1. **Problem:** (a) Given a Poisson random variable $X$ with $P\{X=0\} = e^{-\mu}$, find $E[X]$. 2. **Formula and explanation:** For a Poisson distribution with parameter $\lambda$, the probability mass function is: $$P\{X = k\} = \frac{\lambda^k e^{-\lambda}}{k!}$$ and the expected value is: $$E[X] = \lambda$$ 3. **Solution:** Given $P\{X=0\} = e^{-\mu}$, comparing with the Poisson pmf for $k=0$: $$P\{X=0\} = e^{-\lambda} = e^{-\mu} \implies \lambda = \mu$$ Therefore, $$E[X] = \mu$$ --- (c) **Problem:** $X$ is uniformly distributed over $(1,2)$. Find $z$ such that $$P\{X > z + \mu_x\} = \frac{1}{2}$$ where $\mu_x$ is the mean of $X$. **Solution:** 1. Mean of uniform $(a,b)$ is $\mu_x = \frac{a+b}{2} = \frac{1+2}{2} = 1.5$. 2. We want $P\{X > z + 1.5\} = 0.5$. 3. Since $X \sim U(1,2)$, the CDF is $F_X(x) = \frac{x-1}{1}$ for $1 \le x \le 2$. 4. Then, $$P\{X > t\} = 1 - F_X(t) = 1 - (t-1) = 2 - t$$ 5. Set $t = z + 1.5$: $$2 - (z + 1.5) = 0.5 \implies z + 1.5 = 1.5 \implies z = 0$$ --- (d) **Problem:** $X \sim N(2,1)$, find $P\{|X - 2| < 1\}$. **Solution:** 1. Standardize: $$P\{|X - 2| < 1\} = P(1 < X < 3) = P\left(\frac{1-2}{1} < Z < \frac{3-2}{1}\right) = P(-1 < Z < 1)$$ 2. Using standard normal CDF $\Phi$: $$= \Phi(1) - \Phi(-1) = \Phi(1) - (1 - \Phi(1)) = 2\Phi(1) - 1$$ 3. Numerically, $\Phi(1) \approx 0.8413$, so $$P = 2 \times 0.8413 - 1 = 0.6826$$ --- (e) **Problem:** $X \sim Binomial(n,p)$ with $E[X] = 5$ and $Var[X] = 4$. Find $n$ and $p$. **Solution:** 1. Recall: $$E[X] = np = 5$$ $$Var[X] = np(1-p) = 4$$ 2. From $np=5$, $p = \frac{5}{n}$. 3. Substitute into variance: $$n \times \frac{5}{n} \times \left(1 - \frac{5}{n}\right) = 4 \implies 5 \left(1 - \frac{5}{n}\right) = 4$$ 4. Simplify: $$5 - \frac{25}{n} = 4 \implies \frac{25}{n} = 1 \implies n = 25$$ 5. Then, $$p = \frac{5}{25} = 0.2$$ --- (f) **Problem:** If $E[X] = 10$ and $\sigma_X = 3$, can $X$ have a negative binomial distribution? **Explanation:** For negative binomial with parameters $r, p$: $$E[X] = \frac{r(1-p)}{p}, \quad Var[X] = \frac{r(1-p)}{p^2}$$ Since $Var[X] > E[X]$ always (because $Var[X] = E[X]/p > E[X]$), Check if $Var[X] = 9$ is greater than $E[X] = 10$. Here, $Var[X] = 9 < 10$, which contradicts the property. **Answer:** No, $X$ cannot have a negative binomial distribution with these moments. --- (g) **Problem:** $X$ has negative exponential distribution with mean 2. Find $$P\{X < 1 | X < 2\}$$ **Solution:** 1. Exponential with mean $2$ has parameter: $$\lambda = \frac{1}{2}$$ 2. Conditional probability: $$P\{X < 1 | X < 2\} = \frac{P\{X < 1\}}{P\{X < 2\}} = \frac{1 - e^{-\lambda \times 1}}{1 - e^{-\lambda \times 2}} = \frac{1 - e^{-1/2}}{1 - e^{-1}}$$ --- (h) **Problem:** Name three distributions for which $$P\{X \leq \mu_X\} = \frac{1}{2}$$ **Answer:** 1. Symmetric distributions about the mean, e.g., Normal distribution. 2. Uniform distribution on symmetric intervals. 3. Any continuous distribution where the mean equals the median, e.g., Laplace distribution. --- (i) **Problem:** $X \sim Binomial(n=100, p=0.1)$. Evaluate $$P\{X \leq \mu_X - 3\sigma_X\}$$ **Solution:** 1. Compute mean and std dev: $$\mu_X = np = 10$$ $$\sigma_X = \sqrt{np(1-p)} = \sqrt{100 \times 0.1 \times 0.9} = 3$$ 2. Calculate: $$\mu_X - 3\sigma_X = 10 - 3 \times 3 = 1$$ 3. So, $$P\{X \leq 1\} = P(X=0) + P(X=1)$$ 4. Using binomial pmf: $$P(X=0) = (0.9)^{100} \approx 2.656 \times 10^{-5}$$ $$P(X=1) = 100 \times 0.1 \times (0.9)^{99} \approx 2.95 \times 10^{-4}$$ 5. Sum: $$\approx 3.22 \times 10^{-4}$$ --- (j) **Problem:** If $X$ has Poisson distribution and $P\{X=0\} = p = 1$, find $E[X]$. **Solution:** Since $P\{X=0\} = e^{-\lambda} = 1$, then $\lambda = 0$. Therefore, $$E[X] = \lambda = 0$$ --- (k) **Problem:** For $X \sim Binomial(n,p)$ with fixed $n$, find $p$ maximizing $Var[X]$. **Solution:** Variance: $$Var[X] = np(1-p)$$ Maximize w.r.t $p$: $$\frac{d}{dp} np(1-p) = n(1 - 2p) = 0 \implies p = \frac{1}{2}$$ --- (l) **Problem:** $X$ has negative exponential distribution with parameter $\lambda$. If $$P\{X \leq 1\} = P\{X > 1\}$$ find $Var[X]$. **Solution:** 1. Given: $$1 - e^{-\lambda} = e^{-\lambda} \implies 1 = 2 e^{-\lambda} \implies e^{-\lambda} = \frac{1}{2}$$ 2. So, $$\lambda = \ln 2$$ 3. Variance: $$Var[X] = \frac{1}{\lambda^2} = \frac{1}{(\ln 2)^2}$$ --- (m) **Problem:** $X$ is uniform with mean 1 and variance $\frac{3}{4}$. Find $P\{X < 0\}$. **Solution:** 1. For uniform $(a,b)$: $$\mu = \frac{a+b}{2} = 1$$ $$\sigma^2 = \frac{(b-a)^2}{12} = \frac{3}{4}$$ 2. From variance: $$(b-a)^2 = 9 \implies b - a = 3$$ 3. From mean: $$a + b = 2$$ 4. Solve system: $$b - a = 3$$ $$a + b = 2$$ Add: $$2b = 5 \implies b = 2.5$$ Then, $$a = 2 - b = 2 - 2.5 = -0.5$$ 5. Probability: $$P\{X < 0\} = \frac{0 - a}{b - a} = \frac{0 - (-0.5)}{3} = \frac{0.5}{3} = \frac{1}{6}$$ --- (n) **Problem:** If $X$ has beta distribution, can $E[1/X] = 1$? **Answer:** Yes, if the beta distribution is defined on $(0,1)$ with parameters $\alpha, \beta$ such that $\alpha > 1$ (to ensure $E[1/X]$ exists), it is possible to have $E[1/X] = 1$ by choosing parameters accordingly. --- (o) **Problem:** Can $X$ have the same distribution as $-X$? If so, when? **Answer:** Yes, if $X$ is symmetric about zero, i.e., its distribution satisfies: $$X \stackrel{d}{=} -X$$ Examples include symmetric distributions like Normal(0, $\sigma^2$), uniform on $[-a,a]$, etc. --- (p) **Problem:** If $X$ has moment generating function (MGF): $$M_X(t) = \exp(e^t - 1)$$ Find $E[X]$. **Solution:** 1. Recall: $$E[X] = M_X'(0)$$ 2. Differentiate: $$M_X'(t) = M_X(t) \times e^t$$ 3. Evaluate at $t=0$: $$M_X(0) = \exp(e^0 - 1) = \exp(1 - 1) = 1$$ $$M_X'(0) = 1 \times e^0 = 1$$ Therefore, $$E[X] = 1$$ --- 2 (a) **Problem:** Find the mode of the beta distribution with parameters $\alpha, \beta > 1$. **Solution:** $$\text{Mode} = \frac{\alpha - 1}{\alpha + \beta - 2}$$ (b) **Problem:** Find the mode of the gamma distribution with shape $k > 1$ and scale $\theta$. **Solution:** $$\text{Mode} = (k - 1) \theta$$ --- 3 (a) Parametric family with mean $\geq$ variance: Binomial distribution. (b) Mean $=$ variance: Poisson distribution. (c) Mean $\leq$ variance: Negative binomial distribution. (d) Mean can be less than, equal to, or greater than variance: Normal distribution. --- 4 (a) $X \sim N(2,2)$, express $$P\{|X - 1| \leq 2\}$$ Standardize: $$P(1 - 2 \leq X \leq 1 + 2) = P(-1 \leq X \leq 3)$$ $$= P\left(\frac{-1 - 2}{\sqrt{2}} \leq Z \leq \frac{3 - 2}{\sqrt{2}}\right) = P\left(-\frac{3}{\sqrt{2}} \leq Z \leq \frac{1}{\sqrt{2}}\right)$$ $$= \Phi\left(\frac{1}{\sqrt{2}}\right) - \Phi\left(-\frac{3}{\sqrt{2}}\right)$$ (b) $X \sim N(\mu, \sigma^2 = \mu^2)$, $\mu > 0$, find $$P\{X < -\mu | X < \mu\}$$ Standardize: $$P\{X < -\mu | X < \mu\} = \frac{P\{X < -\mu\}}{P\{X < \mu\}} = \frac{\Phi\left(\frac{-\mu - \mu}{\mu}\right)}{\Phi\left(\frac{\mu - \mu}{\mu}\right)} = \frac{\Phi(-2)}{\Phi(0)} = \Phi(-2)$$ (c) Find $h(\mu)$ so that $P\{X \leq 0\}$ does not depend on $\mu$ for $\mu > 0$. Set: $$Z = \frac{0 - \mu}{\sqrt{h(\mu)}}$$ For $P\{X \leq 0\} = \Phi(Z)$ to be constant, $Z$ must be constant. So, $$\frac{-\mu}{\sqrt{h(\mu)}} = c \implies h(\mu) = \frac{\mu^2}{c^2}$$ Thus, $$\sigma^2 = h(\mu) = k \mu^2$$ for some constant $k$. --- 5 (a) Median of exponential $f_X(x) = \lambda e^{-\lambda x} I_{(0,\infty)}(x)$: Median $m$ satisfies: $$P\{X \leq m\} = 0.5 \implies 1 - e^{-\lambda m} = 0.5 \implies m = \frac{\ln 2}{\lambda}$$ (b) Median of uniform $(\theta_1, \theta_2)$: $$m = \frac{\theta_1 + \theta_2}{2}$$ (c) Binomial $n=4, p=0.5$ median is 2 (since distribution is symmetric). (d) Binomial $n=5, p=0.5$ median is 2 or 3 (both are medians). (e) Binomial $n=2, p=0.9$ median is 2 (since $P(X=2)$ is highest). --- 6 **Problem:** Contractor's low bid $Y \sim U(3C, 2C)$, profit is zero if contractor loses, else profit is bid minus $C$. Find bid maximizing expected profit. **Solution:** 1. Let bid be $b$. 2. Probability contractor wins: $$P(b < Y) = P(Y > b) = \frac{2C - b}{2C - 3C} = \frac{2C - b}{-C}$$ Since $2C - 3C = -C$, interval is reversed, so correct interval is $(2C, 3C)$ or check carefully. Actually, interval $(3C, 2C)$ is invalid since $3C > 2C$; likely typo, assume $(2C, 3C)$. Then, $$P(Y > b) = \frac{3C - b}{3C - 2C} = 3C - b$$ Divided by $C$: $$= \frac{3C - b}{C} = 3 - \frac{b}{C}$$ 3. Expected profit: $$E[\text{profit}] = (b - C) \times P(\text{win}) = (b - C) \left(3 - \frac{b}{C}\right)$$ 4. Maximize w.r.t $b$: $$\frac{d}{db} E = (3 - \frac{b}{C}) + (b - C)(-\frac{1}{C}) = 3 - \frac{b}{C} - \frac{b}{C} + 1 = 4 - \frac{2b}{C}$$ Set derivative zero: $$4 - \frac{2b}{C} = 0 \implies b = 2C$$ --- 7 (a) $X \sim Poisson(4)$, find stock for 95% certainty over 25 days. Total mean: $$\lambda = 4 \times 25 = 100$$ Use normal approximation: $$Z_{0.95} = 1.645$$ Stock needed: $$k = \lambda + Z \sqrt{\lambda} = 100 + 1.645 \times 10 = 116.45 \approx 117$$ (b) Expected days with zero sales: $$P(X=0) = e^{-4}$$ Expected days: $$25 \times e^{-4} \approx 25 \times 0.0183 = 0.4575$$ --- 8 (a) $X \sim Binomial(n,p)$, distribution is binomial with pmf: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$