Exam 1

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Write either True or False for each of the following statements. No justification is required.

Let A, B be any sets. Then |A \cap B| = |A| \cdot |B|.

False.

Let A, B be any events. Then \Pr(A \mid B) = \Pr(B \mid A).

False.

Given a joint distribution for random variables X and Y, then it must be the case that \Pr(X = 0, Y = 0) = \Pr(X = 0)\Pr(Y = 0).

False.

Suppose we flip a coin repeatedly until we first see heads and let T be the number of flips. Then \Pr(T = 2) \geq \Pr(T = 4).

True.

If molecules are observed leaving a cell after 1.5 minutes, 1.8 minutes, 2.1 minutes, 2.2 minutes, and 2.4 minutes, then the maximum likelihood estimation for the rate at which molecules leave the cell is 1/2 per minute.

True.

Suppose \L(\theta) is a likelihood function, and \L(2) = 0.04. Then the probability that \theta = 2 is 0.04.

False.

Consider the sample space S = \{1, 2, 3, 4, 5\} with probability distribution given in the table below. Let A = \{1, 2, 3\} and B = \{1, 3, 5\}. Calculate the following probabilities.

x 1 2 3 4 5 \Pr(x) 0.1 0.2 0.3 0.3 0.1

\Pr(A \cup B)

A\cup B = \{ 1, 2, 3, 5 \}, so \Pr(A\cup B) = 0.1 + 0.2 + 0.3 + 0.1 = \boxed{0.7}

\Pr(A \cap B)

A\cap B = \{ 1, 3 \}, so \Pr(A\cap B) = 0.1 + 0.3 = \boxed{0.4}

\Pr(A \mid B)

\Pr(A\mid B) = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{0.4}{0.1 + 0.3 + 0.1} = \frac{0.4}{0.5} = \boxed{0.8}

Consider the random variable X with probability distribution below.

x 1 2 3 4 5 \Pr(x) 0.2 0.3 0.1 0.25 0.15

Find \E(X).

\E(X) \amp = (1)(0.2) + (2)(0.3) + (3)(0.1) + (4)(0.25) + (5)(0.15) = \boxed{2.85}

Find \Var(X).

\E(X^2) \amp = (1)^2(0.2) + (2)^2(0.3) + (3)^2(0.1) + (4)^2(0.25) + (5)^2(0.15) = 10.05 \Var(X) \amp = \E(X^2) - \left(\E(X)\right)^2 = 10.05 - (2.85)^2 = \boxed{1.9275}

A diagnostic test is developed to detect a disease present in 2.1% of the population. For a patient who has the disease, the test will accurately give a positive result 76% of the time. When the patient does not have the disease, the test will accurately give a negative result 98.2% of the time.

For a patient who receives a negative test, what is the probability they do not have the disease?

Let P be the event of receiving a positive test result and D be the event of having the disease. The given information is: \Pr(D) = 0.021, \Pr(P\mid D) = 0.76, and \Pr(P^c \mid D^c) = 0.982. Then, using Bayes' Theorem: \Pr(D^c\mid P^c) \amp = \frac{\Pr(P^c \mid D^c)\Pr(D^c)}{\Pr(P^c \mid D^c)\Pr(D^c) + \Pr(P^c \mid D)\Pr(D)} \amp = \frac{\Pr(P^c \mid D^c)(1 - \Pr(D))}{\Pr(P^c \mid D^c)(1 - \Pr(D)) + (1 - \Pr(P \mid D))\Pr(D)} \amp = \frac{(0.982)(1 - 0.021)}{(0.982)(1 - 0.021) + (1 - 0.76)(0.021)} \amp \approx \boxed{0.995}

In each of the following scenarios, state the maximum likelihood estimation for the unknown parameter indicated.

A radioactive material is observed for 5 hours. 120 particle emissions are seen. \lambda is the hourly rate of particle emissions.

120 emissions per 5 hours is an hourly rate of \boxed{\widehat{\lambda} = 120/5 = 24 \text{ per hour}}

In each of 5 trials, a coin is flipped until heads is seen. The number of flips in each trial is 3, 4, 3, 5, and 6. p is the probability of the coin coming up heads on a flip.

There are 3 + 4 + 3 + 5 + 6 = 21 total flips, 5 of which are heads, so the MLE is \boxed{\widehat{p} = \frac{5}{21}}

Let X be a continuous random variable with c.d.f. \displaystyle{F(x) = \frac{1}{8}(x^3 + 3x + 4)}, where x \in [-1, 1].

Find the p.d.f. f(x) for X.

f(x) = F'(x) = \frac{1}{8}\left(3x^2 + 3\right) = \frac{3}{8}\left(x^2 + 1\right)

Find \E(X).

\E(X) \amp = \int_{-1}^1 x \cdot f(x)\ dx = \int_{-1}^1 x\cdot \frac{3}{8}\left(x^2 + 1\right)\ dx \amp = \frac{3}{8} \int_{-1}^1 x^3 + x\ dx = \frac{3}{8} \left(\frac{x^4}{4} + \frac{x^2}{2}\right)\bigg|_{-1}^1 \amp = \frac{3}{8} \left[\left(\frac{1}{4} + \frac{1}{2}\right) - \left( \frac{1}{4} + \frac{1}{2}\right)\right] = \boxed{0}

Find \Var(X).

\E(X^2) \amp = \int_{-1}^1 x^2 \cdot f(x)\ dx = \int_{-1}^1 x^2\cdot \frac{3}{8}\left(x^2 + 1\right)\ dx \amp = \frac{3}{8} \int_{-1}^1 x^4 + x^2\ dx = \frac{3}{8} \left(\frac{x^5}{5} + \frac{x^3}{3}\right)\bigg|_{-1}^1 \amp = \frac{3}{8} \left[\left(\frac{1}{5} + \frac{1}{3}\right) - \left( \frac{-1}{5} - \frac{1}{3}\right)\right] = \frac{3}{8} \cdot \frac{16}{15} = \frac{2}{5} \Var(X) \amp = \E(X^2) - (\E(X))^2 = \frac{2}{5} - 0^2 = \boxed{\frac{2}{5}}

Consider the joint distribution for X and Y below. Are X and Y independent?

X = 0 X = 1 X = 2 Y = 1 0.1 0.25 0.2 Y = 2 0.2 0.1 0.15

\Pr(X = 0, Y = 1) \amp = 0.1 \Pr(X = 0) \amp = 0.1 + 0.2 = 0.3 \Pr(Y = 1) \amp = 0.1 + 0.25 + 0.2 = 0.55 \Pr(X = 0)\Pr(Y = 1) \amp = (0.3)(0.55) = 0.165 \neq 0.1, so X, Y are not independent.

Suppose a parameter -1/2 \leq \theta \leq 1 has likelihood function \L(\theta) = \theta^2 - \theta^3. Find the maximum likelihood estimation of \theta.

\L is a continuous function and [-1/2, 1] is a closed interval, so we use the CIM. \L'(\theta) \amp = 2\theta - 3\theta^2 \L'(\theta) = 0 \text{ when } 0 \amp = 2\theta - 3\theta^2 0 \amp =\theta (2 - 3\theta) \theta \amp = 0, \frac{2}{3} Then:

\theta -1/2 0 2/3 1 \L(\theta) 3/8 0 4/27 0

3/8 is the largest value, so the MLE is \boxed{\widehat{\theta} = -1/2}.