Probability Theory Questions and Answers

Answer: C

The false statement is \(\mathrm{Corr}(X, Y) \gt 0\).

With \(E[Y | X] = X^{2}\) and X standard normal, use the tower property:

\[ \begin{aligned} E[XY] &= E[X\,E(Y | X)] = E[X^{3}] = 0 \\ E[Y] &= E[X^{2}] = 1 \end{aligned} \]

So \(\mathrm{cov}(X, Y) = E[XY] - E[X]E[Y] = 0 - 0 = 0\) and the correlation is 0, not positive. Also \(E[X^{2}Y] = E[X^{2} \cdot X^{2}] = E[X^{4}] = 3\). So A, B and D are true, and only C is false: Y depends on X, but symmetrically, through \(X^{2}\).

Answer: B

The false statement is B: the marginal distribution function of X is not \(1 - \tfrac{1}{2}[e^{-x} + 2e^{-2x}]\).

Summing the joint law over \(y = 1, 2\) gives the marginal density of X:

\[ f_{X}(x) = \tfrac{1}{2}[e^{-x} + 2e^{-2x}],\ x \gt 0 \]

which confirms C. Integrating from 0 to \(x\):

\[ \begin{aligned} F(x) &= \tfrac{1}{2}(1 - e^{-x}) + \tfrac{1}{2}(1 - e^{-2x}) \\ &= 1 - \tfrac{1}{2}\left[e^{-x} + e^{-2x}\right] \end{aligned} \]

Option B keeps the factor 2 from the density, which disappears on integration; it even gives \(F(0) = -1/2\). D is true because \(P(Y = y) = \tfrac{1}{2}\int_{0}^{\infty} ye^{-yx}dx = \tfrac{1}{2}\). A is true: Y is discrete and X continuous, so (X, Y) has neither a joint mass function nor a joint density on the plane.

Answer: D

The statement always correct is \(\sum \mu(A_{n}) \le \mu\left(\bigcup A_{n}\right)\).

Let \(A = \bigcup A_{n}\). For each \(N\), finite additivity and monotonicity (which follows from finite additivity and non-negativity) give

\[ \begin{aligned} \sum_{n=1}^{N} \mu(A_{n}) &= \mu\left(\bigcup_{n=1}^{N} A_{n}\right) \\ &\le \mu(A) \end{aligned} \]

Letting \(N \to \infty\) gives the inequality. Equality (option A) is countable additivity, which a merely finitely additive function need not have; so the reverse inequality B can also fail. Option C is wrong because the complements \(A_{n}^{c}\) are not disjoint.

Answer: B

The false statement is \(\lim \int |f_{n}|\,d\mu = \int |f|\,d\mu\).

The uniform bound 100 is not an integrable dominating function unless \(\mu(\Omega) \lt \infty\), and the measure space here is arbitrary. On \(\mathbf{R}\) with Lebesgue measure, \(f_{n} = 1_{[n, n+1]}\) satisfies \(|f_{n}| \le 1\) and \(f_{n} \to 0\) everywhere, yet \(\int |f_{n}| = 1\) for every \(n\), while \(\int |f| = 0\).

The others always hold: A is Fatou's lemma; C holds because \(\limsup|f_{n}| = |f|\) almost everywhere; and D is the monotone convergence theorem applied to the increasing partial sums of \(|f_{k}|\).

Answer: D

The measurable function is \(f(1) = 2\), \(f(2) = 3\), \(f(k) = 10\) for \(k \ge 3\).

The \(\sigma\)-field F is generated by the three atoms {1}, {2} and \(\{3, 4, 5, \ldots\}\), and its sets are the unions of these atoms. A real function is F-measurable exactly when it is constant on each atom, in particular constant on \(\{3, 4, \ldots\}\).

Option D is constant there (value 10), so every preimage is a union of atoms. Option A takes different values 3, 4, ...; option B takes 1 at 3 and 0 beyond; option C alternates between 1 and 2 on the third atom. For instance, in option C the preimage of {1} is the set of odd numbers, which is not in F.

Answer: D

The statement always true is \(\{W_{n}\}\) satisfies the strong law of large numbers.

\(W_{n} = \cos(X_{n}^{2})\) are i.i.d. (functions of i.i.d. variables) and bounded by 1, so \(E|W_{1}| \lt \infty\). Kolmogorov's strong law then gives \(\frac{1}{n}\sum W_{k} \to E W_{1}\) almost surely, whatever the distribution of X.

The \(X_{n}\) themselves need not have a finite mean, for example if they are Cauchy, and then the strong law fails for them. So 'both' and '\(\{X_{n}\}\) satisfies' are not always true, and 'neither' is wrong because \(\{W_{n}\}\) always does.

Answer: D

Statements (II) and (III) only are true.

The marginal density of Y is \(f_{Y}(y) = \int_{0}^{\infty} \tfrac{y}{2}e^{-xy}dx = \tfrac{1}{2}\) on (0, 2). So at \(y = 1\),

\[ \begin{aligned} f(x | 1) &= \frac{f(x, 1)}{f_{Y}(1)} = \frac{\tfrac{1}{2}e^{-x}}{\tfrac{1}{2}} \\ &= e^{-x},\quad x \gt 0 \end{aligned} \]

So (I) is false: \(\tfrac{1}{2}e^{-x}\) is the joint density at \(y = 1\), not the conditional one, and it integrates to 1/2. Given Y = 1, X is exponential with mean 1, so (II) holds, and \(P[X \le 2 | Y = 1] = 1 - e^{-2}\), so (III) holds.

Answer: C

\(E(X) =\) 3/2.

Read the distribution off F. It jumps by 1/4 at 0 and by 1/4 at 1, is flat on [1, 2), and rises linearly with slope 1/2 on [2, 3), from 1/2 to 1. So X is 0 with probability 1/4, 1 with probability 1/4, and uniform on [2, 3] with total probability 1/2.

\[ \begin{aligned} E(X) &= 0 \cdot \tfrac{1}{4} + 1 \cdot \tfrac{1}{4} + \tfrac{1}{2} \cdot 2.5 \\ &= \tfrac{1}{4} + \tfrac{5}{4} = \tfrac{3}{2} \end{aligned} \]

Check with \(E(X) = \int_{0}^{\infty}(1 - F(x))dx\), which is \(\tfrac{3}{4} + \tfrac{1}{2} + \tfrac{1}{4} = \tfrac{3}{2}\). The other values come from treating F as purely continuous or dropping one of the jumps.

Answer: B

The statement that is not correct is E(X) = 1/6.

F rises with slope 1/2 on [0, 1) to the value 1/2, then jumps to 1 at \(x = 1\). So X has density 1/2 on (0, 1) and an atom of mass 1/2 at 1, which makes D correct. Then

\[ \begin{aligned} E(X) &= \int_{0}^{1} \tfrac{x}{2}\,dx + 1 \cdot \tfrac{1}{2} \\ &= \tfrac{1}{4} + \tfrac{1}{2} = \tfrac{3}{4} \end{aligned} \]

so C is correct and B is not. A is also correct: \(P(-1 \lt X \le 1/2) = F(1/2) - F(-1) = 1/4 - 0 = 1/4\).

Answer: C

The inequality is \(P[|X| \ge a] \le E(g(X))/g(a)\).

This is the general Markov (Chebyshev-type) inequality. Since \(g\) is even and non-decreasing on \([0, \infty)\), \(|X| \ge a\) implies \(g(X) = g(|X|) \ge g(a)\). As \(g \ge 0\),

\[ g(X) \ge g(a)\,1_{\{|X| \ge a\}} \]

and taking expectations gives \(E\,g(X) \ge g(a)P[|X| \ge a]\), which is the result (for \(g(a) \gt 0\)). With \(g(x) = x^{2}\) it is Chebyshev's inequality.

The bound controls the tail \(|X| \ge a\), not the central part \(|X| \le a\), and it is an upper bound, not a lower one, so the other three are wrong.