Standard Probability Distributions Questions and Answers

Answer: A

The characteristic function is \(\exp\{it\mu - \sigma|t|\}\).

The standard Cauchy density \(1/(\pi(1 + x^{2}))\) has characteristic function \(e^{-|t|}\). For \(X = \mu + \sigma Z\), \(\phi_{X}(t) = e^{it\mu}\phi_{Z}(\sigma t) = e^{it\mu - \sigma|t|}\).

The absolute value matters: \(e^{it\mu - \sigma t}\) would have modulus \(e^{-\sigma t} \gt 1\) for \(t \lt 0\), impossible for a characteristic function. \(\exp\{it\mu - \sigma^{2}t^{2}/2\}\) is the normal characteristic function, and the last option has no meaning. The kink at \(t = 0\) reflects that the Cauchy distribution has no mean.

Answer: A

The answer is Binomial \((x + y;\ \lambda_{1}/(\lambda_{1} + \lambda_{2}))\), and Poisson with parameter \(\lambda_{2}\) shifted to start at X.

Given \(X + Y = s\), for \(0 \le k \le s\):

\[ \begin{aligned} &P(X = k | X + Y = s) \\ &= \frac{P(X = k)\,P(Y = s - k)}{P(X + Y = s)} \\ &= \binom{s}{k}\frac{\lambda_{1}^{k}\lambda_{2}^{s-k}}{(\lambda_{1} + \lambda_{2})^{s}} \end{aligned} \]

which is \(\binom{s}{k}p^{k}(1 - p)^{s-k}\) with \(p = \lambda_{1}/(\lambda_{1} + \lambda_{2})\). Given X = x, \(X + Y = x + Y\) with Y independent Poisson(\(\lambda_{2}\)), so it is Poisson(\(\lambda_{2}\)) shifted to take the values \(x, x + 1, \ldots\).

The negative binomial, Poisson and geometric descriptions of the first conditional law are all wrong, since given the total the count X cannot exceed \(s\).

Answer: C

\(E[X] = \) 2.

\(\exp[2e^{t} - 2] = \exp[2(e^{t} - 1)]\) is the moment generating function of a Poisson distribution with \(\lambda = 2\), whose mean is 2.

Directly, \(E[X] = M'(0)\):

\[ \begin{aligned} M'(t) &= 2e^{t}\exp[2e^{t} - 2] \\ M'(0) &= 2 \cdot 1 \cdot e^{0} = 2 \end{aligned} \]

0 and 1 come from evaluating the exponent rather than differentiating, and the mean is finite because the MGF exists near 0.

Answer: C

The percentage taller than 1.80 m is 40%.

The normal distribution is symmetric about its mean 1.65. The heights 1.50 and 1.80 are equally far from the mean, 0.15 m below and above. So

\[ P(X \gt 1.80) = P(X \lt 1.50) = 0.40 \]

The value of \(\sigma\) is not needed. 60% is the proportion shorter than 1.80 m, and 80% doubles the wrong tail.

Answer: B

Y is Cauchy with location \(\mu/(\mu^{2} + \sigma^{2})\) and scale \(\sigma/(\mu^{2} + \sigma^{2})\).

Change variables with \(x = 1/y\), \(|dx/dy| = 1/y^{2}\):

\[ \begin{aligned} f_{Y}(y) &= \frac{\sigma/\pi}{y^{2}\left(\sigma^{2} + (1/y - \mu)^{2}\right)} \\ &= \frac{\sigma/\pi}{(\mu^{2} + \sigma^{2})y^{2} - 2\mu y + 1} \end{aligned} \]

Completing the square with \(m = \mu/(\mu^{2} + \sigma^{2})\) and \(s = \sigma/(\mu^{2} + \sigma^{2})\), the denominator is \((\mu^{2} + \sigma^{2})[(y - m)^{2} + s^{2}]\), so \(f_{Y}(y) = \dfrac{s/\pi}{(y - m)^{2} + s^{2}}\), the Cauchy density with location \(m\) and scale \(s\).

With \(\mu = 0, \sigma = 1\) this is the familiar fact that the reciprocal of a standard Cauchy variable is standard Cauchy. The other options keep \(\mu\) or \(\sigma\) unchanged, which fails as soon as \(\mu \ne 0\), or deny that Y is Cauchy.

Answer: B

The probability is \(1 - \exp(-3)\).

Errors occur at an average rate of 1 per 10 seconds, so in 30 seconds the number of errors is Poisson with mean \(\lambda = 3\). Then

\[ P(N \ge 1) = 1 - P(N = 0) = 1 - e^{-3} \]

\(1 - e^{-1}\) uses the rate for 10 seconds instead of 30. \(1 - 4e^{-3}\) is \(P(N \ge 2)\), subtracting both \(P(N = 0) = e^{-3}\) and \(P(N = 1) = 3e^{-3}\), and \(1 - 2e^{-1}\) is the same mistake with \(\lambda = 1\).

Answer: D

The correct statement is \(P[X \ge 5] \gt P[X \ge 6]\).

The mean is 5, so \(P[X \ge 5] = 0.5\). Since 6 lies above the mean, \(P[X \ge 6] = P[Z \ge 1/6] \approx 0.43 \lt 0.5\). In general \(P[X \ge a]\) decreases as \(a\) increases.

The others are false: \(P[X \le 6] \approx 0.57\) is not equal to \(P[X \ge 5] = 0.5\) (A), \(P[X \le 5] = 0.5 \lt P[X \le 6]\) (B), and C reverses the true inequality.

Answer: C

The distribution is the geometric distribution.

Rewrite the m.g.f.:

\[ \begin{aligned} \frac{1}{3 - 2e^{t}} &= \frac{1/3}{1 - \frac{2}{3}e^{t}} \\ &= \sum_{k=0}^{\infty} \frac{1}{3}\left(\frac{2}{3}\right)^{k} e^{tk} \end{aligned} \]

for \(e^{t} \lt 3/2\). This is the m.g.f. of \(P(X = k) = \frac{1}{3}(\frac{2}{3})^{k}\), \(k = 0, 1, 2, \ldots\), a geometric distribution with \(p = 1/3\), whose m.g.f. is \(p/(1 - qe^{t})\).

The exponential has m.g.f. \(\lambda/(\lambda - t)\), with \(t\) not \(e^{t}\); the Cauchy distribution has no m.g.f.; and the logarithmic series m.g.f. involves \(\log(1 - qe^{t})\).

Answer: D

The true statement is the memoryless property \(\bar{F}(t + s) = \bar{F}(t)\bar{F}(s)\).

With mean \(\lambda\), the survival function is \(\bar{F}(x) = e^{-x/\lambda}\), so \(\bar{F}(t + s) = e^{-(t+s)/\lambda} = \bar{F}(t)\bar{F}(s)\) for all \(t, s \ge 0\). The exponential is the only continuous distribution with this property.

The others are wrong. The variance is \(\lambda^{2}\), not \(\lambda\). The m.g.f. \(1/(1 - \lambda t)\) exists only for \(t \lt 1/\lambda\), not for all real \(t\). And \(\int_{0}^{\infty} e^{-\lambda x}\, dx = 1/\lambda\), which is the mean only for the rate parametrisation; here the mean is \(\lambda = \int_{0}^{\infty} e^{-x/\lambda}\, dx\).

Answer: C

Mean and variance are equal for the Poisson distribution.

For a Poisson distribution with parameter \(\lambda\), the probability of k events is \(P(k) = e^{-\lambda}\lambda^{k}/k!\), and both the mean and the variance equal \(\lambda\). This is why it describes counts of rare, random events, such as mutations per generation or colonies per plate.

For a binomial distribution the mean is np and the variance npq, which is smaller than the mean because q is less than 1. A chi-square distribution with n degrees of freedom has mean n and variance 2n. A normal distribution has two independent parameters, the mean and the variance, which need not be equal.