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

Solutions — Derived Distributions, Convolutions, and Covariance¶

Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Fall 2010)

Problem Set 11 Solutions¶

Check book solutions.
(a) To find the MAP estimate, we need to find the value $x$ that maximizes the conditional density $f_{X|Y}(x \mid y)$ by taking its derivative and setting it to 0.

\[f_{X|Y}(x \mid y) = \frac{p_{Y|X}(y \mid x) \cdot f_X(x)}{p_Y(y)}$$ $$= \frac{e^{-x}x^y}{y!} \cdot \mu e^{-\mu x} \cdot \frac{1}{p_Y(y)}$$ $$= \frac{\mu}{y!p_Y(y)} \cdot e^{-(\mu+1)x} x^y\]

\[\frac{d}{dx} f_{X|Y}(x \mid y) = \frac{d}{dx} \left( \frac{\mu}{y!p_Y(y)} \cdot e^{-(\mu+1)x} x^y \right)$$ $$= \frac{\mu}{y!p_Y(y)} x^{y-1} e^{-(\mu+1)x} (y - x(\mu+1))\]

Since the only factor that depends on $x$ which can take on the value 0 is $(y - x(\mu + 1))$, the maximum is achieved at $$\hat{x}_{\text{MAP}}(y) = \frac{y}{1 + \mu}$$

It is easy to check that this value is indeed maximum (the first derivative changes from positive to negative at this value).

(b) i. To show the given identity, we need to use Bayes' rule. We first compute the denominator, $p_Y(y)$

\[p_Y(y) = \int_0^\infty \frac{e^{-x}x^y}{y!} \mu e^{-\mu x} \, dx$$ $$= \frac{\mu}{y!(1+\mu)^{y+1}} \int_0^\infty (1+\mu)^{y+1} x^y e^{-(1+\mu)x} \, dx$$ $$= \frac{\mu}{(1+\mu)^{y+1}}\]

Then, we can substitute into the equation we had derived in part (a)

\[f_{X|Y}(x \mid y) = \frac{\mu}{y!p_Y(y)} x^y e^{-(\mu+1)x}$$ $$= \frac{\mu}{y!} \frac{(1+\mu)^{y+1}}{\mu} x^y e^{-(\mu+1)x}$$ $$= \frac{(1+\mu)^{y+1}}{y!} x^y e^{-(\mu+1)x}\]

Thus, $\lambda = 1 + \mu$.

ii. We first manipulate $x f_{X|Y}(x \mid y)$:

\[x f_{X|Y}(x \mid y) = \frac{(1+\mu)^{y+1}}{y!} x^{y+1} e^{-(\mu+1)x}$$ $$= \frac{y+1}{1+\mu} \frac{(1+\mu)^{y+2}}{(y+1)!} x^{y+1} e^{-(\mu+1)x}$$ $$= \frac{y+1}{1+\mu} f_{X|Y}(x \mid y+1)\]

Now we can find the conditional expectation estimator:

\[\hat{x}_{\text{CE}}(y) = E[X \mid Y = y] = \int_0^\infty x f_{X|Y}(x \mid y) \, dx$$ $$= \int_0^\infty \frac{y+1}{1+\mu} f_{X|Y}(x \mid y+1) \, dx = \frac{y+1}{1+\mu}\]

(c) The conditional expectation estimator is always higher than the MAP estimator by $\frac{1}{1+\mu}$.

(a) The likelihood function is $$\prod_{i=1}^k P_{T_i}(T_i = t_i \mid Q = q) = q^k (1-q)^{\sum_{i=1}^k t_i - k}.$$

To maximize the above probability we set its derivative with respect to $q$ to zero

\[k q^{k-1}(1-q)^{\sum_i^k t_i - k} - \left(\sum_i^k t_i - k\right) q^k (1-q)^{\sum_i^k t_i - k - 1} = 0,\]

or equivalently

\[k(1-q) - \left(\sum_i^k t_i - k\right) q = 0,\]

which yields $\hat{Q}_k = \frac{k}{\sum_{i=1}^k t_i}$. This is not different from the MAP estimate found before. Since the MAP estimate is calculated using a uniform prior, the likelihood function is a 'scaled' version of posterior probability and they can be maximized at the same value of $q$.

(b) Since $\frac{1}{\hat{Q}_k} = \frac{\sum_{i=1}^k T_i}{k}$, and that each $T_i$ is independent identically distributed, it follows that $\frac{1}{\hat{Q}_k}$ is actually a sample mean estimator. The weak law of large numbers says that, when the number of samples increases to infinity, the sample mean estimator converges to the actual mean, which is $\frac{1}{q^*}$ in this case. So we can write the limit of probability as

\[\lim_{k \to \infty} \mathbf{P}\left(\left| \frac{1}{\hat{Q}_k} - \frac{1}{q^*} \right| > \epsilon\right) = \lim_{k \to \infty} \mathbf{P}\left(\left| \frac{\sum_{i=1}^k T_i}{k} - E[T_1] \right| > \epsilon\right) = 0.\]

(c) Chebyshev inequality states that $$\mathbf{P}\left(\left| \frac{\sum_{i=1}^k T_i}{k} - E[T_1] \right| \ge \epsilon\right) \le \frac{\text{var}(T_1)}{k\epsilon^2}.$$

So we have

\[\mathbf{P}\left(\left| \frac{1}{\hat{Q}_k} - \frac{1}{q^*} \right| \le 0.1\right) = \mathbf{P}\left(\left| \frac{\sum_{i=1}^k T_i}{k} - \frac{1}{q^*} \right| \le 0.1\right)$$ $$= 1 - \mathbf{P}\left(\left| \frac{\sum_{i=1}^k T_i}{k} - E[T_1] \right| \ge 0.1\right) \ge 1 - \frac{\text{var}(T_1)}{k * 0.1^2}\]

To ensure the above probability to be greater than 0.95, we need that

\[1 - \frac{\text{var}(T_1)}{k * 0.1^2} = 1 - \frac{\frac{1-q}{q^2}}{k * 0.1^2} \ge 0.95,\]

or

\[k \ge 2000\text{var}(T_1) = 2000 \frac{1-q}{q^2}\]

The number of observations $k$ needed depends on the variance of $T_1$. For $q$ close to 1, the variance is close to 0, and the required number of observations is very small (close to 0). For $q = 1/2$, the variance is maximum ($\text{var}(T_1) = 2$), and we require $k = 4000$. Thus, to guarantee the required accuracy and confidence for all $q$, we need that,

\[k \ge 4000.\]

(a) Normalization of the distribution requires:

\[1 = \sum_{k=0}^\infty p_K(k; \theta) = \sum_{k=0}^\infty \frac{e^{-\frac{k}{\theta}}}{Z(\theta)} = \frac{1}{Z(\theta)} \sum_{k=0}^\infty e^{-\frac{k}{\theta}} = \frac{1}{Z(\theta) \cdot (1 - e^{-\frac{1}{\theta}})},\]

so $Z(\theta) = \frac{1}{1 - e^{-\frac{1}{\theta}}}$.

(b) Rewriting $p_K(k; \theta)$ as:

\[p_K(k; \theta) = \left(e^{-\frac{1}{\theta}}\right)^k \left(1 - e^{-\frac{1}{\theta}}\right), \quad k = 0, 1, \ldots\]

the probability distribution for the photon number is a geometric probability distribution with probability of success $p = 1 - e^{-\frac{1}{\theta}}$, and it is shifted with 1 to the left since it starts with $k = 0$. Therefore the photon number expectation value is

\[\mu_K = \frac{1}{p} - 1 = \frac{1}{1 - e^{-\frac{1}{\theta}}} - 1 = \frac{1}{e^{\frac{1}{\theta}} - 1}\]

and its variance is

\[\sigma_K^2 = \frac{1-p}{p^2} = \frac{e^{-\frac{1}{\theta}}}{(1-e^{-\frac{1}{\theta}})^2} = \mu_K^2 + \mu_K.\]

(c) The joint probability distribution for the $k_i$ is

\[p_K(k_1, \ldots, k_n; \theta) = \frac{1}{Z(\theta)^n} \prod_{i=1}^n e^{-k_i/\theta} = \frac{1}{Z(\theta)^n} e^{-\frac{1}{\theta} \sum_{i=1}^n k_i}.\]

The log likelihood is $-n \log Z(\theta) - 1/\theta \sum_{i=1}^n k_i$. We find the maxima of the log likelihood by setting the derivative with respect to the parameter $\theta$ to zero:

\[\frac{d}{d\theta} \log p_K(k_1, \ldots, k_n; \theta) = -n \cdot \frac{e^{-\frac{1}{\theta}}}{\theta^2(1 - e^{-\frac{1}{\theta}})} + \frac{1}{\theta^2} \sum_{i=1}^n k_i = 0\]

or

\[\frac{1}{e^{\frac{1}{\theta}} - 1} = \frac{1}{n} \sum_{i=1}^n k_i = s_n.\]

For a hot body, $\theta \gg 1$ and $\frac{1}{e^{\frac{1}{\theta}} - 1} \approx \theta$, we obtain

\[\theta \approx \frac{1}{n} \sum_{i=1}^n k_i = s_n.\]

Thus the maximum likelihood estimator $\hat{\Theta}_n$ for the temperature is given in this limit by the sample mean of the photon number

\[\hat{\Theta}_n = \frac{1}{n} \sum_{i=1}^n K_i.\]

(d) According to the central limit theorem, the sample mean approaches for large $n$ a Gaussian distribution with standard deviation our root mean square error

\[\sigma_{\hat{\Theta}_n} = \frac{\sigma_K}{\sqrt{n}}.\]

To allow only for 1% relative root mean square error in the temperature, we need $\frac{\sigma_K}{\sqrt{n}} < 0.01\mu_K$. With $\sigma_K^2 = \mu_K^2 + \mu_K$ it follows that

\[\sqrt{n} > \frac{\sigma_K}{0.01\mu_K} = 100 \frac{\sqrt{\mu_K^2 + \mu_K}}{\mu_K} = 100 \sqrt{1 + \frac{1}{\mu_K}}.\]

In general, for large temperatures, i.e. large mean photon numbers $\mu_K \gg 1$, we need about 10,000 samples.

(e) The 95% confidence interval for the temperature estimate for the situation in part (d), i.e.

\[\sigma_{\hat{\Theta}_n} = \frac{\sigma_K}{\sqrt{n}} = 0.01\mu_K,\]

is

\[[\hat{K} - 1.96\sigma_{\hat{K}}, \hat{K} + 1.96\sigma_{\hat{K}}] = [\hat{K} - 0.0196\mu_K, \hat{K} + 0.0196\mu_K].\]

(a) The sample mean estimator $\hat{\Theta}_n = \frac{W_1 + \dots + W_n}{n}$ in this case is

\[\hat{\Theta}_{1000} = \frac{2340}{1000} = 2.34.\]

From the standard normal table, we have $\Phi(1.96) = 0.975$, so we obtain

\[\mathbf{P}\left( \frac{|\hat{\Theta}_{1000} - \mu|}{\sqrt{\text{var}(W_i)/1000}} \le 1.96 \right) \approx 0.95.\]

Because the variance is less that 4, we have

\[\mathbf{P}\left( \hat{\Theta}_{1000} - \mu \le 1.96\sqrt{\text{var}(W_i)/1000} \right) \le \mathbf{P}\left( \hat{\Theta}_{1000} - \mu \le 1.96\sqrt{4/1000} \right),\]

and letting the right-hand side of the above equation $\approx 0.95$ gives a 95% confidence, i.e.,

\[\left[\hat{\Theta}_{1000} - 1.96\sqrt{4/1000}, \hat{\Theta}_{1000} + 1.96\sqrt{4/1000}\right] = \left[\hat{\Theta}_{1000} - 0.124, \hat{\Theta}_{1000} + 0.124\right] = [2.216, 2.464]\]

(b) The likelihood function is

\[f_W(w; \theta) = \prod_{i=1}^n f_{W_i}(w_i; \theta) = \prod_{i=1}^n \theta e^{-\theta w_i},\]

And the log-likelihood function is

\[\log f_W(w; \theta) = n \log \theta - \theta \sum_{i=1}^n w_i,\]

The derivative with respect to $\theta$ is $\frac{n}{\theta} - \sum_{i=1}^n w_i$, and by setting it to zero, we see that the maximum of $\log f_W(w; \theta)$ over $\theta \ge 0$ is attained at $\hat{\theta}_n = \frac{n}{\sum_{i=1}^n w_i}$. The resulting estimator is

\[\hat{\Theta}_n^{\text{mle}} = \frac{n}{\sum_{i=1}^n W_i}.\]

In this case,

\[\hat{\Theta}_n^{\text{mle}} = \frac{1000}{2340} = 0.4274.\]

(a) Using the regression formulas of Section 9.2, we have

\[\hat{\theta}_1 = \frac{\sum_{i=1}^5 (x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^5 (x_i - \bar{x})^2}, \quad \hat{\theta}_0 = \bar{y} - \hat{\theta}_1 \bar{x},\]

where

\[\bar{x} = \frac{1}{5}\sum_{i=1}^5 x_i = 4.94, \quad \bar{y} = \frac{1}{5}\sum_{i=1}^5 y_i = 134.38.\]

The resulting ML estimates are

\[\hat{\theta}_1 = 40.53, \quad \hat{\theta}_0 = -65.86.\]

(b) Using the same procedure as in part (a), we obtain

\[\hat{\theta}_1 = \frac{\sum_{i=1}^5 (x_i^2 - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^5 (x_i^2 - \bar{x})^2}, \quad \hat{\theta}_0 = \bar{y} - \hat{\theta}_1 \bar{x},\]

where

\[\bar{x} = \frac{1}{5}\sum_{i=1}^5 x_i^2 = 33.60, \quad \bar{y} = \frac{1}{5}\sum_{i=1}^5 y_i = 134.38.\]

which for the given data yields

\[\hat{\theta}_1 = 4.09, \quad \hat{\theta}_0 = -3.07.\]

Figure 1 shows the data points $(x_i, y_i)$, $i = 1, \ldots, 5$, the estimated linear model

\[y = 40.53x - 65.86,\]

and the estimated quadratic model

\[y = 4.09x^2 - 3.07.\]

Regression Plot

Figure 1: Regression Plot

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