Probability Theory Questions and Answers

Answer: A

Both (I) and (II) are true.

The selection probabilities are \(\frac{1}{10}, \frac{3}{10}, \frac{6}{10}\). By the law of total probability,

\[ \begin{aligned} P(R) &= 0.1(0.8) + 0.3(0.5) + 0.6(0.3) \\ &= 0.08 + 0.15 + 0.18 = 0.41 \end{aligned} \]

which is (I). By Bayes' theorem,

\[ \begin{aligned} P(A \text{ or } B \mid R) &= \frac{0.08 + 0.15}{0.41} \\ &= \frac{23}{41} \end{aligned} \]

which is (II). The remaining \(\frac{18}{41}\) is the chance that C is the leader given that R is used.

Answer: B

Only \(G_{1}\) and \(G_{2}\) are distribution functions.

Take F as u running from 0 to 1. \(G_{1} = \frac{pu}{1 - qu}\) and \(G_{2} = \frac{u}{p + qu}\) are increasing in u (their derivatives are \(\frac{p}{(1 - qu)^{2}}\) and \(\frac{p}{(p + qu)^{2}}\), both positive), equal 0 at \(u = 0\), and equal \(\frac{p}{1 - q} = 1\) and \(\frac{1}{p + q} = 1\) at \(u = 1\). Composed with F they are non-decreasing, right-continuous, and tend to 0 and 1 at \(\mp\infty\).

\(G_{3}(x) = p + qF(2)\) is a constant, not depending on x. It cannot tend to 0 at \(-\infty\) (it is at least \(p \gt 0\)), so it is not a distribution function. That rules out every option that includes \(G_{3}\), and 'none' is wrong since \(G_{1}\) and \(G_{2}\) qualify.

Answer: C

The distribution is not symmetric about its median when F is exponential.

The exponential density is \(\lambda e^{-\lambda(x - \theta)}\) for \(x \ge \theta\) and 0 below \(\theta\): it has a long right tail and no left tail at all, so it cannot be a mirror image about its median \(\theta + \frac{\ln 2}{\lambda}\).

The others are symmetric. Binomial(n, ½) has \(P(X = k) = P(X = n - k)\), symmetric about \(\frac{n}{2}\), its median. The uniform on (a, b) is symmetric about \(\frac{a + b}{2}\), and the Cauchy is symmetric about its location parameter \(\theta\), which is its median.

Answer: C

The function that is not a characteristic function is \(\phi(t) = \frac{1}{1 + t^{4}}\).

Near 0, \(\frac{1}{1 + t^{4}} = 1 - t^{4} + \ldots\), so \(\phi'(0) = \phi''(0) = 0\). If it were the characteristic function of X, then \(E(X^{2}) = -\phi''(0) = 0\), forcing \(X = 0\) almost surely, whose characteristic function is identically 1. That contradiction shows it is not a characteristic function.

The others are. \(\phi(t/a)e^{ibt}\) is the characteristic function of \(\frac{X}{a} + b\). \(\phi(-t)\) is that of \(-X\). And a mixture \(\sum a_{j}\phi_{j}\) with weights \(a_{j} \ge 0\) summing to 1 is the characteristic function of the corresponding mixture distribution.

Answer: C

p(x) is a probability mass function when \(k = 1 - p^{2}\).

The probabilities must add to 1 over \(x = 0, 2, 4, \ldots\):

\[ \begin{aligned} k(1 + p^{2} + p^{4} + \ldots) &= 1 \\ \frac{k}{1 - p^{2}} &= 1 \Rightarrow k = 1 - p^{2} \end{aligned} \]

The geometric series converges because \(0 \lt p^{2} \lt 1\).

\(k = 1 - p\) is the constant for support 0, 1, 2, ..., which forgets that only even values are allowed. p and \(p^{2}\) do not make the total 1.

Answer: A

The pgf of Y is \(t^{2} P_{X}(t^{3})\).

By definition \(P_{Y}(t) = E(t^{Y})\), so

\[ \begin{aligned} P_{Y}(t) &= E\left(t^{3X + 2}\right) = t^{2}\,E\left((t^{3})^{X}\right) \\ &= t^{2} P_{X}(t^{3}) \end{aligned} \]

The constant shift 2 becomes the factor \(t^{2}\) and the scale 3 becomes the argument \(t^{3}\).

\(P_{X}(t^{3})\) is the pgf of 3X, without the shift. \(t^{3}P_{X}(t^{2})\) belongs to \(2X + 3\), with the roles swapped, and \(2tP_{X}(t)\) is not a pgf at all, since it equals 2 at \(t = 1\).

Answer: C

E(X) = 0.7 and variance (X) = 0.51.

X is a mixture: with probability 0.7 it equals 1, and with probability 0.3 it is standard normal.

\[ \begin{aligned} E(X) &= 0.7(1) + 0.3(0) = 0.7 \\ E(X^{2}) &= 0.7(1) + 0.3(1) = 1 \\ \mathrm{Var}(X) &= 1 - 0.7^{2} = 0.51 \end{aligned} \]

The normal component contributes \(E(Z^{2}) = 1\), its variance plus its zero mean squared. A variance of 1 forgets to subtract \(0.49\), and 0.30 is only the weighted variance of the normal part, ignoring the spread between the two components.

Answer: B

The FALSE statement is \(f(x - 5) = f(5 - x)\ \forall\, x\).

Rewrite the condition: \(e^{-5t}M(t) = E\,e^{t(X - 5)}\) is the mgf of \(X - 5\), and \(e^{5t}M(-t) = E\,e^{t(5 - X)}\) is the mgf of \(5 - X\). Equal mgfs mean \(X - 5\) and \(5 - X\) have the same distribution, so X is symmetric about 5.

Symmetry about 5 gives \(f(5 + x) = f(5 - x)\) (option A), makes 5 the median (option C), and is exactly option D.

Option B, \(f(x - 5) = f(5 - x)\), says f is an even function, symmetric about 0, which is a different property and false in general (for example for N(5, 1)).

Answer: A

The true statement is E(X) = E(Y); both equal \(\frac{1}{3}\).

\(F_{1}\) has jumps of \(\frac{2}{3}\) at 0 and \(\frac{1}{3}\) at 1, so X is discrete with \(E(X) = 0 \cdot \frac{2}{3} + 1 \cdot \frac{1}{3} = \frac{1}{3}\).

\(F_{2}(x) = \sqrt{x}\) on [0, 1) is continuous with density \(\frac{1}{2\sqrt{x}}\), so

\[ \begin{aligned} E(Y) &= \int_{0}^{1} x \cdot \frac{1}{2\sqrt{x}}\,dx \\ &= \frac{1}{2}\int_{0}^{1} \sqrt{x}\,dx = \frac{1}{3} \end{aligned} \]

Option B takes the jump at 0 as the mean of X. Options C and D have the right E(X) but the wrong E(Y): \(\frac{1}{2}\) is the mean of a uniform variable, and \(\frac{1}{6}\) halves the correct value.

Answer: B

The correct statement is \(\langle Y \rangle = 0\) and \(\langle (Y - \langle Y \rangle)^{2} \rangle = \sigma^{2}\).

Shifting a random variable by a constant shifts its mean by that constant and leaves its spread unchanged:

\[ \begin{aligned} \langle Y \rangle &= \langle X \rangle - \mu = 0 \\ \langle (Y - 0)^{2} \rangle &= \langle (X - \mu)^{2} \rangle = \sigma^{2} \end{aligned} \]

So \(Y\) is the centred version of \(X\), with the same variance. A mean of \(\mu\) would need no shift at all; a variance of 0 would mean \(Y\) is constant; and \(\sigma^{2} - \mu^{2}\) mixes up the variance formula \(\langle X^{2} \rangle - \mu^{2}\) with the variance itself.