Probability Theory Questions and Answers

Q: Let \(X\) be a positive continuous random variable. Consider the transformation \(Y = X^{4}\). Then, the Jacobian of the inverse transformation is

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.

Q: Consider the matrix\[ A = \begin{bmatrix} X_{11} & X_{12} & X_{13} \\ X_{21} & X_{22} & X_{23} \\ X_{31} & X_{32} & X_{33} \end{bmatrix}, \]where \(X_{ij},\ 1 \leq i, j \leq 3\), are independent and identically distributed random variables with the probability mass function\[ f(x) = \begin{cases} \frac{1}{3}, & x = -1, 0, 1, \\ 0, & \text{otherwise.} \end{cases} \]The probability that \(A\) is orthogonal equals

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.

Q: Let \(A, B, C\) be three events such that \(P(C)P(C^{c}) \gt 0\). Consider the following statements.S1: \(P(A|C) \gt P(B|C)\) and \(P(A|C^{c}) \gt P(B|C^{c})\) together imply that \(P(A) \gt P(B)\).S2: \(P(A|C) = P(A|C^{c})\) implies that \(A\) and \(C\) are independent.Then

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 \(C\).

Q: Let the random variables \(X\) and \(Y\) denote the time to failure of component A and component B, respectively, of an instrument. The joint probability density function of \((X, Y)\) is given by\[ f(x, y) = \begin{cases} \frac{1}{10} e^{-\frac{1}{2}(y + 3 - x)}, & 5 \lt x \lt 10,\ y \gt x - 3, \\ 0, & \text{otherwise.} \end{cases} \]The probability, that component B fails before component A, is equal to

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.

Q: Let \(S = \{1, 2, 3, \ldots\}\) and suppose that every subset of \(S\) is an event. Let \(\mathcal{P}(S)\) denote the power set of \(S\). Which of the following statements is/are true?

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.

Q: Let X be a discrete r.v. with the following pmf :\[ P[X = x] = \begin{cases} k; & x = 0, \pm j; \ j = 1, 2, \ldots n \\ 0; & \text{otherwise} \end{cases} \]Let \(Y = X^{2}\), then, \(P[Y = 4]\) is :

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

Q: Let \(X_{1}\) and \(X_{2}\) be jointly distributed with pdf \(f(x_{1}, x_{2})\) given by :\[ \begin{aligned} f(x_{1}, x_{2}) &= \tfrac{1}{4}\{1 + x_{1}x_{2}(x_{1}^{2} - x_{2}^{2})\}; \\ &\qquad |x_{1}| \lt 1 \text{ and } |x_{2}| \lt 1 \\ &= 0; \quad \text{otherwise} \end{aligned} \]Then, the characteristic function of \(X_{1}\) is :

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 \(t = \pi\).

Q: Let X be a r.v. such that variance of X is \(\dfrac{1}{2}\). Then, an upper bound for \(P[|X - FX| \gt 1]\) as given by the Chebychev's inequality is :

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 \(P[|X - E(X)| \lt 1] \ge 1 - 1/2 = 1/2\).

Q: Let \((\Omega, F, P)\) be a probability space. Let \(\{A_{n}\}\) be a sequence of events such that \(P(A_{n}) = 1\) for each \(n \ge 1\). Then \(P\left(\bigcap_{n=1}^{\infty} A_{n}\right)\) =

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.

Q: Let X be a zero-mean unit variance Gaussian random variable. Define the random variable Y as :\[ Y = \begin{cases} X & \text{if } |X| \le \alpha \\ -X & \text{if } |X| \gt \alpha \end{cases} \]\(\alpha\) is some positive real number. What is the distribution of X + Y ?

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.

Let \(X\) be a positive continuous random variable. Consider the transformation \(Y = X^{4}\). Then, the Jacobian of the inverse transformation is

A.\(\frac{3}{4}\, y^{-\frac{3}{4}}\)
B.\(y^{\frac{3}{4}}\)
C.\(\frac{1}{4}\, y^{-\frac{3}{4}}\)
D.\(y^{\frac{1}{4}}\)

Exam Year(s)

IIT JAM - 2026,

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.

Discuss in Forum

Consider the matrix

\[ A = \begin{bmatrix} X_{11} & X_{12} & X_{13} \\ X_{21} & X_{22} & X_{23} \\ X_{31} & X_{32} & X_{33} \end{bmatrix}, \]

where \(X_{ij},\ 1 \leq i, j \leq 3\), are independent and identically distributed random variables with the probability mass function

\[ f(x) = \begin{cases} \frac{1}{3}, & x = -1, 0, 1, \\ 0, & \text{otherwise.} \end{cases} \]

The probability that \(A\) is orthogonal equals

A.\(\frac{2^{4}}{3^{8}}\)
B.\(\frac{1}{3^{5}}\)
C.\(\frac{2^{5}}{3^{9}}\)
D.\(\frac{2^{3}}{3^{7}}\)

Exam Year(s)

IIT JAM - 2026,

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.

Discuss in Forum

Let \(A, B, C\) be three events such that \(P(C)P(C^{c}) \gt 0\). Consider the following statements.

S1: \(P(A|C) \gt P(B|C)\) and \(P(A|C^{c}) \gt P(B|C^{c})\) together imply that \(P(A) \gt P(B)\).

S2: \(P(A|C) = P(A|C^{c})\) implies that \(A\) and \(C\) are independent.

Then

A.S1 is correct and S2 is NOT correct
B.S1 is NOT correct and S2 is correct
C.Both S1 and S2 are correct
D.Neither S1 nor S2 is correct

Exam Year(s)

IIT JAM - 2026,

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 \(C\).

Discuss in Forum

Let the random variables \(X\) and \(Y\) denote the time to failure of component A and component B, respectively, of an instrument. The joint probability density function of \((X, Y)\) is given by

\[ f(x, y) = \begin{cases} \frac{1}{10} e^{-\frac{1}{2}(y + 3 - x)}, & 5 \lt x \lt 10,\ y \gt x - 3, \\ 0, & \text{otherwise.} \end{cases} \]

The probability, that component B fails before component A, is equal to

A.\(\frac{e\sqrt{e} - 1}{e\sqrt{e}}\)
B.\(\frac{e\sqrt{e}}{e\sqrt{e} + 1}\)
C.\(\frac{e\sqrt{e} - 1}{e\sqrt{e} + 1}\)
D.\(\frac{1}{e\sqrt{e}}\)

Exam Year(s)

IIT JAM - 2026,

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.

Discuss in Forum

Let \(S = \{1, 2, 3, \ldots\}\) and suppose that every subset of \(S\) is an event. Let \(\mathcal{P}(S)\) denote the power set of \(S\). Which of the following statements is/are true?

A.There exists a probability function \(P: \mathcal{P}(S) \to [0, \infty)\) such that \(P(\{n\}) = \frac{1}{n+1},\ n \geq 1\)
B.There exists a probability function \(P: \mathcal{P}(S) \to [0, \infty)\) such that \(P(A) = 0\) if \(A\) is a finite set and \(P(A) = 1\) if \(A\) is an infinite set
C.There exists a probability function \(P: \mathcal{P}(S) \to [0, \infty)\) such that \(P(\{1, 2, \ldots, n\}) = \int_{1}^{n} \frac{1}{x}\,dx,\ n \geq 1\)
D.There exists a probability function \(P: \mathcal{P}(S) \to [0, \infty)\) such that \(P(\{1, 2, \ldots, n\}) = \int_{0}^{n} e^{-x}\,dx,\ n \geq 1\)

Exam Year(s)

IIT JAM - 2026,

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.

Discuss in Forum

Let X be a discrete r.v. with the following pmf :

\[ P[X = x] = \begin{cases} k; & x = 0, \pm j; \ j = 1, 2, \ldots n \\ 0; & \text{otherwise} \end{cases} \]

Let \(Y = X^{2}\), then, \(P[Y = 4]\) is :

A.\(\dfrac{1}{2n + 1}\)
B.\(\dfrac{2}{n + 1}\)
C.\(\dfrac{n}{2n + 1}\)
D.\(\dfrac{2}{2n + 1}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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 \(X^{2} = 4\).

Discuss in Forum

Let \(X_{1}\) and \(X_{2}\) be jointly distributed with pdf \(f(x_{1}, x_{2})\) given by :

\[ \begin{aligned} f(x_{1}, x_{2}) &= \tfrac{1}{4}\{1 + x_{1}x_{2}(x_{1}^{2} - x_{2}^{2})\}; \\ &\qquad |x_{1}| \lt 1 \text{ and } |x_{2}| \lt 1 \\ &= 0; \quad \text{otherwise} \end{aligned} \]

Then, the characteristic function of \(X_{1}\) is :

A.\(\dfrac{\sin t}{t}\)
B.\(\dfrac{\cos t}{t}\)
C.\(\dfrac{\sin t}{t^{2}}\)
D.\(\dfrac{t}{\sin t}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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 \(t = \pi\).

Discuss in Forum

Let X be a r.v. such that variance of X is \(\dfrac{1}{2}\). Then, an upper bound for \(P[|X - FX| \gt 1]\) as given by the Chebychev's inequality is :

A.\(\dfrac{1}{4}\)
B.\(\dfrac{1}{2}\)
C.1
D.\(\dfrac{3}{4}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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 \(P[|X - E(X)| \lt 1] \ge 1 - 1/2 = 1/2\).

Discuss in Forum

Let \((\Omega, F, P)\) be a probability space. Let \(\{A_{n}\}\) be a sequence of events such that \(P(A_{n}) = 1\) for each \(n \ge 1\). Then \(P\left(\bigcap_{n=1}^{\infty} A_{n}\right)\) =

A.Zero
B.One
C.Infinity
D.Not defined

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

Discuss in Forum

10.

Let X be a zero-mean unit variance Gaussian random variable. Define the random variable Y as :

\[ Y = \begin{cases} X & \text{if } |X| \le \alpha \\ -X & \text{if } |X| \gt \alpha \end{cases} \]

\(\alpha\) is some positive real number. What is the distribution of X + Y ?

A.Gaussian (0, 1)
B.Gaussian (0, 2)
C.Not Gaussian, but having a distribution which is discontinuous at origin
D.Not Gaussian, but a continuous distribution

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

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