Standard Probability Distributions Questions and Answers

Q: Let \(X\) be a random variable having the moment generating function \(M_{X}(t) = e^{2(e^{t}-1)},\ t \in \mathbb{R}\). Then, \(Var(2X + 3)\) equals

The variance is 8.The moment generating function \(e^{\lambda(e^{t} - 1)}\) belongs to the Poisson distribution with mean \(\lambda\). Here \(\lambda = 2\), so \(X \sim\) Poisson(2) and \(Var(X) = 2\).Adding a constant does not change variance, and multiplying by 2 multiplies it by \(2^{2}\):\[ Var(2X + 3) = 4\,Var(X) = 4 \times 2 = 8 \]16 squares the variance as well as the coefficient. 4 is \(2^{2}\) with \(Var(X)\) forgotten, and 6 adds the constant 3 to \(2\,Var(X)\), but a shift has no effect on spread.

Q: If \(X\) is a random variable with the cumulative distribution function\[ F(x) = \begin{cases} 0, & x \leq 0, \\ 1 - e^{-\frac{x^{2}}{2}}, & x \gt 0, \end{cases} \]then \(E(X)\) equals

The mean is \(\sqrt{\frac{\pi}{2}}\).For a non-negative random variable, \(E(X) = \int_{0}^{\infty} (1 - F(x))\,dx\). Here \(1 - F(x) = e^{-x^{2}/2}\), so\[ \begin{aligned} E(X) &= \int_{0}^{\infty} e^{-\frac{x^{2}}{2}}\,dx \\ &= \frac{1}{2}\sqrt{2\pi} = \sqrt{\frac{\pi}{2}} \end{aligned} \]using half of the Gaussian integral \(\int_{-\infty}^{\infty} e^{-x^{2}/2}dx = \sqrt{2\pi}\). This is the Rayleigh distribution with scale 1, whose mean is about 1.25.\(\sqrt{2\pi}\) is the full Gaussian integral, forgetting that the range is only \((0, \infty)\). \(\sqrt{\frac{2}{\pi}}\) is the reciprocal. 1 is the mode of this distribution, where the density \(xe^{-x^{2}/2}\) peaks, not its mean.

Q: Let X be a normal random variable with mean 1 and variance 1. Define the events :\(A = \{-2 \lt X \lt 1\}\),\(B = \{-1 \lt X \lt 1\}\),\(C = \{0 \lt X \lt 2\}\).Which of the following statements is correct ?

The correct ordering is \(P(B) \lt P(A) \lt P(C)\).Standardise with \(Z = X - 1\), since the mean is 1 and the standard deviation is 1:\(P(A) = P(-3 \lt Z \lt 0) \approx 0.4987\)\(P(B) = P(-2 \lt Z \lt 0) \approx 0.4772\)\(P(C) = P(-1 \lt Z \lt 1) \approx 0.6827\)A and B both end at the mean, and A reaches further into the tail, so \(P(B) \lt P(A)\). C is an interval of the same length as B but centred on the mean, where the density is highest, so it is the largest.The orderings putting C first or making A and B equal ignore that the intervals sit at different distances from the mean of 1, not 0.

Q: If X is \(F(m, n)\) (F-distribution with m and n degrees of freedom) and Y is \(F(n, m)\), given that \(P[X \ge c] = a\) then \(P\left[Y \ge \dfrac{1}{c}\right]\) is :

\(P[Y \ge 1/c] = \)\(1 - a\).If \(X \sim F(m, n)\), then \(1/X \sim F(n, m)\): the F variable is a ratio of two scaled chi-squares, and inverting it swaps numerator and denominator. So Y has the same distribution as \(1/X\), and for \(c \gt 0\)\[ \begin{aligned} P\left[Y \ge \frac{1}{c}\right] &= P\left[\frac{1}{X} \ge \frac{1}{c}\right] \\ &= P[X \le c] \\ &= 1 - P[X \ge c] = 1 - a \end{aligned} \](the F distribution is continuous, so \(P[X = c] = 0\)).This is the identity behind printed F tables: the lower percentage point of \(F(m, n)\) is the reciprocal of the upper point of \(F(n, m)\). Answer a forgets that inverting reverses the inequality, and 1 or \(a/2\) have no basis.

Q: If X is distributed as Binomial \((n, p)\), \(0 \lt p \lt 1\), then :

The true statement is \(n - X\) is distributed \(B(n, 1 - p)\).X counts the successes in n independent trials with success probability p. Then \(n - X\) counts the failures, and each trial is a failure with probability \(1 - p\), independently. So \(n - X \sim B(n, 1 - p)\). Formally, \(P(n - X = k) = P(X = n - k)\), which is \(\binom{n}{n-k}p^{n-k}(1 - p)^{k} = \binom{n}{k}(1 - p)^{k}p^{n-k}\).The others do not make sense as binomial laws. \(-X\) takes negative values, and \(B(-n, p)\) has no meaning. \(X - k\) is a shift that can be negative, while \(B(n - k, p)\) lives on \(0, \ldots, n - k\). \(1/X\) is not integer-valued (and is undefined when X = 0), and \(1/p \gt 1\) cannot be a probability.

Q: Suppose X and Y are independent r.v.'s such that :\[ \begin{aligned} f(x) &= \frac{e^{-x} x^{\alpha - 1}}{\Gamma(\alpha)}; \ x \gt 0, \ \alpha \gt 0 \\ f(y) &= \frac{e^{-y} y^{\beta - 1}}{\Gamma(\beta)}; \ y \gt 0, \ \beta \gt 0 \end{aligned} \]The distribution of \(\dfrac{X}{X + Y}\) is :

\(X/(X + Y)\) has the Beta \((\alpha, \beta)\) distribution.X and Y are independent Gamma variables with shapes \(\alpha\) and \(\beta\) and a common scale 1. Put \(U = X/(X + Y)\) and \(V = X + Y\), so \(X = UV\), \(Y = (1 - U)V\), with Jacobian \(V\). The joint density factorises:\[ \begin{aligned} g(u, v) &= \frac{(uv)^{\alpha - 1}((1 - u)v)^{\beta - 1} e^{-v}}{\Gamma(\alpha)\Gamma(\beta)} v \\ &= u^{\alpha - 1}(1 - u)^{\beta - 1} \cdot v^{\alpha + \beta - 1}e^{-v} \cdot \frac{1}{\Gamma(\alpha)\Gamma(\beta)} \end{aligned} \]So U and V are independent, \(V \sim\) Gamma \((\alpha + \beta, 1)\) and \(U \sim\) Beta \((\alpha, \beta)\) on (0, 1).Gamma \((\alpha + \beta, 1)\) is the distribution of the sum X + Y, not of the ratio, which lies in (0, 1) and so cannot be Gamma at all. Beta \((\alpha + \beta, \alpha - \beta)\) is not even defined when \(\alpha \le \beta\).

Q: Let X be Poisson variate with mean \(\lambda\). Then distribution function of X, evaluated at 1.5 is :

The value is equal to \((\lambda + 1)e^{-\lambda}\).A Poisson variable takes only integer values, so its distribution function is a step function, and \(F(1.5) = P(X \le 1.5) = P(X \le 1)\):\[ \begin{aligned} &P(X = 0) + P(X = 1) \\ &= e^{-\lambda} + \lambda e^{-\lambda} \\ &= (\lambda + 1)e^{-\lambda} \end{aligned} \]It is not 0, since \(P(X = 0) \gt 0\), and it is at least \(e^{-\lambda}\), so B is false. Option D would need it to exceed \(\lambda(\lambda + 1)e^{-\lambda}\), which fails whenever \(\lambda \ge 1\).

Q: A coin is biased so that head is twice as likely to occur as a tail. If a coin is tossed four times, what is the probability of getting two tails and two heads ?

The probability is 8/27.Since a head is twice as likely as a tail, \(P(H) = 2/3\) and \(P(T) = 1/3\). Two heads and two tails in four tosses is a binomial event:\[ \begin{aligned} P &= \binom{4}{2}\left(\tfrac{2}{3}\right)^{2}\left(\tfrac{1}{3}\right)^{2} \\ &= 6 \cdot \tfrac{4}{81} = \tfrac{24}{81} = \tfrac{8}{27} \end{aligned} \]3/8 is the answer for a fair coin (\(6/16\)). 4/27 (that is, 12/81) counts only three of the six orders of HHTT, and 1/8 has no basis here.

Q: Let \(x\) and \(y\) be two independent normal random variables with mean \(\mu_{1}\) and \(\mu_{2}\) respectively and common variance \(\sigma^{2}\). For \(0 \lt \alpha \lt 1\) if the \(100\alpha\)th quantile of \(x\) is larger than \(100\alpha\)th quantile of \(y\), then :

The answer is \(\mu_{1} \gt \mu_{2}\).The \(100\alpha\)th quantile of \(N(\mu, \sigma^{2})\) is \(\mu + \sigma z_{\alpha}\), where \(z_{\alpha}\) is the standard normal quantile. With a common \(\sigma\),\[ \begin{aligned} &\mu_{1} + \sigma z_{\alpha} \gt \mu_{2} + \sigma z_{\alpha} \\ &\iff \mu_{1} \gt \mu_{2} \end{aligned} \]Because the two distributions differ only by a shift, every quantile of \(x\) exceeds the corresponding quantile of \(y\) by \(\mu_{1} - \mu_{2}\). So the means cannot be equal or reversed, and something definite can be said.

Q: The \(p\)th quantile of a standard normal random variable is \(\xi_{p}\). Then, the \(p\)th quantile of a chi-square random variable with 1 degree of freedom is :

The \(p\)th quantile of \(\chi_{1}^{2}\) is \(\xi_{(1 + p)/2}^{2}\).A \(\chi_{1}^{2}\) variable is \(Z^{2}\) with \(Z\) standard normal. For \(c \gt 0\),\[ P(Z^{2} \le c^{2}) = P(-c \le Z \le c) = 2\Phi(c) - 1 \]Setting this equal to \(p\) gives \(\Phi(c) = (1 + p)/2\), so \(c = \xi_{(1 + p)/2}\) and the quantile of \(Z^{2}\) is \(c^{2}\).\(\xi_{p}^{2}\) ignores the two tails (and fails for \(p \lt 1/2\), where \(\xi_{p} \lt 0\)); \(\xi_{1 + p}\) is undefined since \(1 + p \gt 1\); and \(2\xi\) is not even a quantile expression.

Let \(X\) be a random variable having the moment generating function \(M_{X}(t) = e^{2(e^{t}-1)},\ t \in \mathbb{R}\). Then, \(Var(2X + 3)\) equals

A.16
B.4
C.6
D.8

Exam Year(s)

IIT JAM - 2026,

Answer: D

The variance is 8.

The moment generating function \(e^{\lambda(e^{t} - 1)}\) belongs to the Poisson distribution with mean \(\lambda\). Here \(\lambda = 2\), so \(X \sim\) Poisson(2) and \(Var(X) = 2\).

Adding a constant does not change variance, and multiplying by 2 multiplies it by \(2^{2}\):

\[ Var(2X + 3) = 4\,Var(X) = 4 \times 2 = 8 \]

16 squares the variance as well as the coefficient. 4 is \(2^{2}\) with \(Var(X)\) forgotten, and 6 adds the constant 3 to \(2\,Var(X)\), but a shift has no effect on spread.

Discuss in Forum

If \(X\) is a random variable with the cumulative distribution function

\[ F(x) = \begin{cases} 0, & x \leq 0, \\ 1 - e^{-\frac{x^{2}}{2}}, & x \gt 0, \end{cases} \]

then \(E(X)\) equals

A.\(\sqrt{\frac{\pi}{2}}\)
B.\(\sqrt{2\pi}\)
C.\(\sqrt{\frac{2}{\pi}}\)
D.\(1\)

Exam Year(s)

IIT JAM - 2026,

Answer: A

The mean is \(\sqrt{\frac{\pi}{2}}\).

For a non-negative random variable, \(E(X) = \int_{0}^{\infty} (1 - F(x))\,dx\). Here \(1 - F(x) = e^{-x^{2}/2}\), so

\[ \begin{aligned} E(X) &= \int_{0}^{\infty} e^{-\frac{x^{2}}{2}}\,dx \\ &= \frac{1}{2}\sqrt{2\pi} = \sqrt{\frac{\pi}{2}} \end{aligned} \]

using half of the Gaussian integral \(\int_{-\infty}^{\infty} e^{-x^{2}/2}dx = \sqrt{2\pi}\). This is the Rayleigh distribution with scale 1, whose mean is about 1.25.

\(\sqrt{2\pi}\) is the full Gaussian integral, forgetting that the range is only \((0, \infty)\). \(\sqrt{\frac{2}{\pi}}\) is the reciprocal. 1 is the mode of this distribution, where the density \(xe^{-x^{2}/2}\) peaks, not its mean.

Discuss in Forum

Let X be a normal random variable with mean 1 and variance 1. Define the events :

\(A = \{-2 \lt X \lt 1\}\),
\(B = \{-1 \lt X \lt 1\}\),
\(C = \{0 \lt X \lt 2\}\).

Which of the following statements is correct ?

A.\(P(C) \lt P(B) \lt P(A)\)
B.\(P(A) = P(B) \lt P(C)\)
C.\(P(A) = P(B) = P(C)\)
D.\(P(B) \lt P(A) \lt P(C)\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The correct ordering is \(P(B) \lt P(A) \lt P(C)\).

Standardise with \(Z = X - 1\), since the mean is 1 and the standard deviation is 1:

\(P(A) = P(-3 \lt Z \lt 0) \approx 0.4987\)
\(P(B) = P(-2 \lt Z \lt 0) \approx 0.4772\)
\(P(C) = P(-1 \lt Z \lt 1) \approx 0.6827\)

A and B both end at the mean, and A reaches further into the tail, so \(P(B) \lt P(A)\). C is an interval of the same length as B but centred on the mean, where the density is highest, so it is the largest.

The orderings putting C first or making A and B equal ignore that the intervals sit at different distances from the mean of 1, not 0.

Discuss in Forum

If X is \(F(m, n)\) (F-distribution with m and n degrees of freedom) and Y is \(F(n, m)\), given that \(P[X \ge c] = a\) then \(P\left[Y \ge \dfrac{1}{c}\right]\) is :

A.\(a\)
B.1
C.\(\dfrac{a}{2}\)
D.\(1 - a\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

\(P[Y \ge 1/c] = \)\(1 - a\).

If \(X \sim F(m, n)\), then \(1/X \sim F(n, m)\): the F variable is a ratio of two scaled chi-squares, and inverting it swaps numerator and denominator. So Y has the same distribution as \(1/X\), and for \(c \gt 0\)

\[ \begin{aligned} P\left[Y \ge \frac{1}{c}\right] &= P\left[\frac{1}{X} \ge \frac{1}{c}\right] \\ &= P[X \le c] \\ &= 1 - P[X \ge c] = 1 - a \end{aligned} \]

(the F distribution is continuous, so \(P[X = c] = 0\)).

This is the identity behind printed F tables: the lower percentage point of \(F(m, n)\) is the reciprocal of the upper point of \(F(n, m)\). Answer a forgets that inverting reverses the inequality, and 1 or \(a/2\) have no basis.

Discuss in Forum

If X is distributed as Binomial \((n, p)\), \(0 \lt p \lt 1\), then :

A.\(-X\) is distributed \(B(-n, p)\)
B.\(n - X\) is distributed \(B(n, 1 - p)\)
C.\(X - k\) is distributed \(B(n - k, p)\), when k is constant
D.\(\dfrac{1}{X}\) is distributed as \(B\left(n, \dfrac{1}{p}\right)\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The true statement is \(n - X\) is distributed \(B(n, 1 - p)\).

X counts the successes in n independent trials with success probability p. Then \(n - X\) counts the failures, and each trial is a failure with probability \(1 - p\), independently. So \(n - X \sim B(n, 1 - p)\). Formally, \(P(n - X = k) = P(X = n - k)\), which is \(\binom{n}{n-k}p^{n-k}(1 - p)^{k} = \binom{n}{k}(1 - p)^{k}p^{n-k}\).

The others do not make sense as binomial laws. \(-X\) takes negative values, and \(B(-n, p)\) has no meaning. \(X - k\) is a shift that can be negative, while \(B(n - k, p)\) lives on \(0, \ldots, n - k\). \(1/X\) is not integer-valued (and is undefined when X = 0), and \(1/p \gt 1\) cannot be a probability.

Discuss in Forum

Suppose X and Y are independent r.v.'s such that :

\[ \begin{aligned} f(x) &= \frac{e^{-x} x^{\alpha - 1}}{\Gamma(\alpha)}; \ x \gt 0, \ \alpha \gt 0 \\ f(y) &= \frac{e^{-y} y^{\beta - 1}}{\Gamma(\beta)}; \ y \gt 0, \ \beta \gt 0 \end{aligned} \]

The distribution of \(\dfrac{X}{X + Y}\) is :

A.Beta \((\alpha, \beta)\)
B.Gamma \((\alpha + \beta, 1)\)
C.Beta \((\alpha + \beta, \alpha - \beta)\)
D.Gamma \((\alpha\beta, 1)\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

\(X/(X + Y)\) has the Beta \((\alpha, \beta)\) distribution.

X and Y are independent Gamma variables with shapes \(\alpha\) and \(\beta\) and a common scale 1. Put \(U = X/(X + Y)\) and \(V = X + Y\), so \(X = UV\), \(Y = (1 - U)V\), with Jacobian \(V\). The joint density factorises:

\[ \begin{aligned} g(u, v) &= \frac{(uv)^{\alpha - 1}((1 - u)v)^{\beta - 1} e^{-v}}{\Gamma(\alpha)\Gamma(\beta)} v \\ &= u^{\alpha - 1}(1 - u)^{\beta - 1} \cdot v^{\alpha + \beta - 1}e^{-v} \cdot \frac{1}{\Gamma(\alpha)\Gamma(\beta)} \end{aligned} \]

So U and V are independent, \(V \sim\) Gamma \((\alpha + \beta, 1)\) and \(U \sim\) Beta \((\alpha, \beta)\) on (0, 1).

Gamma \((\alpha + \beta, 1)\) is the distribution of the sum X + Y, not of the ratio, which lies in (0, 1) and so cannot be Gamma at all. Beta \((\alpha + \beta, \alpha - \beta)\) is not even defined when \(\alpha \le \beta\).

Discuss in Forum

Let X be Poisson variate with mean \(\lambda\). Then distribution function of X, evaluated at 1.5 is :

A.0
B.less than \(e^{-\lambda}\)
C.equal to \((\lambda + 1)e^{-\lambda}\)
D.greater than \(\lambda(\lambda + 1)e^{-\lambda}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The value is equal to \((\lambda + 1)e^{-\lambda}\).

A Poisson variable takes only integer values, so its distribution function is a step function, and \(F(1.5) = P(X \le 1.5) = P(X \le 1)\):

\[ \begin{aligned} &P(X = 0) + P(X = 1) \\ &= e^{-\lambda} + \lambda e^{-\lambda} \\ &= (\lambda + 1)e^{-\lambda} \end{aligned} \]

It is not 0, since \(P(X = 0) \gt 0\), and it is at least \(e^{-\lambda}\), so B is false. Option D would need it to exceed \(\lambda(\lambda + 1)e^{-\lambda}\), which fails whenever \(\lambda \ge 1\).

Discuss in Forum

A coin is biased so that head is twice as likely to occur as a tail. If a coin is tossed four times, what is the probability of getting two tails and two heads ?

A.\(\dfrac{3}{8}\)
B.\(\dfrac{8}{27}\)
C.\(\dfrac{1}{8}\)
D.\(\dfrac{4}{27}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The probability is 8/27.

Since a head is twice as likely as a tail, \(P(H) = 2/3\) and \(P(T) = 1/3\). Two heads and two tails in four tosses is a binomial event:

\[ \begin{aligned} P &= \binom{4}{2}\left(\tfrac{2}{3}\right)^{2}\left(\tfrac{1}{3}\right)^{2} \\ &= 6 \cdot \tfrac{4}{81} = \tfrac{24}{81} = \tfrac{8}{27} \end{aligned} \]

3/8 is the answer for a fair coin (\(6/16\)). 4/27 (that is, 12/81) counts only three of the six orders of HHTT, and 1/8 has no basis here.

Discuss in Forum

Let \(x\) and \(y\) be two independent normal random variables with mean \(\mu_{1}\) and \(\mu_{2}\) respectively and common variance \(\sigma^{2}\). For \(0 \lt \alpha \lt 1\) if the \(100\alpha\)th quantile of \(x\) is larger than \(100\alpha\)th quantile of \(y\), then :

A.\(\mu_{1} = \mu_{2}\)
B.\(\mu_{1} \gt \mu_{2}\)
C.\(\mu_{1} \lt \mu_{2}\)
D.Nothing can be said about the relationship between \(\mu_{1}\) and \(\mu_{2}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The answer is \(\mu_{1} \gt \mu_{2}\).

The \(100\alpha\)th quantile of \(N(\mu, \sigma^{2})\) is \(\mu + \sigma z_{\alpha}\), where \(z_{\alpha}\) is the standard normal quantile. With a common \(\sigma\),

\[ \begin{aligned} &\mu_{1} + \sigma z_{\alpha} \gt \mu_{2} + \sigma z_{\alpha} \\ &\iff \mu_{1} \gt \mu_{2} \end{aligned} \]

Because the two distributions differ only by a shift, every quantile of \(x\) exceeds the corresponding quantile of \(y\) by \(\mu_{1} - \mu_{2}\). So the means cannot be equal or reversed, and something definite can be said.

Discuss in Forum

10.

The \(p\)th quantile of a standard normal random variable is \(\xi_{p}\). Then, the \(p\)th quantile of a chi-square random variable with 1 degree of freedom is :

A.\(\xi_{p}^{2}\)
B.\(\xi_{1 + p}^{2}\)
C.\(\xi_{(1 + p)/2}^{2}\)
D.\(2\xi\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The \(p\)th quantile of \(\chi_{1}^{2}\) is \(\xi_{(1 + p)/2}^{2}\).

A \(\chi_{1}^{2}\) variable is \(Z^{2}\) with \(Z\) standard normal. For \(c \gt 0\),

\[ P(Z^{2} \le c^{2}) = P(-c \le Z \le c) = 2\Phi(c) - 1 \]

Setting this equal to \(p\) gives \(\Phi(c) = (1 + p)/2\), so \(c = \xi_{(1 + p)/2}\) and the quantile of \(Z^{2}\) is \(c^{2}\).

\(\xi_{p}^{2}\) ignores the two tails (and fails for \(p \lt 1/2\), where \(\xi_{p} \lt 0\)); \(\xi_{1 + p}\) is undefined since \(1 + p \gt 1\); and \(2\xi\) is not even a quantile expression.

Standard Probability Distributions Questions and Answers

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