Probability Theory Questions and Answers

Q: Let \(\Omega = \{1, 2, 3, 4\}\).Consider the two \(\sigma\)-fields :\(\mathbf{F}_{1} = \{\phi, \Omega, \{1\}, \{2\}, \{1, 2\},\) \(\{3, 4\}, \{2, 3, 4\}, \{1, 3, 4\}\}\)\(\mathbf{F}_{2} = \{\phi, \Omega, \{1, 2\}, \{3, 4\}\}\)Let \(f\) denote a function from \((\Omega, \mathbf{F}_{1})\) to \((\Omega, \mathbf{F}_{2})\).Then :

The measurable function is \(f(k) = k\), \(k = 1, 2, 3, 4\).\(f\) is measurable if the inverse image of every set in \(\mathbf{F}_{2}\) lies in \(\mathbf{F}_{1}\). It is enough to check {1, 2} and {3, 4}.B: \(f^{-1}\{1, 2\} = \{1, 2\}\) and \(f^{-1}\{3, 4\} = \{3, 4\}\), both in \(\mathbf{F}_{1}\). Measurable.C: \(f^{-1}\{1, 2\} = \{1, 3\}\), not in \(\mathbf{F}_{1}\).D: \(f^{-1}\{1, 2\} = \{1, 2, 3\}\), not in \(\mathbf{F}_{1}\).So not every \(f\) is measurable (A fails), as C and D show.

Q: Which of the following statements is not correct ? The set of points of discontinuities of a distribution function may be :

The set of discontinuities cannot be uncountable.A distribution function is non-decreasing and bounded between 0 and 1. At each discontinuity it jumps by a positive amount, and at most \(k\) jumps can exceed \(1/k\), since the total of all jumps is at most 1. So the jumps can be listed by size, and the set of discontinuities is countable.All the other possibilities occur: a continuous distribution has none (empty), a binomial has finitely many, and a distribution placing mass \(2^{-n}\) at each \(n\), or at each rational, has countably infinitely many.

Q: Suppose \(\phi(\cdot)\) is a characteristic function of a random variable with distribution function F. Then :

The correct inversion formula is option A:\[ \begin{aligned} F(b) - F(a) &= \lim_{u \to \infty} \frac{1}{2\pi} \\ &\int_{-u}^{u} \frac{e^{-iua} - e^{-iub}}{iu}\phi(u)\, du \end{aligned} \]This is Levy's inversion theorem, valid when \(a\) and \(b\) are continuity points of F. It comes from \(\frac{1}{2\pi}\int_{a}^{b} e^{-iux}\, dx = \frac{e^{-iua} - e^{-iub}}{2\pi iu}\) and Fubini's theorem.Option B lacks the factor \(1/2\pi\), option C has the wrong sign in the exponents, and option D reverses the order of the two terms, which gives \(F(a) - F(b)\).

Q: Suppose a distribution function \(F : \mathbf{R} \to [0, 1]\) is as follows :\[ F(x) = \begin{cases} 0, & \text{if } x \lt 0, \\ 1/4, & \text{if } 0 \le x \lt 1, \\ 1/2, & \text{if } 1 \le x \lt 2, \\ 1/2 + (x - 2)/2, & \text{if } 2 \le x \lt 3, \\ 1, & \text{if } x \ge 3. \end{cases} \]Then \(F(x) = l_{1}F_{1}(x) + l_{2}F_{2}(x)\) where :

The decomposition is \(l_{1} = l_{2} = 1/2\), \(F_{1}\) discrete on {0, 1} with probabilities 1/2, 1/2, and \(F_{2}\) uniform on (2, 3).F jumps by 1/4 at 0 and by 1/4 at 1, so the discrete part has total mass 1/2, split equally between 0 and 1. Normalised, that is the distribution with probabilities 1/2, 1/2 on {0, 1}, with weight \(l_{1} = 1/2\).From 2 to 3, F rises linearly from 1/2 to 1, so the continuous part has mass 1/2 spread uniformly over (2, 3): weight \(l_{2} = 1/2\) and \(F_{2}\) the U(2, 3) distribution function. Check: \(\frac{1}{2}F_{1} + \frac{1}{2}F_{2}\) gives 1/4 on [0, 1), 1/2 on [1, 2) and \(\frac{1}{2} + \frac{x - 2}{2}\) on [2, 3).The options with \(l_{1} = 1/4\) count only one jump, and option D puts the jumps at 1 and 2 instead of 0 and 1.

Q: The waiting time of the customer (in min) is grouped as :Waiting TimeFrequency4-666-888-101210-121012-146The mode of waiting time is :

The mode is 9.33 minutes.The modal class is 8-10, with the highest frequency 12. With \(l = 8\), \(f_{1} = 12\), \(f_{0} = 8\) (class before), \(f_{2} = 10\) (class after) and width \(h = 2\):\[ \begin{aligned} \text{Mode} &= l + \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \times h \\ &= 8 + \frac{12 - 8}{24 - 8 - 10} \times 2 \\ &= 8 + \frac{4}{6} \times 2 = 9.33 \end{aligned} \]The mode leans toward the 6-8 side because that neighbour (8) is smaller than the 10-12 neighbour (10), pulling the peak left of the class midpoint 9. The value 9.67 swaps \(f_{0}\) and \(f_{2}\); 10.33 and 10.67 place the mode in the wrong class.

Q: In how many ways can a lady having 10 dresses, 5 pairs of shoes, and 2 hats be dressed ?

She can be dressed in 100 ways.Choosing an outfit is three independent choices made one after another: a dress (10 ways), a pair of shoes (5 ways) and a hat (2 ways). By the multiplication principle,\[ 10 \times 5 \times 2 = 100 \]The figure 50 forgets the hats, and 500 multiplies by an extra factor of 5. The expression 10!5!2! counts orderings of all the items within each type, which has nothing to do with choosing one of each.

Q: The distribution of a random variable X is given by :\[ F(x) = P[X \le x] = \begin{cases} 0 & \text{if } x \lt 0 \\ \frac{1}{4} & \text{if } 0 \le x \lt \frac{1}{4} \\ \frac{1}{2} & \text{if } \frac{1}{4} \le x \lt \frac{1}{2} \\ \frac{3}{4} & \text{if } \frac{1}{2} \le x \lt \frac{3}{4} \\ \frac{x + 3}{5} & \text{if } \frac{3}{4} \le x \lt 2 \\ 1 & \text{if } x \ge 2 \end{cases} \]Then \(P\left[\frac{1}{2} \le X \le \frac{3}{4}\right]\) is :

The probability is \(\frac{1}{4}\).For a distribution function with jumps, \(P[a \le X \le b] = F(b) - F(a^{-})\), where \(F(a^{-})\) is the left limit at \(a\). Here\[ \begin{aligned} F\left(\tfrac{3}{4}\right) &= \frac{3/4 + 3}{5} = \frac{3.75}{5} = \frac{3}{4} \\ F\left(\tfrac{1}{2}^{-}\right) &= \frac{1}{2} \\ P &= \frac{3}{4} - \frac{1}{2} = \frac{1}{4} \end{aligned} \]This 1/4 is the jump at \(x = 1/2\) (from 1/2 to 3/4); F is continuous at 3/4, since the formula \((x + 3)/5\) gives 3/4 there, so there is no extra mass at the right endpoint.The value 3/4 is \(F(3/4)\) alone, 1/2 forgets to subtract the right amount, and 1/8 does not arise from the distribution function.

Q: Suppose that \(\Omega_{1} = \{1, 2, 3, 4, 5, 6\}\) and that P is a probability measure such that \(P(\{j\}) = 1/6\), for \(j = 1, 2, 3, 4, 5, 6\). Let \(I_{A}\) and \(I_{B}\) be the indicator functions of the subsets \(A = \{1, 3, 5\}\) and \(B = \{2, 3, 4, 6\}\) respectively. The following are two statements :(I) \(I_{A}\) and \(I_{B}\) are identically distributed(II) \(I_{A}\) and \(I_{B}\) are independent random variablesWhich of the following is true ?

Neither (I) nor (II) is true.\(I_{A}\) is 1 with probability \(P(A) = 3/6 = 1/2\), while \(I_{B}\) is 1 with probability \(P(B) = 4/6 = 2/3\). Two indicator variables are identically distributed only when the events have equal probability, so (I) fails.For independence, compare \(P(A \cap B)\) with \(P(A)P(B)\):\[ \begin{aligned} P(A \cap B) &= P(\{3\}) = \frac{1}{6} \\ P(A)P(B) &= \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \end{aligned} \]These differ, so A and B, and hence \(I_{A}\) and \(I_{B}\), are not independent. (II) fails too, which rules out every option asserting either statement.

Q: The cumulative distribution function F of a random variable X is given by :\[ \begin{aligned} F(x) &= 0.2\,\Delta_{0}(x) + 0.4\,\Delta_{1}(x) \\ &\quad + 0.2\,\Delta_{2}(x) + 0.2\,\Delta_{3}(x), \end{aligned} \]where \(\Delta_{a}\) is the degenerate distribution function degenerate at 'a'. The following are two statements :(I) \(P(X = 1 \mid X \le 1.5) = 2/3\)(II) Variance (X) = 0Which of the following is true ?

Only (I) is true.The distribution function puts mass 0.2 at 0, 0.4 at 1, 0.2 at 2 and 0.2 at 3. For (I):\[ \begin{aligned} P(X = 1 \mid X \le 1.5) &= \frac{P(X = 1)}{P(X \in \{0, 1\})} \\ &= \frac{0.4}{0.6} = \frac{2}{3} \end{aligned} \]so (I) holds.For (II), a variance of 0 would mean X is a single constant, but X takes four values. Indeed \(E(X) = 0.4 + 0.4 + 0.6 = 1.4\), \(E(X^{2}) = 0.4 + 0.8 + 1.8 = 3\), so \(\text{Var}(X) = 3 - 1.96 = 1.04\). Statement (II) confuses the components, each of which is degenerate, with the mixture.

Q: Suppose that the conditional distribution of X given that \(N = n\) is binomial with parameters \(n\) and 0.25 and that the marginal distribution of N is geometric with mean 2 and variance 2. Then, variance (X) is equal to :

\(\text{Var}(X) = \frac{8}{16}\).Use the law of total variance, \(\text{Var}(X) = E[\text{Var}(X \mid N)] + \text{Var}(E[X \mid N])\). Given \(N\), X is binomial \((N, 0.25)\), so \(E[X \mid N] = N/4\) and \(\text{Var}(X \mid N) = N \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{3N}{16}\). With \(E(N) = 2\) and \(\text{Var}(N) = 2\):\[ \begin{aligned} \text{Var}(X) &= \frac{3}{16}E(N) + \frac{1}{16}\text{Var}(N) \\ &= \frac{6}{16} + \frac{2}{16} = \frac{8}{16} \end{aligned} \]So the variance can be determined: only the mean and variance of N are needed, not its full distribution. The value 10/16 adds the two terms with the wrong weights, and 1 ignores the thinning by 0.25.

61.

Let \(\Omega = \{1, 2, 3, 4\}\).

Consider the two \(\sigma\)-fields :

\(\mathbf{F}_{1} = \{\phi, \Omega, \{1\}, \{2\}, \{1, 2\},\) \(\{3, 4\}, \{2, 3, 4\}, \{1, 3, 4\}\}\)
\(\mathbf{F}_{2} = \{\phi, \Omega, \{1, 2\}, \{3, 4\}\}\)

Let \(f\) denote a function from \((\Omega, \mathbf{F}_{1})\) to \((\Omega, \mathbf{F}_{2})\).

Then :

A.Any \(f\) is measurable
B.\(f\) defined by \(f(k) = k\), \(k = 1, 2, 3, 4\) is measurable
C.\(f\) defined by \(f(1) = f(3) = 1\) and \(f(2) = f(4) = 4\) is measurable
D.\(f\) defined by \(f(1) = f(2) = f(3) = 1\) and \(f(4) = 4\) is measurable

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The measurable function is \(f(k) = k\), \(k = 1, 2, 3, 4\).

\(f\) is measurable if the inverse image of every set in \(\mathbf{F}_{2}\) lies in \(\mathbf{F}_{1}\). It is enough to check {1, 2} and {3, 4}.

B: \(f^{-1}\{1, 2\} = \{1, 2\}\) and \(f^{-1}\{3, 4\} = \{3, 4\}\), both in \(\mathbf{F}_{1}\). Measurable.
C: \(f^{-1}\{1, 2\} = \{1, 3\}\), not in \(\mathbf{F}_{1}\).
D: \(f^{-1}\{1, 2\} = \{1, 2, 3\}\), not in \(\mathbf{F}_{1}\).

So not every \(f\) is measurable (A fails), as C and D show.

Discuss in Forum

62.

Which of the following statements is not correct ? The set of points of discontinuities of a distribution function may be :

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

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The set of discontinuities cannot be uncountable.

A distribution function is non-decreasing and bounded between 0 and 1. At each discontinuity it jumps by a positive amount, and at most \(k\) jumps can exceed \(1/k\), since the total of all jumps is at most 1. So the jumps can be listed by size, and the set of discontinuities is countable.

All the other possibilities occur: a continuous distribution has none (empty), a binomial has finitely many, and a distribution placing mass \(2^{-n}\) at each \(n\), or at each rational, has countably infinitely many.

Discuss in Forum

63.

Suppose \(\phi(\cdot)\) is a characteristic function of a random variable with distribution function F. Then :

A.\[ \begin{aligned} F(b) - F(a) &= \lim_{u \to \infty} \frac{1}{2\pi} \\ &\int_{-u}^{u} \frac{[\exp(-iua) - \exp(-iub)]\, \phi(u)}{iu}\, du \end{aligned} \]
B.\[ \begin{aligned} F(b) - F(a) &= \lim_{u \to \infty} \\ &\int_{-u}^{u} \frac{[\exp(-iua) - \exp(-iub)]\, \phi(u)}{iu}\, du \end{aligned} \]
C.\[ \begin{aligned} F(b) - F(a) &= \lim_{u \to \infty} \frac{1}{2\pi} \\ &\int_{-u}^{u} \frac{[\exp(iua) - \exp(iub)]\, \phi(u)}{iu}\, du \end{aligned} \]
D.\[ \begin{aligned} F(b) - F(a) &= \lim_{u \to \infty} \frac{1}{2\pi} \\ &\int_{-u}^{u} \frac{[\exp(-iub) - \exp(-iua)]\, \phi(u)}{iu}\, du \end{aligned} \]

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The correct inversion formula is option A:

\[ \begin{aligned} F(b) - F(a) &= \lim_{u \to \infty} \frac{1}{2\pi} \\ &\int_{-u}^{u} \frac{e^{-iua} - e^{-iub}}{iu}\phi(u)\, du \end{aligned} \]

This is Levy's inversion theorem, valid when \(a\) and \(b\) are continuity points of F. It comes from \(\frac{1}{2\pi}\int_{a}^{b} e^{-iux}\, dx = \frac{e^{-iua} - e^{-iub}}{2\pi iu}\) and Fubini's theorem.

Option B lacks the factor \(1/2\pi\), option C has the wrong sign in the exponents, and option D reverses the order of the two terms, which gives \(F(a) - F(b)\).

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64.

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

\[ F(x) = \begin{cases} 0, & \text{if } x \lt 0, \\ 1/4, & \text{if } 0 \le x \lt 1, \\ 1/2, & \text{if } 1 \le x \lt 2, \\ 1/2 + (x - 2)/2, & \text{if } 2 \le x \lt 3, \\ 1, & \text{if } x \ge 3. \end{cases} \]

Then \(F(x) = l_{1}F_{1}(x) + l_{2}F_{2}(x)\) where :

A.\(l_{1} = 1/4\), \(l_{2} = 3/4\), \(F_{1}\) is a distribution function of a discrete random variable with support {0, 1} with respective probabilities 1/4, 3/4 and \(F_{2}\) is a distribution function of uniform U(2, 3) distribution.
B.\(l_{1} = 1/2\), \(l_{2} = 1/2\), \(F_{1}\) is a distribution function of a discrete random variable with support {0, 1} with respective probabilities 1/2, 1/2 and \(F_{2}\) is a distribution function of uniform U(2, 3) distribution.
C.\(l_{1} = 1/4\), \(l_{2} = 3/4\), \(F_{1}\) is a distribution function of a discrete random variable with support {0, 1} with respective probabilities 1/2, 1/2 and \(F_{2}\) is a distribution function of uniform U(2, 3) distribution.
D.\(l_{1} = 1/4\), \(l_{2} = 3/4\), \(F_{1}\) is a distribution function of a discrete random variable with support {1, 2} with respective probabilities 1/4, 3/4 and \(F_{2}\) is a distribution function of uniform U(2, 3) distribution.

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The decomposition is \(l_{1} = l_{2} = 1/2\), \(F_{1}\) discrete on {0, 1} with probabilities 1/2, 1/2, and \(F_{2}\) uniform on (2, 3).

F jumps by 1/4 at 0 and by 1/4 at 1, so the discrete part has total mass 1/2, split equally between 0 and 1. Normalised, that is the distribution with probabilities 1/2, 1/2 on {0, 1}, with weight \(l_{1} = 1/2\).

From 2 to 3, F rises linearly from 1/2 to 1, so the continuous part has mass 1/2 spread uniformly over (2, 3): weight \(l_{2} = 1/2\) and \(F_{2}\) the U(2, 3) distribution function. Check: \(\frac{1}{2}F_{1} + \frac{1}{2}F_{2}\) gives 1/4 on [0, 1), 1/2 on [1, 2) and \(\frac{1}{2} + \frac{x - 2}{2}\) on [2, 3).

The options with \(l_{1} = 1/4\) count only one jump, and option D puts the jumps at 1 and 2 instead of 0 and 1.

Discuss in Forum

65.

The waiting time of the customer (in min) is grouped as :

Waiting Time	Frequency
4-6	6
6-8	8
8-10	12
10-12	10
12-14	6

The mode of waiting time is :

A.9.33
B.9.67
C.10.33
D.10.67

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The mode is 9.33 minutes.

The modal class is 8-10, with the highest frequency 12. With \(l = 8\), \(f_{1} = 12\), \(f_{0} = 8\) (class before), \(f_{2} = 10\) (class after) and width \(h = 2\):

\[ \begin{aligned} \text{Mode} &= l + \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \times h \\ &= 8 + \frac{12 - 8}{24 - 8 - 10} \times 2 \\ &= 8 + \frac{4}{6} \times 2 = 9.33 \end{aligned} \]

The mode leans toward the 6-8 side because that neighbour (8) is smaller than the 10-12 neighbour (10), pulling the peak left of the class midpoint 9. The value 9.67 swaps \(f_{0}\) and \(f_{2}\); 10.33 and 10.67 place the mode in the wrong class.

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66.

In how many ways can a lady having 10 dresses, 5 pairs of shoes, and 2 hats be dressed ?

A.50
B.100
C.500
D.10 ! 5 ! 2 !

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

She can be dressed in 100 ways.

Choosing an outfit is three independent choices made one after another: a dress (10 ways), a pair of shoes (5 ways) and a hat (2 ways). By the multiplication principle,

\[ 10 \times 5 \times 2 = 100 \]

The figure 50 forgets the hats, and 500 multiplies by an extra factor of 5. The expression 10!5!2! counts orderings of all the items within each type, which has nothing to do with choosing one of each.

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67.

The distribution of a random variable X is given by :

\[ F(x) = P[X \le x] = \begin{cases} 0 & \text{if } x \lt 0 \\ \frac{1}{4} & \text{if } 0 \le x \lt \frac{1}{4} \\ \frac{1}{2} & \text{if } \frac{1}{4} \le x \lt \frac{1}{2} \\ \frac{3}{4} & \text{if } \frac{1}{2} \le x \lt \frac{3}{4} \\ \frac{x + 3}{5} & \text{if } \frac{3}{4} \le x \lt 2 \\ 1 & \text{if } x \ge 2 \end{cases} \]

Then \(P\left[\frac{1}{2} \le X \le \frac{3}{4}\right]\) is :

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

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The probability is \(\frac{1}{4}\).

For a distribution function with jumps, \(P[a \le X \le b] = F(b) - F(a^{-})\), where \(F(a^{-})\) is the left limit at \(a\). Here

\[ \begin{aligned} F\left(\tfrac{3}{4}\right) &= \frac{3/4 + 3}{5} = \frac{3.75}{5} = \frac{3}{4} \\ F\left(\tfrac{1}{2}^{-}\right) &= \frac{1}{2} \\ P &= \frac{3}{4} - \frac{1}{2} = \frac{1}{4} \end{aligned} \]

This 1/4 is the jump at \(x = 1/2\) (from 1/2 to 3/4); F is continuous at 3/4, since the formula \((x + 3)/5\) gives 3/4 there, so there is no extra mass at the right endpoint.

The value 3/4 is \(F(3/4)\) alone, 1/2 forgets to subtract the right amount, and 1/8 does not arise from the distribution function.

Discuss in Forum

68.

Suppose that \(\Omega_{1} = \{1, 2, 3, 4, 5, 6\}\) and that P is a probability measure such that \(P(\{j\}) = 1/6\), for \(j = 1, 2, 3, 4, 5, 6\). Let \(I_{A}\) and \(I_{B}\) be the indicator functions of the subsets \(A = \{1, 3, 5\}\) and \(B = \{2, 3, 4, 6\}\) respectively. The following are two statements :

(I) \(I_{A}\) and \(I_{B}\) are identically distributed
(II) \(I_{A}\) and \(I_{B}\) are independent random variables

Which of the following is true ?

A.Both (I) and (II) are true
B.Only (I) is true
C.Neither (I) nor (II) is true
D.Only (II) is true

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

Neither (I) nor (II) is true.

\(I_{A}\) is 1 with probability \(P(A) = 3/6 = 1/2\), while \(I_{B}\) is 1 with probability \(P(B) = 4/6 = 2/3\). Two indicator variables are identically distributed only when the events have equal probability, so (I) fails.

For independence, compare \(P(A \cap B)\) with \(P(A)P(B)\):

\[ \begin{aligned} P(A \cap B) &= P(\{3\}) = \frac{1}{6} \\ P(A)P(B) &= \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \end{aligned} \]

These differ, so A and B, and hence \(I_{A}\) and \(I_{B}\), are not independent. (II) fails too, which rules out every option asserting either statement.

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69.

The cumulative distribution function F of a random variable X is given by :

\[ \begin{aligned} F(x) &= 0.2\,\Delta_{0}(x) + 0.4\,\Delta_{1}(x) \\ &\quad + 0.2\,\Delta_{2}(x) + 0.2\,\Delta_{3}(x), \end{aligned} \]

where \(\Delta_{a}\) is the degenerate distribution function degenerate at 'a'. The following are two statements :

(I) \(P(X = 1 \mid X \le 1.5) = 2/3\)
(II) Variance (X) = 0

Which of the following is true ?

A.Both (I) and (II) are true
B.Only (I) is true
C.Only (II) is true
D.Neither (I) nor (II) is true

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

Only (I) is true.

The distribution function puts mass 0.2 at 0, 0.4 at 1, 0.2 at 2 and 0.2 at 3. For (I):

\[ \begin{aligned} P(X = 1 \mid X \le 1.5) &= \frac{P(X = 1)}{P(X \in \{0, 1\})} \\ &= \frac{0.4}{0.6} = \frac{2}{3} \end{aligned} \]

so (I) holds.

For (II), a variance of 0 would mean X is a single constant, but X takes four values. Indeed \(E(X) = 0.4 + 0.4 + 0.6 = 1.4\), \(E(X^{2}) = 0.4 + 0.8 + 1.8 = 3\), so \(\text{Var}(X) = 3 - 1.96 = 1.04\). Statement (II) confuses the components, each of which is degenerate, with the mixture.

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70.

Suppose that the conditional distribution of X given that \(N = n\) is binomial with parameters \(n\) and 0.25 and that the marginal distribution of N is geometric with mean 2 and variance 2. Then, variance (X) is equal to :

A.1
B.10/16
C.8/16
D.Cannot be determined

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

\(\text{Var}(X) = \frac{8}{16}\).

Use the law of total variance, \(\text{Var}(X) = E[\text{Var}(X \mid N)] + \text{Var}(E[X \mid N])\). Given \(N\), X is binomial \((N, 0.25)\), so \(E[X \mid N] = N/4\) and \(\text{Var}(X \mid N) = N \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{3N}{16}\). With \(E(N) = 2\) and \(\text{Var}(N) = 2\):

\[ \begin{aligned} \text{Var}(X) &= \frac{3}{16}E(N) + \frac{1}{16}\text{Var}(N) \\ &= \frac{6}{16} + \frac{2}{16} = \frac{8}{16} \end{aligned} \]

So the variance can be determined: only the mean and variance of N are needed, not its full distribution. The value 10/16 adds the two terms with the wrong weights, and 1 ignores the thinning by 0.25.

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