Probability Theory Questions and Answers

Q: Let \(\{X_{n}\}\) be a sequence of independent random variables. Define the \(\sigma\)-field :\[ F = \bigcap_{n=1}^{\infty} \sigma(X_{n}, X_{n+1}, \ldots). \]Suppose \(A \in f_{e}\). Then, which of the following statements is more appropriate ?

The appropriate statement is P(A) = 0 or P(A) = 1.F, the intersection of \(\sigma(X_{n}, X_{n+1}, \ldots)\) over all n, is the tail \(\sigma\)-field of the sequence (printed in the question as \(f_{e}\)). Kolmogorov's zero-one law says that when the \(X_{n}\) are independent, every tail event has probability 0 or 1.Examples of tail events are 'the series \(\sum X_{n}\) converges' and 'the limit superior of \(X_{n}\) exceeds 5'. Each is unaffected by changing finitely many terms, and each is either almost certain or almost impossible.Saying always 0 or always 1 is too strong, since both values occur for different tail events (A and its complement are both tail events). \(P(A) = P(A^{c})\) would force \(P(A) = 1/2\), which the law excludes.

Q: If X is a positive random variable with probability density function \(f(x)\), then \(X^{-1}\) has probability density function :

The density of \(X^{-1}\) is \(1/x^{2} f(1/x)\), that is \(\dfrac{1}{x^{2}} f\left(\dfrac{1}{x}\right)\) for \(x \gt 0\).Let \(Y = 1/X\). For \(y \gt 0\), \(Y \le y\) exactly when \(X \ge 1/y\), so\[ F_{Y}(y) = 1 - F_{X}(1/y) \]Differentiating with the chain rule:\[ \begin{aligned} f_{Y}(y) &= -f_{X}(1/y) \cdot \left(-\frac{1}{y^{2}}\right) \\ &= \frac{1}{y^{2}} f_{X}(1/y) \end{aligned} \]The factor \(1/y^{2}\) is the Jacobian \(|dx/dy|\) of the change of variable. \(f(1/x)\) forgets the Jacobian, \(1/x \cdot f(1/x)\) has the wrong power of x (it does not integrate to 1 in general), and \(1/f(x)\) is not a density at all.

Q: Let \(X_{1}\) and \(X_{2}\) be iid random variables with distribution function \(F(x)\). Then, \(P(X_{1} \le X_{2})\) is :

\(P(X_{1} \le X_{2}) = \)1/2.Because \(X_{1}\) and \(X_{2}\) are independent with the same distribution, the pair \((X_{1}, X_{2})\) has the same joint distribution as \((X_{2}, X_{1})\). Hence \(P(X_{1} \lt X_{2}) = P(X_{2} \lt X_{1})\), and\[ \begin{aligned} &P(X_{1} \lt X_{2}) + P(X_{1} = X_{2}) \\ &\quad + P(X_{1} \gt X_{2}) = 1 \end{aligned} \]For a continuous distribution function, \(P(X_{1} = X_{2}) = 0\), so each strict inequality has probability 1/2 and \(P(X_{1} \le X_{2}) = 1/2\). This is the reading the question intends. If F had jumps, ties would have positive probability and the answer would be \(\frac{1}{2} + \frac{1}{2}P(X_{1} = X_{2})\), but that is not among the options either.1/3 and 2/3 have no symmetry basis, and 3/2 exceeds 1, so it is not a probability.

Q: Let X and Y be two random variables defined on the same probability space such that for each \(y \gt 0\), the conditional density function of X given \(Y = y\) is :\[ \begin{aligned} f(x/y) &= \sqrt{\frac{y}{2\pi}}\, e^{-yx^{2}/2}, \\ &y \gt 0,\ x \in R. \end{aligned} \]Let\[ g(y) = \frac{1}{\sqrt{2\pi y}}\, e^{-y/2},\ y \gt 0. \]Which of the following statements is more appropriate ?

The most appropriate statement is E(X|Y) = 0.Given \(Y = y\), X is normal with mean 0 and variance \(1/y\), so \(E(X|Y) = 0\) exactly. The marginal of Y is \(g(y)\), the chi-square density with 1 degree of freedom. Integrating out \(y\):\[ \begin{aligned} f_{X}(x) &= \frac{1}{2\pi}\int_{0}^{\infty} e^{-y(1 + x^{2})/2}dy \\ &= \frac{1}{\pi(1 + x^{2})} \end{aligned} \]So X has the standard Cauchy distribution, and \(E(X)\) does not exist. That rules out \(E(X) = 0\) and \(E(X) \ne E(E(X|Y))\), and makes \(E(E(X|Y)) = 0\) misleading: the iterated expectation is 0 although the expectation it should equal does not exist, since the tower property needs \(E|X| \lt \infty\).

Q: Let Y be a Bernoulli random variable with :\(P(Y = 0) = P(Y = 1) = 1/2\).Let \(X_{n} = (1 - Y/n^{\alpha})^{n}\), \(\alpha \ge 0\), then, which of the following statements are correct ?(i) \(X_{n} \xrightarrow{d} Y\) when \(0 \le \alpha \lt 1\)(ii) \(X_{n} \xrightarrow{d} e^{-Y}\) when \(\alpha = 1\)(iii) \(X_{n} \xrightarrow{d} Y\) when \(\alpha \gt 1\)

The correct statements are (i) and (ii).If \(Y = 0\), \(X_{n} = 1\) for all \(n\). If \(Y = 1\), \(X_{n} = (1 - n^{-\alpha})^{n}\), which tends to 0 when \(0 \le \alpha \lt 1\) (for \(\alpha = 0\) it is exactly 0), to \(e^{-1}\) when \(\alpha = 1\), and to 1 when \(\alpha \gt 1\), since then \(n \cdot n^{-\alpha} \to 0\).(i): for \(\alpha \lt 1\), \(X_{n} \to 1 - Y\), which takes values 1 and 0 with probability 1/2 each, the same distribution as Y. So \(X_{n} \to Y\) in distribution. (ii): for \(\alpha = 1\), \(X_{n} \to e^{-Y}\) (1 if \(Y = 0\), \(e^{-1}\) if \(Y = 1\)). (iii): for \(\alpha \gt 1\), \(X_{n} \to 1\), a constant, which is not distributed like Y.

Q: Let \(\Omega\) be a countably infinite set and let F consists of all subsets of \(\Omega\). Define :\[ \mu(A) = \begin{cases} 0 & \text{if A is finite} \\ \infty & \text{if A is infinite} \end{cases} \]Which of the following statements are true ?(i) \(\mu\) is finitely additive(ii) \(\mu\) is countably additive(iii) \(\{A_{n}\}\) is an \(\uparrow\) sequence of sets with \(\mu(A_{n}) = 0\ \forall\, n\), but \(\mu(\Omega) = \infty\).

The true statements are (i) and (iii).(i): for disjoint A and B, \(A \cup B\) is finite exactly when both are, so \(\mu(A \cup B) = \mu(A) + \mu(B)\) in every case (0 + 0, or \(\infty\) on both sides). \(\mu\) is finitely additive.(ii) fails: \(\Omega\) is the countable disjoint union of its singletons, each of measure 0, yet \(\mu(\Omega) = \infty \ne 0\). So \(\mu\) is not countably additive.(iii) is true and shows the same failure as continuity: list \(\Omega = \{\omega_{1}, \omega_{2}, \ldots\}\) and let \(A_{n} = \{\omega_{1}, \ldots, \omega_{n}\}\). The \(A_{n}\) increase to \(\Omega\), each \(\mu(A_{n}) = 0\), but \(\mu(\Omega) = \infty\).

Q: Let X and Y be two independent zero-mean unit variance Gaussian random variables defined on a common probability space. Define \(U = X + Y\) and \(V = X - Y\). Let \(F = \sigma(X)\). Then, which of the following statements is not true ?

The statement that is not true is E(U|F) and E(V|F) are independent.Given \(F = \sigma(X)\), X is known and Y is independent of it with mean 0, so \(E(Y|F) = 0\). Hence \(E(U|F) = X + 0 = X\) and \(E(V|F) = X - 0 = X\), and \(E(U + V|F) = E(2X|F) = 2X\). Options A, B and C hold almost surely.Both conditional expectations equal the same random variable X, which has variance 1. A non-degenerate random variable is not independent of itself, since \(P(X \le 0, X \le 0) = 1/2 \ne 1/4\). (U and V themselves are independent, being uncorrelated jointly normal variables, which is the trap.)

Q: Let \(X_{1}\) and \(X_{2}\) be two iid random variables with :\(P(X_{1} = 1) = P(X_{1} = -1) = 1/2\).Let \(Z = X_{1} + X_{2}\), \(A_{i} = X_{i}^{-1}(\{1\})\), \(i = 1, 2\).Which of the following is not true on \(Z^{-1}(\{0\})\) ?

The statement that is not true is \(P(A_{1}|Z)P(A_{2}|Z) = P(A_{1} \cap A_{2}|Z)\).On \(\{Z = 0\}\) the outcomes are (1, -1) and (-1, 1), each with conditional probability 1/2. \(A_{1}\) (that \(X_{1} = 1\)) happens in the first, so \(P(A_{1}|Z) = 1/2\), and likewise \(P(A_{2}|Z) = 1/2\). Both cannot happen, since then \(Z = 2\), so \(P(A_{1} \cap A_{2}|Z) = 0\).The product of the conditional probabilities is \(1/4 \ne 0\), so the events, independent unconditionally, are not conditionally independent given \(Z = 0\). Options A, B and C are the true values.

Q: The probability generating function of a particular random variable is given as :\[ P(s) = \frac{1}{2}(s + s^{2}) \]What is the variance of the random variable ?

The variance is 1/4.\(P(s) = \tfrac{1}{2}s + \tfrac{1}{2}s^{2}\) means the variable takes the values 1 and 2, each with probability 1/2. Using the generating function: \(P'(1) = \tfrac{1}{2} + 1 = 3/2\) is the mean and \(P''(1) = 1\) is \(E[X(X - 1)]\), so\[ \begin{aligned} Var(X) &= P''(1) + P'(1) - P'(1)^{2} \\ &= 1 + \tfrac{3}{2} - \tfrac{9}{4} = \tfrac{1}{4} \end{aligned} \]Directly: \(E(X^{2}) = (1 + 4)/2 = 5/2\), and \(5/2 - 9/4 = 1/4\). 1/2 is the probability of each value, not the variance, and 1/3 and 2/3 do not follow from this generating function by any correct route.

Q: Let X be a r.v. with cdf, \(F(x)\), where :\[ F(x) = \begin{cases} 0 & \text{if } x \lt 0 \\ x/2 & \text{if } 0 \le x \lt 1 \\ 3/4 & \text{if } 1 \le x \lt 2 \\ 1 & \text{if } 2 \le x \end{cases} \]Then E(X) is given by :

\(E(X) =\) 1.The distribution is mixed. On [0, 1) the cdf rises continuously as \(x/2\), a density of 1/2 carrying total probability 1/2. At \(x = 1\) it jumps from 1/2 to 3/4, a mass of 1/4 at 1, and at \(x = 2\) it jumps from 3/4 to 1, a mass of 1/4 at 2. So\[ \begin{aligned} E(X) &= \int_{0}^{1} \frac{x}{2}\,dx + 1 \cdot \tfrac{1}{4} + 2 \cdot \tfrac{1}{4} \\ &= \tfrac{1}{4} + \tfrac{1}{4} + \tfrac{1}{2} = 1 \end{aligned} \]3/8 and 1/2 come from missing one or both point masses, and 3/2 from counting the mass at 2 twice.

11.

Let \(\{X_{n}\}\) be a sequence of independent random variables. Define the \(\sigma\)-field :

\[ F = \bigcap_{n=1}^{\infty} \sigma(X_{n}, X_{n+1}, \ldots). \]

Suppose \(A \in f_{e}\). Then, which of the following statements is more appropriate ?

A.\(P(A) = 0\)
B.\(P(A) = 1\)
C.\(P(A) = P(A^{C})\)
D.\(P(A) = 0\) or \(P(A) = 1\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The appropriate statement is P(A) = 0 or P(A) = 1.

F, the intersection of \(\sigma(X_{n}, X_{n+1}, \ldots)\) over all n, is the tail \(\sigma\)-field of the sequence (printed in the question as \(f_{e}\)). Kolmogorov's zero-one law says that when the \(X_{n}\) are independent, every tail event has probability 0 or 1.

Examples of tail events are 'the series \(\sum X_{n}\) converges' and 'the limit superior of \(X_{n}\) exceeds 5'. Each is unaffected by changing finitely many terms, and each is either almost certain or almost impossible.

Saying always 0 or always 1 is too strong, since both values occur for different tail events (A and its complement are both tail events). \(P(A) = P(A^{c})\) would force \(P(A) = 1/2\), which the law excludes.

Discuss in Forum

12.

If X is a positive random variable with probability density function \(f(x)\), then \(X^{-1}\) has probability density function :

A.\(1/f(x)\)
B.\(f(1/x)\)
C.\(1/x^{2} f(1/x)\)
D.\(1/x f(1/x)\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The density of \(X^{-1}\) is \(1/x^{2} f(1/x)\), that is \(\dfrac{1}{x^{2}} f\left(\dfrac{1}{x}\right)\) for \(x \gt 0\).

Let \(Y = 1/X\). For \(y \gt 0\), \(Y \le y\) exactly when \(X \ge 1/y\), so

\[ F_{Y}(y) = 1 - F_{X}(1/y) \]

Differentiating with the chain rule:

\[ \begin{aligned} f_{Y}(y) &= -f_{X}(1/y) \cdot \left(-\frac{1}{y^{2}}\right) \\ &= \frac{1}{y^{2}} f_{X}(1/y) \end{aligned} \]

The factor \(1/y^{2}\) is the Jacobian \(|dx/dy|\) of the change of variable. \(f(1/x)\) forgets the Jacobian, \(1/x \cdot f(1/x)\) has the wrong power of x (it does not integrate to 1 in general), and \(1/f(x)\) is not a density at all.

Discuss in Forum

13.

Let \(X_{1}\) and \(X_{2}\) be iid random variables with distribution function \(F(x)\). Then, \(P(X_{1} \le X_{2})\) is :

A.1/3
B.2/3
C.1/2
D.3/2

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

\(P(X_{1} \le X_{2}) = \)1/2.

Because \(X_{1}\) and \(X_{2}\) are independent with the same distribution, the pair \((X_{1}, X_{2})\) has the same joint distribution as \((X_{2}, X_{1})\). Hence \(P(X_{1} \lt X_{2}) = P(X_{2} \lt X_{1})\), and

\[ \begin{aligned} &P(X_{1} \lt X_{2}) + P(X_{1} = X_{2}) \\ &\quad + P(X_{1} \gt X_{2}) = 1 \end{aligned} \]

For a continuous distribution function, \(P(X_{1} = X_{2}) = 0\), so each strict inequality has probability 1/2 and \(P(X_{1} \le X_{2}) = 1/2\). This is the reading the question intends. If F had jumps, ties would have positive probability and the answer would be \(\frac{1}{2} + \frac{1}{2}P(X_{1} = X_{2})\), but that is not among the options either.

1/3 and 2/3 have no symmetry basis, and 3/2 exceeds 1, so it is not a probability.

Discuss in Forum

14.

Let X and Y be two random variables defined on the same probability space such that for each \(y \gt 0\), the conditional density function of X given \(Y = y\) is :

\[ \begin{aligned} f(x/y) &= \sqrt{\frac{y}{2\pi}}\, e^{-yx^{2}/2}, \\ &y \gt 0,\ x \in R. \end{aligned} \]

Let

\[ g(y) = \frac{1}{\sqrt{2\pi y}}\, e^{-y/2},\ y \gt 0. \]

Which of the following statements is more appropriate ?

A.E(X|Y) = 0
B.E(E(X|Y)) = 0
C.E(X) = 0
D.\(E(X) \ne E(E(X|Y))\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The most appropriate statement is E(X|Y) = 0.

Given \(Y = y\), X is normal with mean 0 and variance \(1/y\), so \(E(X|Y) = 0\) exactly. The marginal of Y is \(g(y)\), the chi-square density with 1 degree of freedom. Integrating out \(y\):

\[ \begin{aligned} f_{X}(x) &= \frac{1}{2\pi}\int_{0}^{\infty} e^{-y(1 + x^{2})/2}dy \\ &= \frac{1}{\pi(1 + x^{2})} \end{aligned} \]

So X has the standard Cauchy distribution, and \(E(X)\) does not exist. That rules out \(E(X) = 0\) and \(E(X) \ne E(E(X|Y))\), and makes \(E(E(X|Y)) = 0\) misleading: the iterated expectation is 0 although the expectation it should equal does not exist, since the tower property needs \(E|X| \lt \infty\).

Discuss in Forum

15.

Let Y be a Bernoulli random variable with :

\(P(Y = 0) = P(Y = 1) = 1/2\).

Let \(X_{n} = (1 - Y/n^{\alpha})^{n}\), \(\alpha \ge 0\), then, which of the following statements are correct ?

(i) \(X_{n} \xrightarrow{d} Y\) when \(0 \le \alpha \lt 1\)
(ii) \(X_{n} \xrightarrow{d} e^{-Y}\) when \(\alpha = 1\)
(iii) \(X_{n} \xrightarrow{d} Y\) when \(\alpha \gt 1\)

A.(i) and (ii)
B.(ii) and (iii)
C.(i) and (iii)
D.Only (i)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The correct statements are (i) and (ii).

If \(Y = 0\), \(X_{n} = 1\) for all \(n\). If \(Y = 1\), \(X_{n} = (1 - n^{-\alpha})^{n}\), which tends to 0 when \(0 \le \alpha \lt 1\) (for \(\alpha = 0\) it is exactly 0), to \(e^{-1}\) when \(\alpha = 1\), and to 1 when \(\alpha \gt 1\), since then \(n \cdot n^{-\alpha} \to 0\).

(i): for \(\alpha \lt 1\), \(X_{n} \to 1 - Y\), which takes values 1 and 0 with probability 1/2 each, the same distribution as Y. So \(X_{n} \to Y\) in distribution. (ii): for \(\alpha = 1\), \(X_{n} \to e^{-Y}\) (1 if \(Y = 0\), \(e^{-1}\) if \(Y = 1\)). (iii): for \(\alpha \gt 1\), \(X_{n} \to 1\), a constant, which is not distributed like Y.

Discuss in Forum

16.

Let \(\Omega\) be a countably infinite set and let F consists of all subsets of \(\Omega\). Define :

\[ \mu(A) = \begin{cases} 0 & \text{if A is finite} \\ \infty & \text{if A is infinite} \end{cases} \]

Which of the following statements are true ?

(i) \(\mu\) is finitely additive
(ii) \(\mu\) is countably additive
(iii) \(\{A_{n}\}\) is an \(\uparrow\) sequence of sets with \(\mu(A_{n}) = 0\ \forall\, n\), but \(\mu(\Omega) = \infty\).

A.(i) and (ii)
B.(i) and (iii)
C.(ii) and (iii)
D.(i), (ii) and (iii)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The true statements are (i) and (iii).

(i): for disjoint A and B, \(A \cup B\) is finite exactly when both are, so \(\mu(A \cup B) = \mu(A) + \mu(B)\) in every case (0 + 0, or \(\infty\) on both sides). \(\mu\) is finitely additive.

(ii) fails: \(\Omega\) is the countable disjoint union of its singletons, each of measure 0, yet \(\mu(\Omega) = \infty \ne 0\). So \(\mu\) is not countably additive.

(iii) is true and shows the same failure as continuity: list \(\Omega = \{\omega_{1}, \omega_{2}, \ldots\}\) and let \(A_{n} = \{\omega_{1}, \ldots, \omega_{n}\}\). The \(A_{n}\) increase to \(\Omega\), each \(\mu(A_{n}) = 0\), but \(\mu(\Omega) = \infty\).

Discuss in Forum

17.

Let X and Y be two independent zero-mean unit variance Gaussian random variables defined on a common probability space. Define \(U = X + Y\) and \(V = X - Y\). Let \(F = \sigma(X)\). Then, which of the following statements is not true ?

A.E(U|F) = X a.s.
B.E(V|F) = X a.s.
C.E(U + V|F) = 2X a.s.
D.E(U|F) and E(V|F) are independent

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The statement that is not true is E(U|F) and E(V|F) are independent.

Both conditional expectations equal the same random variable X, which has variance 1. A non-degenerate random variable is not independent of itself, since \(P(X \le 0, X \le 0) = 1/2 \ne 1/4\). (U and V themselves are independent, being uncorrelated jointly normal variables, which is the trap.)

Discuss in Forum

18.

Let \(X_{1}\) and \(X_{2}\) be two iid random variables with :

\(P(X_{1} = 1) = P(X_{1} = -1) = 1/2\).

Let \(Z = X_{1} + X_{2}\), \(A_{i} = X_{i}^{-1}(\{1\})\), \(i = 1, 2\).

Which of the following is not true on \(Z^{-1}(\{0\})\) ?

A.\(P(A_{1}|Z) = 1/2\)
B.\(P(A_{2}|Z) = 1/2\)
C.\(P(A_{1} \cap A_{2}|Z) = 0\)
D.\(P(A_{1}|Z)\,P(A_{2}|Z) = P(A_{1} \cap A_{2}|Z)\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The statement that is not true is \(P(A_{1}|Z)P(A_{2}|Z) = P(A_{1} \cap A_{2}|Z)\).

On \(\{Z = 0\}\) the outcomes are (1, -1) and (-1, 1), each with conditional probability 1/2. \(A_{1}\) (that \(X_{1} = 1\)) happens in the first, so \(P(A_{1}|Z) = 1/2\), and likewise \(P(A_{2}|Z) = 1/2\). Both cannot happen, since then \(Z = 2\), so \(P(A_{1} \cap A_{2}|Z) = 0\).

The product of the conditional probabilities is \(1/4 \ne 0\), so the events, independent unconditionally, are not conditionally independent given \(Z = 0\). Options A, B and C are the true values.

Discuss in Forum

19.

The probability generating function of a particular random variable is given as :

\[ P(s) = \frac{1}{2}(s + s^{2}) \]

What is the variance of the random variable ?

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

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The variance is 1/4.

\(P(s) = \tfrac{1}{2}s + \tfrac{1}{2}s^{2}\) means the variable takes the values 1 and 2, each with probability 1/2. Using the generating function: \(P'(1) = \tfrac{1}{2} + 1 = 3/2\) is the mean and \(P''(1) = 1\) is \(E[X(X - 1)]\), so

\[ \begin{aligned} Var(X) &= P''(1) + P'(1) - P'(1)^{2} \\ &= 1 + \tfrac{3}{2} - \tfrac{9}{4} = \tfrac{1}{4} \end{aligned} \]

Directly: \(E(X^{2}) = (1 + 4)/2 = 5/2\), and \(5/2 - 9/4 = 1/4\). 1/2 is the probability of each value, not the variance, and 1/3 and 2/3 do not follow from this generating function by any correct route.

Discuss in Forum

20.

Let X be a r.v. with cdf, \(F(x)\), where :

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

Then E(X) is given by :

A.3/8
B.1/2
C.3/2
D.1

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

\(E(X) =\) 1.

The distribution is mixed. On [0, 1) the cdf rises continuously as \(x/2\), a density of 1/2 carrying total probability 1/2. At \(x = 1\) it jumps from 1/2 to 3/4, a mass of 1/4 at 1, and at \(x = 2\) it jumps from 3/4 to 1, a mass of 1/4 at 2. So

\[ \begin{aligned} E(X) &= \int_{0}^{1} \frac{x}{2}\,dx + 1 \cdot \tfrac{1}{4} + 2 \cdot \tfrac{1}{4} \\ &= \tfrac{1}{4} + \tfrac{1}{4} + \tfrac{1}{2} = 1 \end{aligned} \]

3/8 and 1/2 come from missing one or both point masses, and 3/2 from counting the mass at 2 twice.

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