Probability Theory Questions and Answers

Answer: B

\(E(X|Y = y) =\) \(y/2\).

The marginal of Y is \(f_{Y}(y) = \int_{0}^{y} 2\,dx = 2y\) for \(0 \lt y \lt 1\). The conditional density of X given \(Y = y\) is \(2/(2y) = 1/y\) on \((0, y)\), the uniform distribution on \((0, y)\). Its mean is the midpoint:

\[ E(X|Y = y) = \int_{0}^{y} \frac{x}{y}\,dx = \frac{y}{2} \]

\(2y\) is the marginal density of Y, not a conditional mean. \(y\) would be the upper end of the range, and 1 would put X at the top of the triangle, outside \(x \lt y\).

Answer: C

The characteristic function is \(\exp(itc)\).

The characteristic function is \(\varphi(t) = E(e^{itX})\). A degenerate variable takes the single value \(c\) with probability 1, so

\[ \varphi(t) = e^{itc} \]

Its modulus is 1 for every \(t\), and \(\varphi(0) = 1\) as for any characteristic function, which rules out the constants \(c\) and 0. \(\exp(-itc)\) is the complex conjugate, the characteristic function of \(-X\), or of X under the opposite sign convention, which is not the standard definition.

Answer: C

The probability is \(\frac{1}{6}\).

Two dice give \(6 \times 6 = 36\) equally likely outcomes. Both show the same number in exactly 6 of them: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). So

\[ P = \frac{6}{36} = \frac{1}{6} \]

Equivalently, whatever the first die shows, the second matches it with probability 1/6. The value 1/36 is the probability of one particular double, such as two sixes. The value 11/36 is the chance that at least one die shows a given number, such as at least one six, and 1/3 corresponds to no natural count here.

Answer: B

The probability is 0.25.

Mutually exclusive means \(A \cap B = \emptyset\), so whenever A occurs B does not. Hence

\[ P(A \cap B^{c}) = P(A) - P(A \cap B) = 0.25 - 0 = 0.25 \]

0.1 is \(P(B) - P(A)\), 0.6 is \(P(A \cup B)\), and 0.4 is \(1 - P(A \cup B)\), the chance that neither occurs. None of these is the event 'A but not B'.

Answer: C

The probability is 1/3.

Treat the four children as distinct. There are \(4! = 24\) orders. An alternating row is BGBG or GBGB, and each pattern can be filled in \(2! \times 2! = 4\) ways, so there are 8 alternating orders:

\[ P = \frac{8}{24} = \frac{1}{3} \]

Counting patterns only gives the same: of the \(\binom{4}{2} = 6\) equally likely boy-girl patterns, 2 alternate. 1/2 and 2/3 come from counting just one pattern or its complement wrongly.

Answer: D

\(\mathrm{cov}(5x - 2, 2y - 5) =\) 20.

Covariance ignores added constants and is bilinear in scale. From \(\mathrm{cov}(2x, y) = 2\,\mathrm{cov}(x, y) = 4\), \(\mathrm{cov}(x, y) = 2\). Then

\[ \begin{aligned} &\mathrm{cov}(5x - 2, 2y - 5) \\ &= 5 \cdot 2 \cdot \mathrm{cov}(x, y) = 10 \cdot 2 = 20 \end{aligned} \]

40 multiplies the given value 4 by 10 without first dividing by 2, 10 forgets that \(\mathrm{cov}(x, y)\) is 2 rather than 1, and 0 would wrongly treat the shifts as removing the relationship.

Answer: D

The answer is \(P^{2}(B) \le P(A)\).

\(P(A | B) = P(A \cap B)/P(B) = P(B)\) gives \(P(A \cap B) = P^{2}(B)\). Since \(A \cap B \subseteq A\),

\[ P^{2}(B) = P(A \cap B) \le P(A) \]

The others need not hold. Take \(P(A) = 0.7\), \(P(B) = 0.4\) and \(P(A \cap B) = 0.16\), which is consistent and gives \(P(A | B) = 0.4 = P(B)\). Then \(P(B | A) = 0.16/0.7 \approx 0.23 \ne P(A)\); A and B are not independent, since \(P(A)P(B) = 0.28 \ne 0.16\); and \(P^{2}(A) = 0.49 \gt P(B)\).

Answer: B

The feasible value is E[X] = 2.1.

The probabilities must sum to 1, so \(p + q = 1 - 0.3 = 0.7\). Then

\[ \begin{aligned} E[X] &= 0.1 + 2p + 3q \\ &= 0.1 + 2(0.7) + q = 1.5 + q \end{aligned} \]

With \(0 \le q \le 0.7\), \(E[X]\) ranges over [1.5, 2.2]. 2.1 is reached with \(q = 0.6\), \(p = 0.1\). 1.2 is below the minimum, and 2.8 and 3.4 exceed the maximum (3.4 is even above the largest value X can take).

Answer: C

The probability is 0.2.

Given \(X_{2} = 1\), the event \(X_{1} + X_{2} = 0\) is the event \(X_{1} = -1\). The variables are independent, so conditioning on \(X_{2}\) and \(X_{3}\) does not change the distribution of \(X_{1}\):

\[ \begin{aligned} &P[X_{1} = -1 \mid X_{2} = 1, X_{3} = 2] \\ &= P[X_{1} = -1] = 0.2 \end{aligned} \]

The condition on \(X_{3}\) is irrelevant. 0.3 is \(P[X = 0]\), the value for \(X_{1} + X_{2} = 0\) only if \(X_{2}\) were 0, and 0.11 and 0.13 come from multiplying probabilities that conditioning removes.

Answer: A

The answer is \(M_{Z}(t) = M_{X}(t)M_{Y}(t)\) and \(M_{W}(t) = \int M_{Y}(tx)f_{X}(x)\,dx\).

For the sum, independence gives \(E[e^{t(X + Y)}] = E[e^{tX}]E[e^{tY}]\). For the product, condition on X:

\[ \begin{aligned} M_{W}(t) &= E[e^{tXY}] = E\big[E[e^{(tX)Y} \mid X]\big] \\ &= E[M_{Y}(tX)] \end{aligned} \]

which is \(\int M_{Y}(tx)f_{X}(x)\,dx\) when X has a density. (Strictly, this needs \(M_{Y}(tx)\) to be finite for the relevant \(x\).)

The mgf of a product is not the sum or the product of the mgfs, and the mgf of a sum is the product, not the sum, so B, C and D each contain a false half.