Statistics Graduate Level Quiz: Conditional Probability And Expectation

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Conditional Probability And ExpectationQuestion 1 of 10

Let $U_1$ and $U_2$ be independent uniform random variables on $[0,1]$ , and let $M=\max(U_1,U_2)$ . What is $E[M\mid M>1/2]$ ?

\frac{2}{3}

\frac{3}{4}

\frac{7}{9}

\frac{5}{6}

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Conditional Probability And Expectation

Practice Conditional Probability And Expectation in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Conditional Probability And Expectation, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let $U_1$ and $U_2$ be independent uniform random variables on $[0,1]$ , and let $M=\max(U_1,U_2)$ . What is $E[M\mid M>1/2]$ ?

$\frac{2}{3}$
$\frac{3}{4}$
$\frac{7}{9}$ (correct answer)
$\frac{5}{6}$

Explanation: When you see a conditional expectation involving order statistics, your instinct should be to work directly from the conditional distribution. Here,

M = \max(U_1, U_2)

has CDF

F_M(m) = m^2

and PDF

f_M(m) = 2m

[0,1]

. To find

E[M \mid M > 1/2]

, you need the conditional PDF of

M

given

M > 1/2

. First, compute

P(M > 1/2) = 1 - (1/2)^2 = 3/4

. The conditional PDF is then:

f_{M \mid M > 1/2}(m) = \frac{f_M(m)}{P(M > 1/2)} = \frac{2m}{3/4} = \frac{8m}{3}, \quad m \in (1/2, 1]

Now compute the conditional expectation:

E[M \mid M > 1/2] = \int_{1/2}^{1} m \cdot \frac{8m}{3}\, dm = \frac{8}{3}\int_{1/2}^{1} m^2\, dm = \frac{8}{3} \cdot \left[\frac{m^3}{3}\right]_{1/2}^{1} = \frac{8}{3} \cdot \frac{7}{24} = \frac{7}{9}

This confirms the answer is C. Choice A,

2/3

, is the unconditional mean of

M

— a classic trap when you forget to apply the conditioning. Choice B,

3/4

, is

P(M > 1/2)

itself, which students sometimes confuse with the conditional mean. Choice D,

5/6

, likely comes from incorrectly using a uniform conditional density on

(1/2, 1)

rather than the correct triangular-shaped conditional PDF. The key study tip: always re-derive the conditional PDF by truncating and renormalizing the original density. Never assume the conditional distribution inherits the same functional form as the unconditional one.

Question 2

The joint density of $X$ and $Y$ is proportional to $x+y$ on the region $0<x<y<1$ and is zero elsewhere. What is $P(X<1/2\mid Y>1/2)$ ?

$\frac{1}{2}$
$\frac{7}{16}$
$\frac{4}{7}$ (correct answer)
$\frac{2}{3}$

Explanation: When working with joint densities on non-rectangular regions, your first instinct should be to find the normalizing constant, then carefully set up conditional probability using the definition

P(A \mid B) = P(A \cap B)/P(B)

. Start by finding

c

such that

c\int_0^1\int_0^y (x+y)\,dx\,dy = 1

. The inner integral gives

\int_0^y(x+y)\,dx = \frac{y^2}{2}+y^2 = \frac{3y^2}{2}

, so the double integral equals

\int_0^1 \frac{3y^2}{2}\,dy = \frac{1}{2}

. Thus

c = 2

and

f(x,y) = 2(x+y)

0 < x < y < 1

. Now compute

P(Y > 1/2)

. Integrate over

1/2 < y < 1

0 < x < y

P(Y>1/2) = 2\int_{1/2}^1\int_0^y(x+y)\,dx\,dy = 2\int_{1/2}^1\frac{3y^2}{2}\,dy = 3\int_{1/2}^1 y^2\,dy = 3\left[\frac{y^3}{3}\right]_{1/2}^1 = 1 - \frac{1}{8} = \frac{7}{8}.

Next,

P(X < 1/2,\, Y > 1/2)

requires splitting into regions:

1/2 < y < 1,\, 0 < x < \min(y, 1/2)

. For

y > 1/2

x

runs from

0

1/2

2\int_{1/2}^1\int_0^{1/2}(x+y)\,dx\,dy = 2\int_{1/2}^1\left(\frac{1}{8}+\frac{y}{2}\right)dy = 2\left[\frac{y}{8}+\frac{y^2}{4}\right]_{1/2}^1 = 2\cdot\frac{1}{4} = \frac{1}{2}.

P(X<1/2\mid Y>1/2) = \frac{1/2}{7/8} = \frac{4}{7}

, confirming C. Choice A (

1/2

) ignores conditioning entirely. Choice B (

7/16

) likely comes from forgetting the factor of 2 or misidentifying the integration region. Choice D (

2/3

) results from computing

P(Y>1/2)

incorrectly as

3/4

. Study tip: Always sketch the support region before integrating — on triangular domains like

0<x<y<1

, the limits change depending on which event you're conditioning on, and a diagram prevents costly region errors.

Question 3

A screening program operates in two regions. A randomly selected participant comes from Region A with probability $0.40$ and Region B with probability $0.60$ . In Region A, disease prevalence is $0.10$ , test sensitivity is $0.90$ , and specificity is $0.95$ . In Region B, disease prevalence is $0.02$ , sensitivity is $0.80$ , and specificity is $0.98$ .

Given that the randomly selected participant tests positive, what is the probability that the participant came from Region A?

$0.717$ (correct answer)
$0.400$
$0.790$
$0.667$

Explanation: Whenever you see a question involving conditional probability across multiple subgroups, reach for Bayes' theorem combined with the law of total probability. Here, you need to find the probability a positive test came from Region A — meaning you must work backward from an observed outcome. Start by computing the probability of testing positive in each region. For Region A:

P(+|A) = P(D)\cdot\text{Sens} + P(\bar{D})\cdot(1-\text{Spec}) = (0.10)(0.90) + (0.90)(0.05) = 0.09 + 0.045 = 0.135

. For Region B:

P(+|B) = (0.02)(0.80) + (0.98)(0.02) = 0.016 + 0.0196 = 0.0356

. Now apply total probability:

P(+) = (0.40)(0.135) + (0.60)(0.0356) = 0.054 + 0.02136 = 0.07536

. Finally, Bayes' theorem gives:

P(A|+) = \frac{(0.40)(0.135)}{0.07536} = \frac{0.054}{0.07536} \approx 0.717

, confirming A is correct. Choice B (0.400) is the trap of simply reporting the prior probability of Region A — ignoring the diagnostic information entirely. Choice C (0.790) likely results from omitting the false-positive term when computing

P(+|B)

, underestimating Region B's contribution to positive tests. Choice D (0.667) resembles an incorrect ratio using only the true-positive components without properly weighting by region priors. Strategy tip: In multi-group Bayes problems, always compute the full

P(+|\text{region})

using both sensitivity and false-positive rate — forgetting false positives is the single most common error on these problems.

Question 4

Conditional on an unknown response probability $P$ , patient outcomes are independent Bernoulli trials. The prior distribution is $P\sim\operatorname{Beta}(2,3)$ . Four patients are observed, of whom three respond and one does not.

What is the posterior predictive probability that exactly one of the next two patients responds?

$\frac{4}{9}$ (correct answer)
$\frac{40}{81}$
$\frac{5}{9}$
$\frac{2}{5}$

Explanation: When you see a Bayesian problem asking for a posterior predictive probability, your goal is to average the likelihood of future observations over the posterior distribution of the unknown parameter — not plug in a point estimate. Start by updating the prior. With

P \sim \text{Beta}(2,3)

and observing 3 successes and 1 failure, the posterior is

P \mid \text{data} \sim \text{Beta}(2+3,\, 3+1) = \text{Beta}(5,4)

, giving posterior mean

\frac{5}{9}

. Now compute the posterior predictive probability that exactly 1 of the next 2 patients responds. You need:

\Pr(X=1) = \binom{2}{1}\mathbb{E}[P(1-P)] = 2\,\mathbb{E}[P - P^2]

Using moments of

\text{Beta}(5,4)

\mathbb{E}[P] = \frac{5}{9}

and

\mathbb{E}[P^2] = \frac{\alpha(\alpha+1)}{(\alpha+\beta)(\alpha+\beta+1)} = \frac{5 \cdot 6}{9 \cdot 10} = \frac{30}{90} = \frac{1}{3}

. So:

\Pr(X=1) = 2\!\left(\frac{5}{9} - \frac{1}{3}\right) = 2 \cdot \frac{2}{9} = \frac{4}{9}

This confirms A is correct. Choice B,

\frac{40}{81}

, is the trap of plugging the posterior mean directly into the binomial formula:

2 \cdot \frac{5}{9} \cdot \frac{4}{9} = \frac{40}{81}

— this ignores parameter uncertainty. Choice C,

\frac{5}{9}

, is simply the posterior mean itself, mistaking it for a predictive probability. Choice D,

\frac{2}{5}

, corresponds to using the prior mean

\frac{2}{5}

and never updating. Key tip: Never substitute the posterior mean into a binomial formula for predictive probabilities — always integrate over the full posterior using

\mathbb{E}[P^k]

moments.

Question 5

Two independent failure times satisfy $T_1\sim\operatorname{Exponential}(2)$ and $T_2\sim\operatorname{Exponential}(1)$ , where the parameters are rates.

What is $E[T_1\mid T_1>1,\ T_2>1,\ T_1<T_2]$ ?

$\frac{1}{3}$
$\frac{4}{3}$ (correct answer)
$\frac{3}{2}$
$2$

Explanation: When conditioning on multiple events simultaneously, your first instinct should be to identify the joint conditional distribution carefully — especially when competing exponentials are involved. Here, you need

E[T_1 \mid T_1 > 1,\ T_2 > 1,\ T_1 < T_2]

. By the memoryless property of exponentials, conditioning on

T_1 > 1

and

T_2 > 1

shifts both clocks: let

S_1 = T_1 - 1 \mid T_1 > 1 \sim \text{Exp}(2)

and

S_2 = T_2 - 1 \mid T_2 > 1 \sim \text{Exp}(1)

, independently. The condition

T_1 < T_2

becomes

S_1 < S_2

. So the problem reduces to finding

E[1 + S_1 \mid S_1 < S_2] = 1 + E[S_1 \mid S_1 < S_2]

. For independent exponentials with rates

\lambda_1 = 2

and

\lambda_2 = 1

, the conditional expectation

E[S_1 \mid S_1 < S_2]

equals

\frac{1}{\lambda_1 + \lambda_2} = \frac{1}{3}

. This is a standard result: given that

S_1

"wins" the race, the winning time is

\text{Exp}(\lambda_1 + \lambda_2)

, giving mean

\frac{1}{3}

. Therefore

E[T_1 \mid \cdots] = 1 + \frac{1}{3} = \frac{4}{3}

, confirming B. Choice A (

\frac{1}{3}

) forgets to add back the shifted baseline of 1 after applying the memoryless property. Choice C (

\frac{3}{2}

) likely confuses the combined rate with

\lambda_1 + \lambda_2 = 3

but adds

1

incorrectly, or misapplies the unconditional mean of

T_1

. Choice D (

2

) is simply the unconditional mean

1/\lambda_1 = 1/2

... wait — it may reflect treating the problem as just

E[T_1] = 1/\lambda_1

with the wrong rate, ignoring all conditioning. Study tip: Always apply the memoryless property first to simplify conditioning on survival past a threshold, then use the competing-exponentials "race" formula

E[\min] = \frac{1}{\sum \lambda_i}

for the residual.

Question 6

A study enrolls a random number $N$ of subjects, where $N$ has a Poisson distribution with mean $2$ . Conditional on $N$ , the outcome for each enrolled subject has conditional mean $E[X_i\mid N]=1+N/2$ . No conditional independence assumption is made. Define the total outcome as $S=\sum_{i=1}^N X_i$ , with $S=0$ when $N=0$ .

What is $E[S]$ ?

$2$
$3$
$4$
$5$ (correct answer)

Explanation: When a sum has a random number of terms, the key tool is the law of total expectation:

E[S] = E[E[S \mid N]]

. Start by conditioning on

N

, then average over the randomness in

N

itself. Given

N = n

, the conditional expectation of the sum is:

E[S \mid N = n] = \sum_{i=1}^{n} E[X_i \mid N = n] = n \cdot \left(1 + \frac{n}{2}\right) = n + \frac{n^2}{2}

Notice that no conditional independence assumption is needed here — you only need the conditional mean of each

X_i

, not anything about their joint distribution. Now apply the law of total expectation:

E[S] = E\!\left[N + \frac{N^2}{2}\right] = E[N] + \frac{1}{2}E[N^2]

Since

N \sim \text{Poisson}(2)

, you have

E[N] = 2

and

\text{Var}(N) = 2

, so

E[N^2] = \text{Var}(N) + (E[N])^2 = 2 + 4 = 6

. Therefore:

E[S] = 2 + \frac{1}{2}(6) = 2 + 3 = 5

The correct answer is D. Choice A (

2

) likely comes from computing

E[N]

alone and ignoring the

N

-dependent mean structure. Choice B (

3

) might arise from computing

\frac{1}{2}E[N^2]

without adding

E[N]

. Choice C (

4

) could result from forgetting to account for the variance term when computing

E[N^2]

, mistakenly using

E[N]^2 = 4

instead of

E[N^2] = 6

. The study tip: whenever you see

E[N^2]

in a Poisson problem, always decompose it as

\text{Var}(N) + (E[N])^2

. Forgetting the variance term is one of the most common computational errors on graduate-level probability exams.

Question 7

Independent event counts satisfy $X\sim\operatorname{Poisson}(2)$ and $Y\sim\operatorname{Poisson}(3)$ .

Independent event counts satisfy $X\sim\operatorname{Poisson}(2)$ and $Y\sim\operatorname{Poisson}(3)$ . What is $E[X\mid X+Y=4,\ X\geq 1]$ ?

$\frac{8}{5}$
$\frac{125}{68}$ (correct answer)
$\frac{544}{625}$
$2$

Explanation: When you see a conditional expectation involving Poisson random variables with a sum constraint, reach for the key theorem: given

X \sim \text{Poisson}(\lambda)

Y \sim \text{Poisson}(\mu)

independent, the conditional distribution of

X \mid X+Y=n

\text{Binomial}\!\left(n, \frac{\lambda}{\lambda+\mu}\right)

. Here that means

X \mid X+Y=4 \sim \text{Binomial}(4, 2/5)

. But you also need

X \geq 1

, so you're working with a truncated distribution. The conditional expectation becomes:

E[X \mid X+Y=4,\ X\geq 1] = \frac{E[X \cdot \mathbf{1}_{X\geq 1} \mid X+Y=4]}{P(X\geq 1 \mid X+Y=4)}

Let

p = 2/5

n=4

. Then

P(X=0 \mid X+Y=4) = (3/5)^4 = 81/625

, so

P(X \geq 1) = 1 - 81/625 = 544/625

. The unconditional binomial mean is

E[X \mid X+Y=4] = 4 \cdot (2/5) = 8/5

, and since

X \cdot \mathbf{1}_{X=0} = 0

, we have

E[X \cdot \mathbf{1}_{X\geq1}] = 8/5

. Therefore:

E[X \mid X+Y=4,\ X\geq 1] = \frac{8/5}{544/625} = \frac{8}{5} \cdot \frac{625}{544} = \frac{5000}{2720} = \frac{125}{68}

This confirms B. Choice A,

8/5

, is the binomial mean without truncation — forgetting the

X \geq 1

constraint. Choice C,

544/625

, is actually

P(X \geq 1 \mid X+Y=4)

, a probability mistaken for an expectation. Choice D,

2

, is simply

E[X]

with no conditioning at all. When conditioning on multiple events, always decompose using

E[X \mid A \cap B] = E[X \cdot \mathbf{1}_B \mid A] / P(B \mid A)

— truncation changes the mean, and that denominator is the step students most often drop.

Question 8

Let $Y$ be standard normal, and suppose an integrable random variable $X$ satisfies $E[X\mid Y]=Y^2$ . Use $\phi(1)=0.2420$ and $1-\Phi(1)=0.1587$ .

What is the approximate value of $E[X\mid |Y|>1]$ ?

$1.000$
$1.525$
$2.525$ (correct answer)
$3.050$

Explanation: Whenever you see a conditional expectation problem involving a truncated normal distribution, your first instinct should be to use the law of total expectation combined with direct computation via the normal density. Since

E[X \mid Y] = Y^2

, the tower property gives

E[X \mid |Y| > 1] = E[Y^2 \mid |Y| > 1]

. This is because conditioning further on the event

\{|Y|>1\}

yields

E[E[X|Y] \mid |Y|>1] = E[Y^2 \mid |Y|>1]

. So the problem reduces to computing the second moment of a standard normal truncated to

|Y|>1

. Using the truncated normal formula:

E[Y^2 \mid |Y|>1] = \frac{E[Y^2 \cdot \mathbf{1}_{|Y|>1}]}{P(|Y|>1)}

The denominator is

2(1-\Phi(1)) = 2(0.1587) = 0.3174

. For the numerator, use integration by parts:

E[Y^2 \cdot \mathbf{1}_{|Y|>1}] = 2\int_1^\infty y^2 \phi(y)\,dy

. Since

\frac{d}{dy}[-\phi(y)] = y\phi(y)

, we get

\int_1^\infty y^2\phi(y)\,dy = \phi(1) + (1-\Phi(1)) = 0.2420 + 0.1587 = 0.4007

. So the numerator is

2(0.4007) = 0.8014

. Therefore

E[Y^2 \mid |Y|>1] = \frac{0.8014}{0.3174} \approx 2.525

, confirming C. Choice A (

1.000

) ignores the truncation entirely — that's

E[Y^2]

for an untruncated standard normal. Choice B (

1.525

) likely arises from computing only the numerator term

\phi(1)

and mishandling the denominator. Choice D (

3.050

) doubles the correct answer, a sign of forgetting to divide by the tail probability. Study tip: Memorize that

\int_a^\infty y^2\phi(y)\,dy = a\phi(a) + (1-\Phi(a))

— this identity appears repeatedly in truncated normal calculations on graduate-level exams.

Question 9

A random variable $X$ takes values $-2,-1,1,3$ with probabilities $0.1,0.2,0.3,0.4$ , respectively. Let $\mathcal{G}$ be the sigma-field generated by the event $\{X>0\}$ . What is $E[(X-E[X\mid\mathcal{G}])^2]$ ?

$\frac{79}{105}$ (correct answer)
$\frac{329}{100}$
$\frac{787}{210}$
$0$

Explanation: When you see a conditional expectation with respect to a sigma-field generated by a single event, recognize that

\mathcal{G} = \sigma(\{X > 0\})

partitions the sample space into exactly two atoms:

\{X > 0\} = \{X = 1 \text{ or } 3\}

and

\{X \leq 0\} = \{X = -2 \text{ or } -1\}

. The conditional expectation

E[X \mid \mathcal{G}]

is therefore a step function — constant on each atom. First, compute

E[X]

E[X] = (-2)(0.1) + (-1)(0.2) + (1)(0.3) + (3)(0.4) = -0.2 - 0.2 + 0.3 + 1.2 = 1.1

. On

\{X \leq 0\}

(probability 0.3):

E[X \mid \mathcal{G}] = \frac{(-2)(0.1) + (-1)(0.2)}{0.3} = \frac{-0.4}{0.3} = -\frac{4}{3}

. On

\{X > 0\}

(probability 0.7):

E[X \mid \mathcal{G}] = \frac{(1)(0.3) + (3)(0.4)}{0.7} = \frac{1.5}{0.7} = \frac{15}{7}

. Now compute

E[(X - E[X \mid \mathcal{G}])^2]

by summing over all four values. For

X = -2

(-2 + \frac{4}{3})^2 = \frac{4}{9}

; for

X = -1

(-1 + \frac{4}{3})^2 = \frac{1}{9}

; for

X = 1

(1 - \frac{15}{7})^2 = \frac{64}{49}

; for

X = 3

(3 - \frac{15}{7})^2 = \frac{36}{49}

. Weighting:

(0.1)\frac{4}{9} + (0.2)\frac{1}{9} + (0.3)\frac{64}{49} + (0.4)\frac{36}{49} = \frac{6}{90} + \frac{33.6}{49 \cdot 10} = \frac{79}{105}

, confirming A. Choice B (

\frac{329}{100}

) is

E[X^2] - (E[X])^2 = \text{Var}(X)

, a common confusion — you're asked for variance around the conditional mean, not the unconditional variance. Choice C mixes intermediate calculations incorrectly. Choice D would only be correct if

\mathcal{G}

were the full sigma-field generated by

X

itself, making the residual identically zero. Remember:

E[(X - E[X\mid\mathcal{G}])^2]

measures the unexplained variance after conditioning — it equals

\text{Var}(X) - \text{Var}(E[X\mid\mathcal{G}])

by the law of total variance, which you can use as a check.

Question 10

Suppose $X$ and $Y$ are square-integrable random variables satisfying $E[X\mid Y]=E[X]$ almost surely. Which statement must hold for every bounded measurable function $g$ ?

$X$ and $Y$ are necessarily independent random variables.
$E[Y\mid X]=E[Y]$ must also hold almost surely.
$\operatorname{Var}(X\mid Y)=\operatorname{Var}(X)$ must hold almost surely.
$\operatorname{Cov}(X,g(Y))=0$ whenever the covariance is defined. (correct answer)

Explanation: When you see a condition like

E[X \mid Y] = E[X]

, your instinct should be to explore what it implies about covariances — not independence or symmetry. This condition means

X

is "mean-independent" of

Y

, which is weaker than full independence but still has powerful covariance consequences. Here's why D is correct. For any bounded measurable

g

, you can compute:

\operatorname{Cov}(X, g(Y)) = E[Xg(Y)] - E[X]E[g(Y)]

By the tower property (iterated expectations):

E[Xg(Y)] = E\!\left[E[Xg(Y) \mid Y]\right] = E\!\left[g(Y)\,E[X \mid Y]\right] = E\!\left[g(Y)\cdot E[X]\right] = E[X]\,E[g(Y)]

\operatorname{Cov}(X, g(Y)) = 0

— exactly as D claims. Now for the distractors. A is wrong because mean-independence does not imply full independence. You can construct examples where

E[X \mid Y] = E[X]

yet the joint distribution is not a product measure. B is wrong because the conditional expectation condition is not symmetric — knowing

E[X \mid Y] = E[X]

says nothing about

E[Y \mid X]

. These are distinct properties, and neither implies the other in general. C is wrong because while

\operatorname{Var}(X) = E[\operatorname{Var}(X \mid Y)] + \operatorname{Var}(E[X \mid Y])

and the second term vanishes (since

E[X \mid Y]

is constant), this only tells you

E[\operatorname{Var}(X \mid Y)] = \operatorname{Var}(X)

— not that

\operatorname{Var}(X \mid Y)

equals

\operatorname{Var}(X)

almost surely. Study tip: The tower property is your best tool whenever conditional expectations appear — always ask whether you can condition inside an expectation to exploit a given constraint.