Probability Theory Questions and Answers

Q: Suppose X is a continuous random variable having the following probability density function :\(f(x) = C(4x - 2x^{2}),\ 0 \lt x \lt 2\)\(= 0,\) otherwiseWhat is the value of \(P[X \gt 1]\) ?

\(P[X \gt 1] = \) 1/2.The density is proportional to \(4x - 2x^{2} = 2x(2 - x)\) on (0, 2). Replacing \(x\) by \(2 - x\) leaves it unchanged, so the density is symmetric about \(x = 1\), and half the probability lies on each side of 1.Directly: \(\int_{0}^{2} (4x - 2x^{2})dx = 8 - 16/3 = 8/3\), so \(C = 3/8\), and\[ \begin{aligned} P[X \gt 1] &= \tfrac{3}{8}\int_{1}^{2} (4x - 2x^{2})dx \\ &= \tfrac{3}{8}\left(6 - \tfrac{14}{3}\right) = \tfrac{1}{2} \end{aligned} \]C/2 leaves the constant unevaluated, and 2/3 and 1/4 do not respect the symmetry.

Q: For a data-set of \(n\) observations the sample mean, median and mode are 63, 60 and 58 respectively. Hence the data are :

The data are skewed to the right.Here mean (63) \(\gt\) median (60) \(\gt\) mode (58). In a positively skewed (right-skewed) distribution, a long tail of large values pulls the mean above the median, and the median above the mode. Pearson's coefficient of skewness, (mean - mode)/s, is positive.For symmetric data the three measures would coincide, and left skew would reverse the order. A single reported mode says nothing about bimodality.

Q: Suppose \(E[X] = 5\). Which of the following is always false ?

The statement that is always false is \(E[X^{2}] \lt (E[X])^{2}\).\(\mathrm{Var}(X) = E[X^{2}] - (E[X])^{2} \ge 0\), so \(E[X^{2}] \ge (E[X])^{2} = 25\) for every random variable. The strict reverse inequality can never hold.The other statements are not always false. (C) is always true. (B) is true: \(E[X^{2}] = 25\) means variance 0, so X equals 5 with probability 1. (D) is usually true by Jensen's inequality (exp is strictly convex), and fails only in the degenerate case.

Q: Suppose X and Y are independent r.v.s. each with mean 1.Let \(\phi_{X}(t)\) denote the characteristic function of a random variable. Which of the following is false ?

The false statement is \(E[XY \mid X] = 1\).Given X, the factor X is known and Y is independent of X, so\[ E[XY \mid X] = X\,E[Y] = X \cdot 1 = X \]which is a random variable, not the constant 1. (Its expectation is 1.)The others are true by independence: the characteristic function of a sum of independent variables is the product; \(X^{2}\) and \(Y^{2}\) are independent, so the same holds for them; and functions of independent variables, such as \(1 - X^{2}\) and \(\exp(Y)\), are independent.

Q: Let the joint distribution of (X, Y) be specified by :Y \ X123411/163/321/85/3223/321/85/323/16Then \(P[X = 2Y]\) is :

\(P[X = 2Y] = \) 9/32.\(X = 2Y\) happens only at \((X, Y) = (2, 1)\) and \((4, 2)\):\[ \begin{aligned} \frac{3}{32} + \frac{3}{16} &= \frac{3}{32} + \frac{6}{32} \\ &= \frac{9}{32} \end{aligned} \](The table's total is 1, as it should be.) 0 ignores these cells, and 11/32 or 1/2 add other cells, such as (2, 2) or (4, 1), where \(X \ne 2Y\).

Q: Let \(\mathbf{B}\) be the \(\sigma\)-field of subsets of R generated by \(\{(a, b) \mid a, b \in R\}\). Which of the following is not true ?

The statement that is not true is all uncountable subsets of R belong to \(\mathbf{B}\).\(\mathbf{B}\) is the Borel \(\sigma\)-field. It has the cardinality of the continuum, while there are \(2^{\mathfrak{c}}\) subsets of R, so most subsets, and in particular many uncountable ones, are not Borel. (A Vitali set is even non-Lebesgue-measurable.)The others are true. Singletons are closed, \(\{a\} = \bigcap (a - 1/n, a + 1/n)\), so countable sets are countable unions of Borel sets. Closed intervals are complements of open sets, and \([a, b) = \{a\} \cup (a, b)\).

Q: Let R be the set of real numbers, \(\mathbf{B}\) the Borel \(\sigma\)-field. Let '\(m\)' denotes the Lebesgue measure and \(\mu\) the measure defined by the function \(F(x) = (1 - e^{-x})\) if \(x \gt 0\) and \(F(x) = 0\) elsewhere.Consider the statements :(I) \(\mu \lt\lt m\), (i.e. \(\mu\) is absolutely continuous w.r.t. \(m\))(II) \(m \lt\lt \mu\)(III) \(\dfrac{d\mu}{dm}(x) = e^{-x}\ x \gt 0\), \(\dfrac{d\mu}{dm}(x) = 0\), \(x \le 0\)Which of the above statements are true ?

The true statements are (I) and (III) only.\(\mu\) is the exponential distribution, with density \(e^{-x}\) on \(x \gt 0\) and 0 elsewhere. Whenever \(m(A) = 0\), \(\mu(A) = \int_{A} e^{-x}1_{x \gt 0}\,dx = 0\), so \(\mu \ll m\) (I). By the Radon-Nikodym theorem the derivative \(d\mu/dm\) is that density, \(e^{-x}\) for \(x \gt 0\) and 0 for \(x \le 0\) (III).(II) fails: \(\mu((-\infty, 0]) = 0\) but \(m((-\infty, 0]) = \infty\), so a \(\mu\)-null set need not be \(m\)-null.

Q: Suppose a distribution function \(F : \mathbf{R} \to [0, 1]\) of a random variable X is as follows :\[ F(x) = \begin{cases} 0, & \text{if } x \lt -2, \\ 1/3, & \text{if } -2 \le x \lt 0, \\ 1/2, & \text{if } 0 \le x \lt 5, \\ 1/2 + (x - 5)^{2}/2, & \text{if } 5 \le x \lt 6, \\ 1, & \text{if } x \ge 6. \end{cases} \]Then E(X) is :

\(E(X) = \) 13/6.F has jumps of \(1/3\) at \(x = -2\) and \(1/2 - 1/3 = 1/6\) at \(x = 0\), and on [5, 6) a density \(F'(x) = x - 5\), which carries the remaining probability 1/2 (F reaches 1 at 6 without a jump). So\[ \begin{aligned} E(X) &= -2 \cdot \tfrac{1}{3} + 0 \cdot \tfrac{1}{6} + \int_{5}^{6} x(x - 5)\,dx \\ &= -\tfrac{2}{3} + \left(\tfrac{1}{3} + \tfrac{5}{2}\right) = \tfrac{13}{6} \end{aligned} \](The integral is \(\int_{0}^{1}(t + 5)t\,dt = 1/3 + 5/2\).) The other values come from mishandling the jumps, for instance dropping the mass at -2 gives 17/6, not offered, while 11/6 and 125/6 do not use the mixture correctly.

Q: Which of the following statements is correct ?The set of points of continuities of a distribution function is :

The set of continuity points is uncountable.A distribution function is non-decreasing, and a monotone function has at most countably many discontinuities: each jump contains a distinct rational number, or more directly, there are at most \(n\) jumps of size greater than \(1/n\), since the total increase is at most 1.So the discontinuities form a countable set, and the continuity points are the complement of a countable subset of \(\mathbf{R}\), which is uncountable (and dense). It is never countable, finite or empty.

Q: Suppose X and Y are two random variables whose third order moments exist. Then \((E(|X + Y|^{3}))^{1/3}\) is less than or equal to :

The bound is \((E|X|^{3})^{1/3} + (E|Y|^{3})^{1/3}\).This is Minkowski's inequality in \(L^{3}\): the \(L^{p}\) norm \(\|Z\|_{p} = (E|Z|^{p})^{1/p}\) satisfies the triangle inequality for \(p \ge 1\), so\[ \|X + Y\|_{3} \le \|X\|_{3} + \|Y\|_{3} \]The product form (A) fails, for example with \(Y = 0\): the left side is \(\|X\|_{3}\) but the right side is 0. The forms with first absolute moments do not hold in general either, since \(L^{3}\) norms can be much larger than \(L^{1}\) ones.

41.

Suppose X is a continuous random variable having the following probability density function :

\(f(x) = C(4x - 2x^{2}),\ 0 \lt x \lt 2\)
\(= 0,\) otherwise

What is the value of \(P[X \gt 1]\) ?

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

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

\(P[X \gt 1] = \) 1/2.

The density is proportional to \(4x - 2x^{2} = 2x(2 - x)\) on (0, 2). Replacing \(x\) by \(2 - x\) leaves it unchanged, so the density is symmetric about \(x = 1\), and half the probability lies on each side of 1.

Directly: \(\int_{0}^{2} (4x - 2x^{2})dx = 8 - 16/3 = 8/3\), so \(C = 3/8\), and

\[ \begin{aligned} P[X \gt 1] &= \tfrac{3}{8}\int_{1}^{2} (4x - 2x^{2})dx \\ &= \tfrac{3}{8}\left(6 - \tfrac{14}{3}\right) = \tfrac{1}{2} \end{aligned} \]

C/2 leaves the constant unevaluated, and 2/3 and 1/4 do not respect the symmetry.

Discuss in Forum

42.

For a data-set of \(n\) observations the sample mean, median and mode are 63, 60 and 58 respectively. Hence the data are :

A.Symmetric about sample median
B.Skewed to the left
C.Skewed to the right
D.Bimodal

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The data are skewed to the right.

Here mean (63) \(\gt\) median (60) \(\gt\) mode (58). In a positively skewed (right-skewed) distribution, a long tail of large values pulls the mean above the median, and the median above the mode. Pearson's coefficient of skewness, (mean - mode)/s, is positive.

For symmetric data the three measures would coincide, and left skew would reverse the order. A single reported mode says nothing about bimodality.

Discuss in Forum

43.

Suppose \(E[X] = 5\). Which of the following is always false ?

A.\(E[X^{2}] \lt (E[X])^{2}\)
B.If \(E[X^{2}] = 25\), then \(X = 5\) wp 1
C.\(E[X^{2}] \ge 25\)
D.\(E[\exp|X|] \gt \exp(E[|X|])\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The statement that is always false is \(E[X^{2}] \lt (E[X])^{2}\).

\(\mathrm{Var}(X) = E[X^{2}] - (E[X])^{2} \ge 0\), so \(E[X^{2}] \ge (E[X])^{2} = 25\) for every random variable. The strict reverse inequality can never hold.

The other statements are not always false. (C) is always true. (B) is true: \(E[X^{2}] = 25\) means variance 0, so X equals 5 with probability 1. (D) is usually true by Jensen's inequality (exp is strictly convex), and fails only in the degenerate case.

Discuss in Forum

44.

Suppose X and Y are independent r.v.s. each with mean 1.

Let \(\phi_{X}(t)\) denote the characteristic function of a random variable. Which of the following is false ?

A.\(\phi_{X+Y}(t) = \phi_{X}(t) . \phi_{Y}(t)\ \forall t\)
B.\(E[XY \mid X] = 1\)
C.\(\phi_{X^{2}+Y^{2}}(t) = \phi_{X^{2}}(t) . \phi_{Y^{2}}(t) \forall t\)
D.\((1 - X^{2})\) and \(\exp(Y)\) are independent

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The false statement is \(E[XY \mid X] = 1\).

Given X, the factor X is known and Y is independent of X, so

\[ E[XY \mid X] = X\,E[Y] = X \cdot 1 = X \]

which is a random variable, not the constant 1. (Its expectation is 1.)

The others are true by independence: the characteristic function of a sum of independent variables is the product; \(X^{2}\) and \(Y^{2}\) are independent, so the same holds for them; and functions of independent variables, such as \(1 - X^{2}\) and \(\exp(Y)\), are independent.

Discuss in Forum

45.

Let the joint distribution of (X, Y) be specified by :

Y \ X	1	2	3	4
1	1/16	3/32	1/8	5/32
2	3/32	1/8	5/32	3/16

Then \(P[X = 2Y]\) is :

A.0
B.\(\dfrac{9}{32}\)
C.\(\dfrac{11}{32}\)
D.\(\dfrac{1}{2}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

\(P[X = 2Y] = \) 9/32.

\(X = 2Y\) happens only at \((X, Y) = (2, 1)\) and \((4, 2)\):

\[ \begin{aligned} \frac{3}{32} + \frac{3}{16} &= \frac{3}{32} + \frac{6}{32} \\ &= \frac{9}{32} \end{aligned} \]

(The table's total is 1, as it should be.) 0 ignores these cells, and 11/32 or 1/2 add other cells, such as (2, 2) or (4, 1), where \(X \ne 2Y\).

Discuss in Forum

46.

Let \(\mathbf{B}\) be the \(\sigma\)-field of subsets of R generated by \(\{(a, b) \mid a, b \in R\}\). Which of the following is not true ?

A.Every countable subset of \(R \in \mathbf{B}\)
B.All closed intervals \(\in \mathbf{B}\)
C.All uncountable subsets of \(R \in \mathbf{B}\)
D.Intervals of the form \([a, b) \in \mathbf{B}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The statement that is not true is all uncountable subsets of R belong to \(\mathbf{B}\).

\(\mathbf{B}\) is the Borel \(\sigma\)-field. It has the cardinality of the continuum, while there are \(2^{\mathfrak{c}}\) subsets of R, so most subsets, and in particular many uncountable ones, are not Borel. (A Vitali set is even non-Lebesgue-measurable.)

The others are true. Singletons are closed, \(\{a\} = \bigcap (a - 1/n, a + 1/n)\), so countable sets are countable unions of Borel sets. Closed intervals are complements of open sets, and \([a, b) = \{a\} \cup (a, b)\).

Discuss in Forum

47.

Let R be the set of real numbers, \(\mathbf{B}\) the Borel \(\sigma\)-field. Let '\(m\)' denotes the Lebesgue measure and \(\mu\) the measure defined by the function \(F(x) = (1 - e^{-x})\) if \(x \gt 0\) and \(F(x) = 0\) elsewhere.

Consider the statements :

(I) \(\mu \lt\lt m\), (i.e. \(\mu\) is absolutely continuous w.r.t. \(m\))
(II) \(m \lt\lt \mu\)
(III) \(\dfrac{d\mu}{dm}(x) = e^{-x}\ x \gt 0\), \(\dfrac{d\mu}{dm}(x) = 0\), \(x \le 0\)

Which of the above statements are true ?

A.(I) and (II) only
B.(I) and (III) only
C.(II) and (III) only
D.All the three

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The true statements are (I) and (III) only.

\(\mu\) is the exponential distribution, with density \(e^{-x}\) on \(x \gt 0\) and 0 elsewhere. Whenever \(m(A) = 0\), \(\mu(A) = \int_{A} e^{-x}1_{x \gt 0}\,dx = 0\), so \(\mu \ll m\) (I). By the Radon-Nikodym theorem the derivative \(d\mu/dm\) is that density, \(e^{-x}\) for \(x \gt 0\) and 0 for \(x \le 0\) (III).

(II) fails: \(\mu((-\infty, 0]) = 0\) but \(m((-\infty, 0]) = \infty\), so a \(\mu\)-null set need not be \(m\)-null.

Discuss in Forum

48.

Suppose a distribution function \(F : \mathbf{R} \to [0, 1]\) of a random variable X is as follows :

\[ F(x) = \begin{cases} 0, & \text{if } x \lt -2, \\ 1/3, & \text{if } -2 \le x \lt 0, \\ 1/2, & \text{if } 0 \le x \lt 5, \\ 1/2 + (x - 5)^{2}/2, & \text{if } 5 \le x \lt 6, \\ 1, & \text{if } x \ge 6. \end{cases} \]

Then E(X) is :

A.\(\dfrac{11}{6}\)
B.\(\dfrac{1}{6}\)
C.\(\dfrac{125}{6}\)
D.\(\dfrac{13}{6}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

\(E(X) = \) 13/6.

F has jumps of \(1/3\) at \(x = -2\) and \(1/2 - 1/3 = 1/6\) at \(x = 0\), and on [5, 6) a density \(F'(x) = x - 5\), which carries the remaining probability 1/2 (F reaches 1 at 6 without a jump). So

\[ \begin{aligned} E(X) &= -2 \cdot \tfrac{1}{3} + 0 \cdot \tfrac{1}{6} + \int_{5}^{6} x(x - 5)\,dx \\ &= -\tfrac{2}{3} + \left(\tfrac{1}{3} + \tfrac{5}{2}\right) = \tfrac{13}{6} \end{aligned} \]

(The integral is \(\int_{0}^{1}(t + 5)t\,dt = 1/3 + 5/2\).) The other values come from mishandling the jumps, for instance dropping the mass at -2 gives 17/6, not offered, while 11/6 and 125/6 do not use the mixture correctly.

Discuss in Forum

49.

Which of the following statements is correct ?

The set of points of continuities of a distribution function is :

A.uncountable
B.countable
C.finite
D.empty

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The set of continuity points is uncountable.

A distribution function is non-decreasing, and a monotone function has at most countably many discontinuities: each jump contains a distinct rational number, or more directly, there are at most \(n\) jumps of size greater than \(1/n\), since the total increase is at most 1.

So the discontinuities form a countable set, and the continuity points are the complement of a countable subset of \(\mathbf{R}\), which is uncountable (and dense). It is never countable, finite or empty.

Discuss in Forum

50.

Suppose X and Y are two random variables whose third order moments exist. Then \((E(|X + Y|^{3}))^{1/3}\) is less than or equal to :

A.\((E(|X|^{3}))^{1/3}\ (E(|Y|^{3}))^{1/3}\)
B.\((E(|X|^{3}))^{1/3} + (E(|Y|^{3}))^{1/3}\)
C.\((E(|X|) + E(|Y|))^{1/3}\)
D.\((E|X| . E|Y|)^{1/3}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The bound is \((E|X|^{3})^{1/3} + (E|Y|^{3})^{1/3}\).

This is Minkowski's inequality in \(L^{3}\): the \(L^{p}\) norm \(\|Z\|_{p} = (E|Z|^{p})^{1/p}\) satisfies the triangle inequality for \(p \ge 1\), so

\[ \|X + Y\|_{3} \le \|X\|_{3} + \|Y\|_{3} \]

The product form (A) fails, for example with \(Y = 0\): the left side is \(\|X\|_{3}\) but the right side is 0. The forms with first absolute moments do not hold in general either, since \(L^{3}\) norms can be much larger than \(L^{1}\) ones.

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