Probability Theory Questions and Answers

Answer: C

The Jacobian is \(\frac{1}{4}\, y^{-\frac{3}{4}}\).

Since \(X \gt 0\), the inverse transformation is \(x = y^{\frac{1}{4}}\), and its Jacobian is the derivative of \(x\) with respect to \(y\):

\[ \begin{aligned} J = \frac{dx}{dy} &= \frac{1}{4}\, y^{\frac{1}{4} - 1} \\ &= \frac{1}{4}\, y^{-\frac{3}{4}} \end{aligned} \]

This is the factor used in \(f_{Y}(y) = f_{X}(y^{1/4}) \cdot \frac{1}{4} y^{-3/4}\). The option \(y^{\frac{1}{4}}\) is the inverse transformation itself, not its derivative. \(\frac{3}{4} y^{-\frac{3}{4}}\) uses the wrong constant, and \(y^{\frac{3}{4}}\) has the wrong sign in the exponent.

Answer: A

The probability is \(\frac{2^{4}}{3^{8}}\).

Each row of an orthogonal matrix has length 1. With entries from \(\{-1, 0, 1\}\), a row of length 1 has exactly one entry \(\pm 1\) and the rest 0. Rows must also be mutually orthogonal, so their non-zero entries sit in different columns. So \(A\) is orthogonal exactly when it is a signed permutation matrix.

There are \(3! = 6\) ways to place the non-zero entries and \(2^{3} = 8\) choices of sign, giving 48 such matrices. All \(3^{9}\) matrices are equally likely, so

\[ \begin{aligned} P &= \frac{48}{3^{9}} = \frac{2^{4} \cdot 3}{3^{9}} \\ &= \frac{2^{4}}{3^{8}} \end{aligned} \]

\(\frac{2^{5}}{3^{9}}\) counts 32 matrices and \(\frac{2^{3}}{3^{7}}\) counts 72, so both miscount the signed permutations; \(\frac{1}{3^{5}}\) corresponds to 81.

Answer: C

Both S1 and S2 are correct.

Both follow from the theorem of total probability, which applies because \(P(C)\) and \(P(C^{c})\) are both positive.

S1:

\[ \begin{aligned} P(A) &= P(A|C)P(C) + P(A|C^{c})P(C^{c}) \\ &\gt P(B|C)P(C) + P(B|C^{c})P(C^{c}) = P(B) \end{aligned} \]

The strict inequality survives because each weight is positive. (Simpson's paradox concerns comparisons where the weights differ between the two groups; here both events are weighted by the same \(P(C)\) and \(P(C^{c})\).)

S2: if \(P(A|C) = P(A|C^{c}) = p\), then \(P(A) = p\,P(C) + p\,P(C^{c}) = p\), so \(P(A|C) = P(A)\), which is independence of \(A\) and \(C\).

Answer: A

The probability is \(\frac{e\sqrt{e} - 1}{e\sqrt{e}}\), that is \(1 - e^{-3/2}\).

For fixed \(x\) in \((5, 10)\), the density in \(y\) is \(\frac{1}{10}e^{-\frac{1}{2}(y - (x - 3))}\) for \(y \gt x - 3\). Integrating over \(y\) gives \(\frac{1}{5}\), so \(X \sim U(5, 10)\), and given \(X = x\), \(Y - (x - 3)\) is exponential with mean 2.

Component B fails first when \(Y \lt X\), that is \(Y - (x - 3) \lt 3\):

\[ \begin{aligned} P(Y \lt X \mid X = x) &= 1 - e^{-\frac{3}{2}} \\ P(Y \lt X) &= \int_{5}^{10} \frac{1}{5}\left(1 - e^{-\frac{3}{2}}\right)dx = 1 - e^{-\frac{3}{2}} \end{aligned} \]

Since \(e^{3/2} = e\sqrt{e}\), this is \(\frac{e\sqrt{e} - 1}{e\sqrt{e}}\), about 0.78. \(\frac{1}{e\sqrt{e}}\) is the complementary probability that A fails first, and the two options with \(e\sqrt{e} + 1\) in the denominator have no basis in the integral.

Answer: D

The only true statement is the one with \(P(\{1, 2, \ldots, n\}) = \int_{0}^{n} e^{-x}\,dx\).

A probability function must be non-negative, give \(P(S) = 1\), and be countably additive. Check each proposal against these:

• \(P(\{n\}) = \frac{1}{n+1}\): countable additivity forces \(P(S) = \sum \frac{1}{n+1}\), a divergent harmonic series, not 1. Impossible.
• Finite sets 0, infinite sets 1: \(S\) is the countable union of the singletons \(\{n\}\), each of probability 0, so countable additivity gives \(P(S) = 0\), not 1. Impossible.
• \(P(\{1, \ldots, n\}) = \int_{1}^{n} \frac{dx}{x} = \log_{e} n\): this exceeds 1 once \(n \geq 3\), which no probability can. Impossible.
• \(P(\{1, \ldots, n\}) = 1 - e^{-n}\): set \(P(\{1\}) = 1 - e^{-1}\) and \(P(\{n\}) = e^{-(n-1)} - e^{-n} \gt 0\) for \(n \geq 2\). These are positive and sum to \(\lim(1 - e^{-n}) = 1\), so they define a valid probability.

Answer: D

\(P[Y = 4] = \)\(\dfrac{2}{2n + 1}\).

X takes the \(2n + 1\) values \(0, \pm 1, \pm 2, \ldots, \pm n\), each with the same probability \(k\). The probabilities must add to 1, so \((2n + 1)k = 1\) and \(k = 1/(2n + 1)\).

\(Y = X^{2} = 4\) happens exactly when \(X = 2\) or \(X = -2\) (both values exist once \(n \ge 2\)), so

\[ P[Y = 4] = P[X = 2] + P[X = -2] = \frac{2}{2n + 1} \]

\(1/(2n + 1)\) counts only one of the two values. \(2/(n + 1)\) comes from miscounting the support as \(n + 1\) points, and \(n/(2n + 1)\) is roughly the probability that X is positive, not that \(X^{2} = 4\).

Answer: A

The characteristic function of \(X_{1}\) is \(\dfrac{\sin t}{t}\).

First find the marginal density of \(X_{1}\) by integrating out \(x_{2}\) over \((-1, 1)\):

\[ \begin{aligned} f_{1}(x_{1}) &= \frac{1}{4}\int_{-1}^{1} dx_{2} \\ &\quad + \frac{x_{1}}{4}\int_{-1}^{1} x_{2}(x_{1}^{2} - x_{2}^{2})\, dx_{2} \end{aligned} \]

In the second integral, \(x_{2}(x_{1}^{2} - x_{2}^{2})\) is an odd function of \(x_{2}\), so it integrates to 0 over the symmetric interval. What remains is \(\frac{1}{4} \cdot 2 = \frac{1}{2}\) for \(|x_{1}| \lt 1\): \(X_{1}\) is uniform on \((-1, 1)\).

\[ \begin{aligned} \phi(t) &= \frac{1}{2}\int_{-1}^{1} e^{itx}\, dx \\ &= \frac{e^{it} - e^{-it}}{2it} = \frac{\sin t}{t} \end{aligned} \]

The other expressions cannot be characteristic functions: every characteristic function equals 1 at \(t = 0\), while \(\cos t/t\) and \(\sin t/t^{2}\) blow up there, and \(t/\sin t\) exceeds 1 in absolute value near \(t = \pi\).

Answer: B

Chebyshev's inequality gives the bound \(\dfrac{1}{2}\).

Here FX stands for the expectation \(E(X)\). Chebyshev's inequality states that for any \(k \gt 0\),

\[ P[|X - E(X)| \ge k] \le \frac{\text{Var}(X)}{k^{2}} \]

With \(\text{Var}(X) = 1/2\) and \(k = 1\), the bound is \((1/2)/1^{2} = 1/2\). The event with strict inequality, \(|X - E(X)| \gt 1\), is contained in the one with \(\ge\), so the same bound applies.

1/4 is what you get by squaring the variance instead of dividing by \(k^{2}\). 1 is the trivial bound every probability satisfies and uses none of the information given. 3/4 does not follow from the inequality with these numbers at all; the complementary form gives \(P[|X - E(X)| \lt 1] \ge 1 - 1/2 = 1/2\).

Answer: B

The probability is One.

Pass to complements. Each \(A_{n}^{c}\) has probability 0, and by countable subadditivity

\[ \begin{aligned} P\left(\bigcup_{n} A_{n}^{c}\right) &\le \sum_{n} P(A_{n}^{c}) \\ &= 0 \end{aligned} \]

By De Morgan's law, \(\bigcap_{n} A_{n}\) is the complement of \(\bigcup_{n} A_{n}^{c}\), so its probability is \(1 - 0 = 1\). A countable intersection of almost sure events is almost sure.

The intersection is an event in F because F is a \(\sigma\)-field, so its probability is defined; it cannot be infinity, since probabilities lie in [0, 1]; and zero contradicts the calculation above.

Answer: C

X + Y is not Gaussian, with a distribution that is discontinuous at the origin.

Split on the event in the definition of Y:

• if \(|X| \le \alpha\), then \(Y = X\) and \(X + Y = 2X\);
• if \(|X| \gt \alpha\), then \(Y = -X\) and \(X + Y = 0\).

So X + Y equals 0 with probability \(P(|X| \gt \alpha) = 2(1 - \Phi(\alpha)) \gt 0\). A point mass at 0 makes the distribution function jump there, so it is discontinuous at the origin, and no Gaussian distribution has an atom. Away from 0, X + Y is spread over \([-2\alpha, 2\alpha]\) only, which is bounded.

The point of the question is that Y itself is N(0, 1), by the symmetry of X, yet X + Y is not Gaussian: two normal variables need not be jointly normal. Gaussian (0, 2) is what independence would give, and 'continuous distribution' misses the atom.