): $$ \text{var}(X) = \mathbf{E}[\text{var}(X \mid Y)] + \text{var}(\mathbf{E}[X \mid Y]). $$ Recall that the variance of a uniform random variable distributed over \([a, b]\) is \((b - a)^2/12\). Since \(Y\) is uniformly distributed over \([0, \ell]\), we have $$ \begin{aligned} \text{var}(Y) &= \frac{\ell^2}{12}, \ \text{var}(X \mid Y) &= \frac{Y^2}{12}. \end{aligned} $$ We know that \(\mathbf{E}[X \mid Y] = Y/2\), and so $$ \text{var}(\mathbf{E}[X \mid Y]) = \text{var}(Y/2) = \frac{1}{4} \text{var}(Y) = \frac{\ell^2}{48}. $$ Also, $$ \begin{aligned} \mathbf{E}[\text{var}(X \mid Y)] &= \mathbf{E}\left[\frac{Y^2}{12}\right] \ &= \int_0^\ell \frac{y^2}{12} f_Y(y) dy \ &= \frac{1}{12} \cdot \frac{1}{\ell} \int_0^\ell y^2 dy \ &= \frac{\ell^2}{36}. \end{aligned} $$ Combining these results, we obtain $$ \text{var}(X) = \mathbf{E}[\text{var}(X \mid Y)] + \text{var}(\mathbf{E}[X \mid Y]) = \frac{\ell^2}{36} + \frac{\ell^2}{48} = \frac{7\ell^2}{144}. $$

  1. Let \(X_i\) denote the number of widgets in the \(i^{\text{th}}\) box. Then \(T = \sum_{i=1}^N X_i\). $$ \begin{aligned} \mathbf{E}[T] &= \mathbf{E}\left[\mathbf{E}\left[\sum_{i=1}^N X_i \Big| N\right]\right] \ &= \mathbf{E}\left[\sum_{i=1}^N \mathbf{E}[X_i \mid N]\right] \ &= \mathbf{E}\left[\sum_{i=1}^N \mathbf{E}[X]\right] \ &= \mathbf{E}[X] \cdot \mathbf{E}[N] = 100. \end{aligned} $$

and, $$ \begin{aligned} \text{var}(T) &= \mathbf{E}[\text{var}(T \mid N)] + \text{var}(\mathbf{E}[T \mid N]) \ &= \mathbf{E}\left[\text{var}\left(\sum_{i=1}^N X_i \Big| N\right)\right] + \text{var}\left(\mathbf{E}\left[\sum_{i=1}^N X_i \Big| N\right]\right) \ &= \mathbf{E}[N \text{var}(X)] + \text{var}(N \mathbf{E}[X]) \ &= (\text{var}(X)) \mathbf{E}[N] + (\mathbf{E}[X])^2 \text{var}(N) \ &= 16 \cdot 10 + 100 \cdot 16 = 1760. \end{aligned} $$

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