Probability Theory Questions and Answers

Q: Suppose \(\phi(t) = \exp(-|t|)\) is a characteristic function of a random variable X. Then the distribution of X is :

X has Student's t distribution with 1 degree of freedom, that is, the standard Cauchy distribution.The standard Cauchy density \(f(x) = \frac{1}{\pi(1 + x^{2})}\) has characteristic function \(e^{-|t|}\), and the t distribution with one degree of freedom is exactly this Cauchy law. The non-differentiability of \(e^{-|t|}\) at 0 matches the fact that the Cauchy distribution has no mean.The other options are the usual mix-up of roles. The normal has characteristic function \(e^{-t^{2}/2}\). The Laplace density \(\frac{1}{2}e^{-|x|}\) has characteristic function \(1/(1 + t^{2})\): density and characteristic function are swapped relative to the Cauchy. A gamma law has \((1 - it/\beta)^{-\alpha}\), which is complex-valued, while \(e^{-|t|}\) is real, as it must be for a symmetric law.

Q: Suppose X is a real random variable on a probability space \((\Omega, A, P)\) and \(\{A_{n}, n \ge 1\}\) is a sequence of events in A such that :\[ A_{n} \to \Omega \text{ as } n \to \infty. \]Which of the following is true ?

\(X I_{A_{n}} \xrightarrow{P} X\).\(A_{n} \to \Omega\) means that every outcome eventually belongs to \(A_{n}\) (for the limit set to be \(\Omega\), \(\liminf A_{n} = \Omega\)). So for every \(\omega\), \(I_{A_{n}}(\omega) = 1\) for all large \(n\), and \(X I_{A_{n}}(\omega) = X(\omega)\) from some point on. That is almost sure, indeed pointwise, convergence to X, which implies convergence in probability.In probability terms, \(P(|X I_{A_{n}} - X| \gt \varepsilon) \le P(A_{n}^{c}) \to 0\) by continuity of P.The limit is not 0 (in probability or in law) unless X itself is 0, and it is not the constant 1 unless X is 1.

Q: Suppose that X is a non-negative integer valued random variable having \(\varphi\) as its probability generating function. Then, variance (X) = 1, when :

\(\text{Var}(X) = 1\) for \(\varphi(s) = (s + s^{3})/2\).That pgf puts probability 1/2 on each of the values 1 and 3, so \(E(X) = 2\) and \(\text{Var}(X) = \frac{1}{2}(1 - 2)^{2} + \frac{1}{2}(3 - 2)^{2} = 1\). Equivalently \(\varphi'(1) = 2\), \(\varphi''(1) = 3\), and \(\text{Var} = \varphi''(1) + \varphi'(1) - \varphi'(1)^{2} = 3 + 2 - 4 = 1\).The others give different variances:\((1 + s)/2\) is Bernoulli(1/2), variance 1/4.\(1/(2 - s) = \frac{1/2}{1 - s/2}\) is geometric on {0, 1, 2, ...} with \(p = 1/2\), variance \(q/p^{2} = 2\).\(s/(3 - 2s) = \frac{s/3}{1 - 2s/3}\) is geometric on {1, 2, ...} with \(p = 1/3\), variance \(q/p^{2} = 6\).

Q: Consider the following probability density function :P(x) = 0 \(x \lt -2\) and \(x \gt 3\)P(x) = a \(-2 \lt x \lt 2\)P(x) = \(\frac{1}{12}\) \(2 \lt x \lt 3\)The value of a is :

The constant is \(a = \frac{11}{48}\).A probability density must integrate to 1 over all x. The density is zero outside -2 to 3, equal to a over the 4 units from -2 to 2, and 1/12 over the 1 unit from 2 to 3:\[ \begin{aligned} 4a + \frac{1}{12} &= 1 \\ 4a &= \frac{11}{12} \\ a &= \frac{11}{48} \end{aligned} \]The distractor 11/12 forgets to divide by the width 4 of the interval, and 1/12 and 1/48 do not satisfy the normalisation at all.

Q: The ogive more than and ogive less than of the following grouped dataClassFrequency0-888-201220-301830-351035-404intersect at which of the following points ?

The ogives intersect at 23.33, the median.The 'less than' and 'more than' ogives cross at the median. The total frequency is N = 52, so N/2 = 26. The cumulative frequencies are 8, 20, 38, 48, 52, so the median class is 20-30, with lower limit 20, cumulative frequency before it 20, frequency 18 and width 10:\[ \begin{aligned} \text{Median} &= 20 + \frac{26 - 20}{18} \times 10 \\ &= 20 + 3.33 = 23.33 \end{aligned} \]The nearby values 23.00, 23.67 and 24.00 come from rounding or from using the wrong cumulative frequency or class width.

Q: Which of the following is the smallest field containing a non-empty subset A of \(\Omega\) ?

The smallest field containing A is {A, AC, φ, Ω}.A field (algebra) of sets must contain \(\Omega\) and \(\phi\) and be closed under complements and finite unions. Any field containing A therefore contains \(A^{C}\), \(\phi\) and \(\Omega\). These four sets are already closed under complement and union (for instance \(A \cup A^{C} = \Omega\)), so they form a field, and it is the smallest one.The trivial field \(\{\phi, \Omega\}\) does not contain A unless A is \(\Omega\). The power set is a field containing A, but generally a much larger one, so it is not the smallest.

Q: Which of the following statements cannot be always true ?

The statement that need not hold is B ∩ C = φ implies X(B) ∩ X(C) = φ.Direct images do not preserve disjointness. If X is constant, say \(X(w) = 1\) for all w, then for disjoint non-empty B and C, \(X(B) = X(C) = \{1\}\), which intersect.The others are standard results. Inverse images preserve complements and countable unions, so \(X^{-1}(\tau_{j})\) is a \(\sigma\)-field; the \(\sigma\)-field generated by \(X^{-1}(\tau)\) equals \(X^{-1}(\sigma(\tau))\); and if \(X \ge 0\) with \(E(X) = 0\), then \(P(X \gt 1/n) \le nE(X) = 0\) for every n, so X = 0 almost surely.

Q: Let \(\Omega = \{1, 2, 3, \ldots, 10\}\), A = {1, 2} and \(\mathbf{F} = \{\phi, A, A^{C}, \Omega\}\), consider the following functions :\[ X(w) = 100\ \forall\, w \in \Omega \]\[ Y(w) = \begin{cases} 100 & \text{if } w \in A \\ 400 & \text{if } w \notin A \end{cases} \]\[ Z(w) = w\ \forall\, w \in \Omega \]Then which of the following is true ?

Only X and Y are F-measurable.A function is F-measurable when the set where it takes each value (more generally each inverse image of a Borel set) lies in F = {φ, A, AC, Ω}.X is constant, so its inverse images are \(\phi\) or \(\Omega\): measurable. Y takes the value 100 on A and 400 on \(A^{C}\), so its inverse images are \(\phi\), A, \(A^{C}\) or \(\Omega\): measurable.Z(w) = w takes ten different values. The set {Z = 3} = {3} is not in F, so Z is not measurable. That rules out 'all three', 'only X' and 'none'.

Q: Suppose that probabilities are assigned to the subsets of \(\Omega = \{a, b, c, d, e, f\}\) according to the classical definition of probability. Which of the following statements is/are true for the sets A = {a, b}, B = {b, d} and C = {a, c, d} ?(I) C and A are independent(II) C and B are independent(III) C and \(A \cup B\) are independent

Only (I) and (II) are true.With six equally likely outcomes, \(P(A) = \frac{2}{6}\), \(P(B) = \frac{2}{6}\) and \(P(C) = \frac{3}{6}\).(I) \(A \cap C = \{a\}\), so \(P(A \cap C) = \frac{1}{6} = \frac{1}{3} \cdot \frac{1}{2}\): independent.(II) \(B \cap C = \{d\}\), so \(P(B \cap C) = \frac{1}{6} = \frac{1}{3} \cdot \frac{1}{2}\): independent.(III) \(A \cup B = \{a, b, d\}\) with probability \(\frac{1}{2}\), and \(C \cap (A \cup B) = \{a, d\}\) with probability \(\frac{1}{3}\). Since \(\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \neq \frac{1}{3}\), these are not independent.So (I) and (II) hold and (III) fails.

71.

If X is a non-negative random variable, which of the following statements is not true ?

A.\(E(X^{2}) \ge (E(X))^{2}\)
B.\(E(X^{3}) \ge (E(X))^{3}\)
C.\(E(X^{1/4}) \ge (E(X))^{1/4}\)
D.\(E(X^{-2}) \ge (E(X))^{-2}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The inequality that is not true in general is \(E(X^{1/4}) \ge (E(X))^{1/4}\).

Jensen's inequality says \(E\,\phi(X) \ge \phi(E X)\) for convex \(\phi\), and the reverse for concave \(\phi\). On \([0, \infty)\), \(x^{2}\), \(x^{3}\) and \(x^{-2}\) (for \(x \gt 0\)) are convex, so the first, second and fourth inequalities hold. But \(x^{1/4}\) is concave, so

\[ E(X^{1/4}) \le (E X)^{1/4} \]

with strict inequality unless X is constant. For example, if X is 0 or 16 with probability 1/2 each, \(E(X^{1/4}) = 1\) while \((E X)^{1/4} = 8^{1/4} \approx 1.68\).

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72.

Suppose \(\phi(t) = \exp(-|t|)\) is a characteristic function of a random variable X. Then the distribution of X is :

A.Normal
B.Laplace
C.Student's t with 1 degree of freedom
D.Gamma

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

X has Student's t distribution with 1 degree of freedom, that is, the standard Cauchy distribution.

The standard Cauchy density \(f(x) = \frac{1}{\pi(1 + x^{2})}\) has characteristic function \(e^{-|t|}\), and the t distribution with one degree of freedom is exactly this Cauchy law. The non-differentiability of \(e^{-|t|}\) at 0 matches the fact that the Cauchy distribution has no mean.

The other options are the usual mix-up of roles. The normal has characteristic function \(e^{-t^{2}/2}\). The Laplace density \(\frac{1}{2}e^{-|x|}\) has characteristic function \(1/(1 + t^{2})\): density and characteristic function are swapped relative to the Cauchy. A gamma law has \((1 - it/\beta)^{-\alpha}\), which is complex-valued, while \(e^{-|t|}\) is real, as it must be for a symmetric law.

Discuss in Forum

73.

Suppose X is a real random variable on a probability space \((\Omega, A, P)\) and \(\{A_{n}, n \ge 1\}\) is a sequence of events in A such that :

\[ A_{n} \to \Omega \text{ as } n \to \infty. \]

Which of the following is true ?

A.\(X I_{A_{n}} \xrightarrow{P} 0\)
B.\(X I_{A_{n}} \xrightarrow{L} 0\)
C.\(X I_{A_{n}} \xrightarrow{P} X\)
D.\(X I_{A_{n}} \xrightarrow{P} 1\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

\(X I_{A_{n}} \xrightarrow{P} X\).

\(A_{n} \to \Omega\) means that every outcome eventually belongs to \(A_{n}\) (for the limit set to be \(\Omega\), \(\liminf A_{n} = \Omega\)). So for every \(\omega\), \(I_{A_{n}}(\omega) = 1\) for all large \(n\), and \(X I_{A_{n}}(\omega) = X(\omega)\) from some point on. That is almost sure, indeed pointwise, convergence to X, which implies convergence in probability.

In probability terms, \(P(|X I_{A_{n}} - X| \gt \varepsilon) \le P(A_{n}^{c}) \to 0\) by continuity of P.

The limit is not 0 (in probability or in law) unless X itself is 0, and it is not the constant 1 unless X is 1.

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74.

Suppose that X is a non-negative integer valued random variable having \(\varphi\) as its probability generating function. Then, variance (X) = 1, when :

A.\(\varphi(s) = s/(3 - 2s),\ 0 \le s \le 1\)
B.\(\varphi(s) = 1/(2 - s),\ 0 \le s \le 1\)
C.\(\varphi(s) = (1 + s)/2,\ 0 \le s \le 1\)
D.\(\varphi(s) = (s + s^{3})/2,\ 0 \le s \le 1\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

\(\text{Var}(X) = 1\) for \(\varphi(s) = (s + s^{3})/2\).

That pgf puts probability 1/2 on each of the values 1 and 3, so \(E(X) = 2\) and \(\text{Var}(X) = \frac{1}{2}(1 - 2)^{2} + \frac{1}{2}(3 - 2)^{2} = 1\). Equivalently \(\varphi'(1) = 2\), \(\varphi''(1) = 3\), and \(\text{Var} = \varphi''(1) + \varphi'(1) - \varphi'(1)^{2} = 3 + 2 - 4 = 1\).

The others give different variances:

\((1 + s)/2\) is Bernoulli(1/2), variance 1/4.
\(1/(2 - s) = \frac{1/2}{1 - s/2}\) is geometric on {0, 1, 2, ...} with \(p = 1/2\), variance \(q/p^{2} = 2\).
\(s/(3 - 2s) = \frac{s/3}{1 - 2s/3}\) is geometric on {1, 2, ...} with \(p = 1/3\), variance \(q/p^{2} = 6\).

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75.

Consider the following probability density function :

P(x) = 0 \(x \lt -2\) and \(x \gt 3\)
P(x) = a \(-2 \lt x \lt 2\)
P(x) = \(\frac{1}{12}\) \(2 \lt x \lt 3\)

The value of a is :

A.\(\frac{11}{12}\)
B.\(\frac{1}{12}\)
C.\(\frac{11}{48}\)
D.\(\frac{1}{48}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The constant is \(a = \frac{11}{48}\).

A probability density must integrate to 1 over all x. The density is zero outside -2 to 3, equal to a over the 4 units from -2 to 2, and 1/12 over the 1 unit from 2 to 3:

\[ \begin{aligned} 4a + \frac{1}{12} &= 1 \\ 4a &= \frac{11}{12} \\ a &= \frac{11}{48} \end{aligned} \]

The distractor 11/12 forgets to divide by the width 4 of the interval, and 1/12 and 1/48 do not satisfy the normalisation at all.

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76.

The ogive more than and ogive less than of the following grouped data

Class	Frequency
0-8	8
8-20	12
20-30	18
30-35	10
35-40	4

intersect at which of the following points ?

A.23.00
B.23.33
C.23.67
D.24.00

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The ogives intersect at 23.33, the median.

The 'less than' and 'more than' ogives cross at the median. The total frequency is N = 52, so N/2 = 26. The cumulative frequencies are 8, 20, 38, 48, 52, so the median class is 20-30, with lower limit 20, cumulative frequency before it 20, frequency 18 and width 10:

\[ \begin{aligned} \text{Median} &= 20 + \frac{26 - 20}{18} \times 10 \\ &= 20 + 3.33 = 23.33 \end{aligned} \]

The nearby values 23.00, 23.67 and 24.00 come from rounding or from using the wrong cumulative frequency or class width.

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77.

Which of the following is the smallest field containing a non-empty subset A of \(\Omega\) ?

A.Trivial field
B.\(\{A, A^{C}, \phi, \Omega\}\)
C.Power set of \(\Omega\)
D.None of the above

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The smallest field containing A is {A, A^C, φ, Ω}.

A field (algebra) of sets must contain \(\Omega\) and \(\phi\) and be closed under complements and finite unions. Any field containing A therefore contains \(A^{C}\), \(\phi\) and \(\Omega\). These four sets are already closed under complement and union (for instance \(A \cup A^{C} = \Omega\)), so they form a field, and it is the smallest one.

The trivial field \(\{\phi, \Omega\}\) does not contain A unless A is \(\Omega\). The power set is a field containing A, but generally a much larger one, so it is not the smallest.

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78.

Which of the following statements cannot be always true ?

A.Let \(X : \Omega \to \Omega^{*}\) and if \(\tau_{j}\) be the \(\sigma\)-field of subsets of \(\Omega^{*}\), then \(X^{-1}(\tau_{j})\) is also \(\sigma\)-field
B.\(\sigma\{X^{-1}(\tau)\} = X^{-1}\{\sigma(\tau)\}\) where, \(\tau\) denote the class of subsets of \(\Omega^{*}\) and \(X : \Omega \to \Omega^{*}\)
C.If \(B \cap C = \phi\), then \(X(B) \cap X(C) = \phi\) where \(X(A) = \{X(w) \mid w \in A\}\)
D.\(X \ge 0\) and \(E(X) = 0\), then X = 0 a.s.

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The statement that need not hold is B ∩ C = φ implies X(B) ∩ X(C) = φ.

Direct images do not preserve disjointness. If X is constant, say \(X(w) = 1\) for all w, then for disjoint non-empty B and C, \(X(B) = X(C) = \{1\}\), which intersect.

The others are standard results. Inverse images preserve complements and countable unions, so \(X^{-1}(\tau_{j})\) is a \(\sigma\)-field; the \(\sigma\)-field generated by \(X^{-1}(\tau)\) equals \(X^{-1}(\sigma(\tau))\); and if \(X \ge 0\) with \(E(X) = 0\), then \(P(X \gt 1/n) \le nE(X) = 0\) for every n, so X = 0 almost surely.

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79.

Let \(\Omega = \{1, 2, 3, \ldots, 10\}\), A = {1, 2} and \(\mathbf{F} = \{\phi, A, A^{C}, \Omega\}\), consider the following functions :

\[ X(w) = 100\ \forall\, w \in \Omega \]

\[ Y(w) = \begin{cases} 100 & \text{if } w \in A \\ 400 & \text{if } w \notin A \end{cases} \]

\[ Z(w) = w\ \forall\, w \in \Omega \]

Then which of the following is true ?

A.Only X and Y are F-measurable r.v.s
B.All the functions X, Y and Z are F-measurable r.v.s.
C.Only X is a F-measurable r.v.
D.All the functions X, Y, Z are not F-measurable

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

Only X and Y are F-measurable.

A function is F-measurable when the set where it takes each value (more generally each inverse image of a Borel set) lies in F = {φ, A, A^C, Ω}.

X is constant, so its inverse images are \(\phi\) or \(\Omega\): measurable. Y takes the value 100 on A and 400 on \(A^{C}\), so its inverse images are \(\phi\), A, \(A^{C}\) or \(\Omega\): measurable.

Z(w) = w takes ten different values. The set {Z = 3} = {3} is not in F, so Z is not measurable. That rules out 'all three', 'only X' and 'none'.

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80.

Suppose that probabilities are assigned to the subsets of \(\Omega = \{a, b, c, d, e, f\}\) according to the classical definition of probability. Which of the following statements is/are true for the sets A = {a, b}, B = {b, d} and C = {a, c, d} ?

(I) C and A are independent
(II) C and B are independent
(III) C and \(A \cup B\) are independent

A.Only (I) and (II) are true
B.Only (I) and (III) are true
C.Only (II) and (III) are true
D.All the three are false

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

Only (I) and (II) are true.

With six equally likely outcomes, \(P(A) = \frac{2}{6}\), \(P(B) = \frac{2}{6}\) and \(P(C) = \frac{3}{6}\).

(I) \(A \cap C = \{a\}\), so \(P(A \cap C) = \frac{1}{6} = \frac{1}{3} \cdot \frac{1}{2}\): independent.
(II) \(B \cap C = \{d\}\), so \(P(B \cap C) = \frac{1}{6} = \frac{1}{3} \cdot \frac{1}{2}\): independent.
(III) \(A \cup B = \{a, b, d\}\) with probability \(\frac{1}{2}\), and \(C \cap (A \cup B) = \{a, d\}\) with probability \(\frac{1}{3}\). Since \(\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \neq \frac{1}{3}\), these are not independent.

So (I) and (II) hold and (III) fails.

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