Standard Probability Distributions Questions and Answers

Answer: B

X is binomial with mean 6.

The probability generating function of a binomial \((n, p)\) variable is \((q + pt)^{n}\) with \(q = 1 - p\). Matching \((0.4 + 0.6t)^{10}\) gives \(n = 10\), \(p = 0.6\), \(q = 0.4\), so

\[ \begin{aligned} E(X) &= np = 10 \times 0.6 = 6 \\ P_{X}'(1) &= 10 \times 0.6 \times 1^{9} = 6 \end{aligned} \]

Mean 4 is \(nq\), the expected number of failures, which comes from reading 0.4 as the success probability. A negative binomial pgf has the form \(p^{r}(1 - qt)^{-r}\), with a negative exponent, so neither negative binomial option fits a polynomial pgf.

Answer: D

The true statement is the mean deviation from the mean is the same as the mean deviation from the median.

A normal distribution is symmetric, so its mean, median and mode coincide at \(\mu\). Mean deviation about the mean and about the median are then the same quantity, \(E|X - \mu| = \sigma\sqrt{2/\pi} \approx 0.8\sigma\).

The other statements have the standard values swapped or wrong. Symmetry makes the skewness \(\beta_{1} = \mu_{3}^{2}/\mu_{2}^{3} = 0\), not 3. The kurtosis \(\beta_{2} = \mu_{4}/\mu_{2}^{2} = 3\) (the normal is mesokurtic); it is the excess kurtosis \(\beta_{2} - 3\) that is 0. And \(\mu \pm \sigma\) holds about 68.27% of the area; 95% needs about \(\mu \pm 1.96\sigma\).

Answer: C

\(E\{\max(X, Y)\} = \frac{1}{\sqrt{\pi}}\).

Use \(\max(X, Y) = \frac{X + Y}{2} + \frac{|X - Y|}{2}\). The first term has mean 0. The difference \(X - Y\) is \(N(0, 2)\), and for \(Z \sim N(0, \sigma^{2})\), \(E|Z| = \sigma\sqrt{2/\pi}\). So

\[ \begin{aligned} E|X - Y| &= \sqrt{2} \cdot \sqrt{\frac{2}{\pi}} = \frac{2}{\sqrt{\pi}} \\ E \max(X, Y) &= \frac{1}{2} \cdot \frac{2}{\sqrt{\pi}} = \frac{1}{\sqrt{\pi}} \approx 0.564 \end{aligned} \]

The value \(1/\sqrt{2\pi}\) is what you get by treating \(X - Y\) as having variance 1 instead of 2. The answers 1 and 3 are far too large: the maximum of two standard normals has a mean well below one standard deviation.

Answer: A

The false statement is that \(X_{1}\) and \(X_{2}\) are independent binomial variables.

Each margin of a multinomial is binomial, so \(X_{1} \sim B(n, p_{1})\) and \(X_{2} \sim B(n, p_{2})\), but they are not independent: they compete for the same \(n\) trials, and \(X_{1} + X_{2} \le n\). Their covariance is \(-np_{1}p_{2} \neq 0\).

The other statements are true. \(\text{Cov}(X_{i}, X_{j}) = -np_{i}p_{j}\) for \(i \neq j\), so \(\text{Cov}(X_{1}, X_{3}) = -np_{1}p_{3}\). The margin \(X_{3}\) is \(B(n, p_{3})\), with variance \(np_{3}(1 - p_{3})\). And merging categories 1 and 2 gives a binomial with success probability \(p_{1} + p_{2} = 1 - p_{3}\), so \(X_{1} + X_{2} \sim B(n, 1 - p_{3})\).

Answer: D

\(E(S_{1} \mid S_{2} = 4) = 2\).

For two i.i.d. exponential variables, given their sum \(S_{2} = s\), the first one \(X_{1}\) is uniformly distributed on [0, s]: the joint density \(\lambda^{2}e^{-\lambda(x_{1} + x_{2})}\) is constant along the line \(x_{1} + x_{2} = s\). Alternatively, by symmetry \(E(X_{1} \mid S_{2}) = E(X_{2} \mid S_{2})\), and the two add to \(S_{2}\).

\[ \begin{aligned} E(S_{1} \mid S_{2} = 4) &= E(X_{1} \mid X_{1} + X_{2} = 4) \\ &= \frac{4}{2} = 2 \end{aligned} \]

The answer does not involve \(\lambda\) at all, which rules out \(\lambda/2\), \(2\lambda\) and \(2/\lambda\). The last is \(2E(X_{1})\), the unconditional mean of \(S_{2}\), not a conditional mean of \(S_{1}\).

Answer: C

The false statement is the distribution of the nth order statistic is exponential.

For n independent exponential variables with rate \(\lambda\), the largest one has

\[ P(X_{(n)} \le x) = (1 - e^{-\lambda x})^{n} \]

which is not an exponential distribution function for \(n \ge 2\).

The other statements are true. The exponential is memoryless, \(P(X \gt s + t \mid X \gt s) = P(X \gt t)\); its hazard rate \(f(x)/(1 - F(x)) = \lambda\) is constant; and the minimum satisfies \(P(X_{(1)} \gt x) = e^{-n\lambda x}\), so it is exponential with rate \(n\lambda\).

Answer: B

V(X) < V(Y).

Equal means give \(np = \lambda\). Then

\[ V(X) = np(1 - p) = \lambda(1 - p) \lt \lambda = V(Y) \]

because \(0 \lt p \lt 1\) makes \(1 - p \lt 1\). So the binomial variance is always strictly smaller than the Poisson variance with the same mean, never equal or larger.

Answer: D

max(X, Y) has the Beta(2, 1) distribution.

For \(0 \lt z \lt 1\),

\[ P(\max(X, Y) \le z) = P(X \le z)P(Y \le z) = z^{2} \]

so the density is \(2z\) on (0, 1). The Beta(a, b) density is proportional to \(z^{a-1}(1 - z)^{b-1}\), and \(2z\) matches \(a = 2\), \(b = 1\).

Beta(1, 2), with density \(2(1 - z)\), is the distribution of the minimum. Beta(1, 1) is the uniform, and Beta(2, 2) has density \(6z(1 - z)\).

Answer: D

The jump at 2 is 6/27.

For the geometric distribution \(P(X = k) = q^{k-1}p\) on {1, 2, ...}, the variance is \(q/p^{2}\). Setting it equal to 6:

\[ \begin{aligned} \frac{1 - p}{p^{2}} &= 6 \\ 6p^{2} + p - 1 &= 0 \\ (3p - 1)(2p + 1) &= 0 \end{aligned} \]

so \(p = \frac{1}{3}\). The jump of the distribution function at 2 is \(P(X = 2) = qp = \frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9} = \frac{6}{27}\).

8/27 would be \(P(X \gt 3) = q^{3}\), and 2/27 has no matching event, so the value can be determined and is 6/27.

Answer: D

The conditional probability is \(\binom{n}{j}\binom{n}{k-j} / \binom{2n}{k}\), a hypergeometric distribution (printed as option D).

X and Y are independent Binomial(n, p), so \(X + Y\) is Binomial(2n, p). Then

\[ \begin{aligned} &P(X = j \mid X + Y = k) \\ &= \frac{P(X = j)\,P(Y = k - j)}{P(X + Y = k)} \\ &= \frac{\binom{n}{j}p^{j}q^{n-j} \cdot \binom{n}{k-j}p^{k-j}q^{n-k+j}}{\binom{2n}{k}p^{k}q^{2n-k}} \end{aligned} \]

All the powers of p and q cancel, leaving \(\binom{n}{j}\binom{n}{k-j}/\binom{2n}{k}\), free of p: given the total, it is like drawing k items from 2n of which n are 'X' items. The option prints the denominator as \(\binom{2n}{n}\); it is the only option with the hypergeometric numerator.

Option B still depends on p, which the conditioning removes. Option C, Binomial(k, ½), is the answer for two i.i.d. Poisson variables, not binomial ones, and \(j/k\) is not a probability distribution in j.