Probability Theory Questions and Answers

Answer: D

The appropriate statement is P(A) = 0 or P(A) = 1.

F, the intersection of \(\sigma(X_{n}, X_{n+1}, \ldots)\) over all n, is the tail \(\sigma\)-field of the sequence (printed in the question as \(f_{e}\)). Kolmogorov's zero-one law says that when the \(X_{n}\) are independent, every tail event has probability 0 or 1.

Examples of tail events are 'the series \(\sum X_{n}\) converges' and 'the limit superior of \(X_{n}\) exceeds 5'. Each is unaffected by changing finitely many terms, and each is either almost certain or almost impossible.

Saying always 0 or always 1 is too strong, since both values occur for different tail events (A and its complement are both tail events). \(P(A) = P(A^{c})\) would force \(P(A) = 1/2\), which the law excludes.

Answer: C

The density of \(X^{-1}\) is \(1/x^{2} f(1/x)\), that is \(\dfrac{1}{x^{2}} f\left(\dfrac{1}{x}\right)\) for \(x \gt 0\).

Let \(Y = 1/X\). For \(y \gt 0\), \(Y \le y\) exactly when \(X \ge 1/y\), so

\[ F_{Y}(y) = 1 - F_{X}(1/y) \]

Differentiating with the chain rule:

\[ \begin{aligned} f_{Y}(y) &= -f_{X}(1/y) \cdot \left(-\frac{1}{y^{2}}\right) \\ &= \frac{1}{y^{2}} f_{X}(1/y) \end{aligned} \]

The factor \(1/y^{2}\) is the Jacobian \(|dx/dy|\) of the change of variable. \(f(1/x)\) forgets the Jacobian, \(1/x \cdot f(1/x)\) has the wrong power of x (it does not integrate to 1 in general), and \(1/f(x)\) is not a density at all.

Answer: C

\(P(X_{1} \le X_{2}) = \)1/2.

Because \(X_{1}\) and \(X_{2}\) are independent with the same distribution, the pair \((X_{1}, X_{2})\) has the same joint distribution as \((X_{2}, X_{1})\). Hence \(P(X_{1} \lt X_{2}) = P(X_{2} \lt X_{1})\), and

\[ \begin{aligned} &P(X_{1} \lt X_{2}) + P(X_{1} = X_{2}) \\ &\quad + P(X_{1} \gt X_{2}) = 1 \end{aligned} \]

For a continuous distribution function, \(P(X_{1} = X_{2}) = 0\), so each strict inequality has probability 1/2 and \(P(X_{1} \le X_{2}) = 1/2\). This is the reading the question intends. If F had jumps, ties would have positive probability and the answer would be \(\frac{1}{2} + \frac{1}{2}P(X_{1} = X_{2})\), but that is not among the options either.

1/3 and 2/3 have no symmetry basis, and 3/2 exceeds 1, so it is not a probability.

Answer: A

The most appropriate statement is E(X|Y) = 0.

Given \(Y = y\), X is normal with mean 0 and variance \(1/y\), so \(E(X|Y) = 0\) exactly. The marginal of Y is \(g(y)\), the chi-square density with 1 degree of freedom. Integrating out \(y\):

\[ \begin{aligned} f_{X}(x) &= \frac{1}{2\pi}\int_{0}^{\infty} e^{-y(1 + x^{2})/2}dy \\ &= \frac{1}{\pi(1 + x^{2})} \end{aligned} \]

So X has the standard Cauchy distribution, and \(E(X)\) does not exist. That rules out \(E(X) = 0\) and \(E(X) \ne E(E(X|Y))\), and makes \(E(E(X|Y)) = 0\) misleading: the iterated expectation is 0 although the expectation it should equal does not exist, since the tower property needs \(E|X| \lt \infty\).

Answer: A

The correct statements are (i) and (ii).

If \(Y = 0\), \(X_{n} = 1\) for all \(n\). If \(Y = 1\), \(X_{n} = (1 - n^{-\alpha})^{n}\), which tends to 0 when \(0 \le \alpha \lt 1\) (for \(\alpha = 0\) it is exactly 0), to \(e^{-1}\) when \(\alpha = 1\), and to 1 when \(\alpha \gt 1\), since then \(n \cdot n^{-\alpha} \to 0\).

(i): for \(\alpha \lt 1\), \(X_{n} \to 1 - Y\), which takes values 1 and 0 with probability 1/2 each, the same distribution as Y. So \(X_{n} \to Y\) in distribution. (ii): for \(\alpha = 1\), \(X_{n} \to e^{-Y}\) (1 if \(Y = 0\), \(e^{-1}\) if \(Y = 1\)). (iii): for \(\alpha \gt 1\), \(X_{n} \to 1\), a constant, which is not distributed like Y.

Answer: B

The true statements are (i) and (iii).

(i): for disjoint A and B, \(A \cup B\) is finite exactly when both are, so \(\mu(A \cup B) = \mu(A) + \mu(B)\) in every case (0 + 0, or \(\infty\) on both sides). \(\mu\) is finitely additive.

(ii) fails: \(\Omega\) is the countable disjoint union of its singletons, each of measure 0, yet \(\mu(\Omega) = \infty \ne 0\). So \(\mu\) is not countably additive.

(iii) is true and shows the same failure as continuity: list \(\Omega = \{\omega_{1}, \omega_{2}, \ldots\}\) and let \(A_{n} = \{\omega_{1}, \ldots, \omega_{n}\}\). The \(A_{n}\) increase to \(\Omega\), each \(\mu(A_{n}) = 0\), but \(\mu(\Omega) = \infty\).

Answer: D

The statement that is not true is E(U|F) and E(V|F) are independent.

Given \(F = \sigma(X)\), X is known and Y is independent of it with mean 0, so \(E(Y|F) = 0\). Hence \(E(U|F) = X + 0 = X\) and \(E(V|F) = X - 0 = X\), and \(E(U + V|F) = E(2X|F) = 2X\). Options A, B and C hold almost surely.

Both conditional expectations equal the same random variable X, which has variance 1. A non-degenerate random variable is not independent of itself, since \(P(X \le 0, X \le 0) = 1/2 \ne 1/4\). (U and V themselves are independent, being uncorrelated jointly normal variables, which is the trap.)

Answer: D

The statement that is not true is \(P(A_{1}|Z)P(A_{2}|Z) = P(A_{1} \cap A_{2}|Z)\).

On \(\{Z = 0\}\) the outcomes are (1, -1) and (-1, 1), each with conditional probability 1/2. \(A_{1}\) (that \(X_{1} = 1\)) happens in the first, so \(P(A_{1}|Z) = 1/2\), and likewise \(P(A_{2}|Z) = 1/2\). Both cannot happen, since then \(Z = 2\), so \(P(A_{1} \cap A_{2}|Z) = 0\).

The product of the conditional probabilities is \(1/4 \ne 0\), so the events, independent unconditionally, are not conditionally independent given \(Z = 0\). Options A, B and C are the true values.

Answer: B

The variance is 1/4.

\(P(s) = \tfrac{1}{2}s + \tfrac{1}{2}s^{2}\) means the variable takes the values 1 and 2, each with probability 1/2. Using the generating function: \(P'(1) = \tfrac{1}{2} + 1 = 3/2\) is the mean and \(P''(1) = 1\) is \(E[X(X - 1)]\), so

\[ \begin{aligned} Var(X) &= P''(1) + P'(1) - P'(1)^{2} \\ &= 1 + \tfrac{3}{2} - \tfrac{9}{4} = \tfrac{1}{4} \end{aligned} \]

Directly: \(E(X^{2}) = (1 + 4)/2 = 5/2\), and \(5/2 - 9/4 = 1/4\). 1/2 is the probability of each value, not the variance, and 1/3 and 2/3 do not follow from this generating function by any correct route.

Answer: D

\(E(X) =\) 1.

The distribution is mixed. On [0, 1) the cdf rises continuously as \(x/2\), a density of 1/2 carrying total probability 1/2. At \(x = 1\) it jumps from 1/2 to 3/4, a mass of 1/4 at 1, and at \(x = 2\) it jumps from 3/4 to 1, a mass of 1/4 at 2. So

\[ \begin{aligned} E(X) &= \int_{0}^{1} \frac{x}{2}\,dx + 1 \cdot \tfrac{1}{4} + 2 \cdot \tfrac{1}{4} \\ &= \tfrac{1}{4} + \tfrac{1}{4} + \tfrac{1}{2} = 1 \end{aligned} \]

3/8 and 1/2 come from missing one or both point masses, and 3/2 from counting the mass at 2 twice.