Statistics Graduate Level Quiz: Conditional Distributions And Marginalization

10 questions · exam conditions

0:00

Conditional Distributions And MarginalizationQuestion 1 of 10

Suppose $(X,Y)$ is jointly normal with zero means, $\operatorname{Var}(X)=4$ , $\operatorname{Var}(Y)=9$ , and $\operatorname{Cov}(X,Y)=3$ . Define $W=X+Y$ . Which distribution is the conditional distribution of $Y$ given $W=w$ ?

N\left(\frac{12}{19}w,\frac{27}{19}\right)

N\left(\frac{3}{4}w,\frac{27}{4}\right)

N\left(\frac{9}{19}w,\frac{108}{19}\right)

N\left(\frac{12}{19}w,\frac{171}{19}\right)

← Back to quizzes

Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Conditional Distributions And Marginalization

Practice Conditional Distributions And Marginalization 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 Distributions And Marginalization, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

How to use this quiz

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

$N\left(\frac{12}{19}w,\frac{27}{19}\right)$ (correct answer)
$N\left(\frac{3}{4}w,\frac{27}{4}\right)$
$N\left(\frac{9}{19}w,\frac{108}{19}\right)$
$N\left(\frac{12}{19}w,\frac{171}{19}\right)$

Explanation: When you see a joint normality problem asking for a conditional distribution, reach for the bivariate (or multivariate) normal conditional formulas: if

(X,Y)

is jointly normal, then

Y \mid (W=w)

is normal with mean

\mu_{Y|W} = \mu_Y + \frac{\text{Cov}(Y,W)}{\text{Var}(W)}(w - \mu_W)

and variance

\sigma^2_{Y|W} = \text{Var}(Y) - \frac{[\text{Cov}(Y,W)]^2}{\text{Var}(W)}

. First, compute the needed quantities. Since

W = X+Y

\text{Var}(W) = \text{Var}(X) + 2\text{Cov}(X,Y) + \text{Var}(Y) = 4 + 6 + 9 = 19

. And

\text{Cov}(Y,W) = \text{Cov}(Y, X+Y) = \text{Cov}(Y,X) + \text{Var}(Y) = 3 + 9 = 12

. All means are zero, so the conditional mean is

\frac{12}{19}w

. The conditional variance is

9 - \frac{144}{19} = \frac{171 - 144}{19} = \frac{27}{19}

. This confirms answer A,

N\!\left(\frac{12}{19}w,\, \frac{27}{19}\right)

. Answer B uses

\text{Var}(W) = 16

(perhaps treating

\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y)

and ignoring the covariance term entirely, then using only

\text{Var}(X)=4

), producing incorrect regression coefficients. Answer C gets

\text{Cov}(Y,W) = 9

(forgetting to add the cross-covariance), yielding the wrong mean

\frac{9}{19}w

and wrong variance. Answer D computes the conditional mean correctly as

\frac{12}{19}w

but reports

\text{Var}(W) = 19

as the conditional variance — a classic trap of confusing the marginal variance of

W

with the residual variance. Your strategy: always build a checklist — compute

\text{Var}(W)

(don't drop the covariance!), compute

\text{Cov}(Y,W)

, then plug into the two formulas. Most distractors target exactly one of these steps.

Question 2

A pair of continuous random variables has joint density $f_{X,Y}(x,y)=c(x+y)$ on the region $0<y<x<1$ and density zero elsewhere. For a fixed $y\in(0,1)$ , what is $P\left(X>\frac{1+y}{2}\mid Y=y\right)$ ?

$\frac{3+2y-5y^2}{4}$
$\frac{3+5y}{4(1+3y)}$ (correct answer)
$\frac{1+3y}{3+5y}$
$\frac{3+5y}{4(1+y)}$

Explanation: When working with conditional distributions of continuous random variables, your first instinct should be to derive the conditional density

f_{X|Y}(x|y)

by dividing the joint density by the marginal density of

Y

. Finding the marginal and conditional densities. For fixed

y \in (0,1)

x

ranges from

y

1

(since

y < x < 1

). The marginal is:

f_Y(y) = \int_y^1 c(x+y)\,dx = c\left[\frac{x^2}{2}+yx\right]_y^1 = c\left(\frac{1}{2}+y-\frac{y^2}{2}-y^2\right) = c\cdot\frac{1+2y-3y^2}{2}

The conditional density is then:

f_{X|Y}(x|y) = \frac{x+y}{\int_y^1(x+y)\,dx} = \frac{2(x+y)}{1+2y-3y^2}

Computing the conditional probability. Now integrate from

\frac{1+y}{2}

1

P\!\left(X>\tfrac{1+y}{2}\mid Y=y\right) = \frac{2}{1+2y-3y^2}\int_{(1+y)/2}^{1}(x+y)\,dx

The integral evaluates to

\frac{(1-y)^2}{8}(3+5y)/(1-y)^2 \cdot \frac{(1-y)^2}{4} \cdot \frac{3+5y}{4}

. More carefully:

\int_{(1+y)/2}^1(x+y)\,dx = \frac{(1-y)^2}{8}+\frac{y(1-y)}{2} = \frac{(1-y)(3+5y)}{8}

. Since

1+2y-3y^2=(1-y)(1+3y)

, the result is:

\frac{2(1-y)(3+5y)/8}{(1-y)(1+3y)} = \frac{3+5y}{4(1+3y)}

This confirms answer B. Choice A omits the normalization by the marginal entirely — a classic error. Choice C inverts the fraction. Choice D uses

(1+y)

instead of

(1+3y)

in the denominator, suggesting an algebraic mistake in computing the marginal. Study tip: Always carefully factor expressions like

1+2y-3y^2=(1-y)(1+3y)

— messy algebra in the denominator is where most errors occur on these problems.

Question 3

A latent binary state satisfies $P(X=1)=0.4$ . An intermediate binary measurement has probabilities $P(Y=1\mid X=1)=0.8$ and $P(Y=1\mid X=0)=0.3$ . A final binary test depends only on the intermediate measurement, with $P(Z=1\mid Y=1)=0.9$ and $P(Z=1\mid Y=0)=0.2$ .

After marginalizing over the unobserved intermediate measurement, what is $P(X=1\mid Z=1)$ ?

$\frac{76}{117}$
$\frac{16}{25}$
$\frac{152}{275}$ (correct answer)
$\frac{2}{5}$

Explanation: When you see a latent variable problem like this, recognize it as a Bayesian network with a hidden middle node:

X \to Y \to Z

. Because

Z

is conditionally independent of

X

given

Y

, you must marginalize over

Y

to connect

X

and

Z

directly — then apply Bayes' theorem. Start by computing

P(Z=1 \mid X=x)

for each state of

X

, using the law of total probability over

Y

P(Z=1 \mid X=1) = P(Z=1\mid Y=1)P(Y=1\mid X=1) + P(Z=1\mid Y=0)P(Y=0\mid X=1)

= (0.9)(0.8) + (0.2)(0.2) = 0.72 + 0.04 = 0.76

P(Z=1 \mid X=0) = (0.9)(0.3) + (0.2)(0.7) = 0.27 + 0.14 = 0.41

Now compute the marginal

P(Z=1)

P(Z=1) = (0.76)(0.4) + (0.41)(0.6) = 0.304 + 0.246 = 0.550

Applying Bayes' theorem:

P(X=1 \mid Z=1) = \frac{(0.76)(0.4)}{0.550} = \frac{0.304}{0.550} = \frac{304}{550} = \frac{152}{275}

This confirms C is correct. Choice A,

\frac{76}{117}

, likely arises from an arithmetic error in computing

P(Z=1)

— perhaps forgetting to weight

P(X=0)=0.6

correctly. Choice B,

\frac{16}{25}

, corresponds to a clean but incorrect simplification, suggesting someone skipped the marginalization step entirely and used

P(Y=1\mid X=1)

directly. Choice D,

\frac{2}{5}

, is simply the prior

P(X=1)

, the classic error of ignoring the evidence

Z=1

altogether. Your takeaway: in any chain

X \to Y \to Z

with

Y

unobserved, always marginalize over

Y

first, then apply Bayes. Skipping the marginalization step is the most common trap on latent-variable problems.

Question 4

A latent event rate has a Gamma distribution with shape $3$ and rate $2$ . Conditional on this rate, observed counts in disjoint equal-length intervals are independent Poisson random variables with that rate.

If the count in one interval is $X=2$ , what is the conditional probability that the count $Y$ in a new interval equals zero?

$\left(\frac{3}{4}\right)^5$ (correct answer)
$\left(\frac{2}{3}\right)^3$
$\exp\left(-\frac{5}{3}\right)$
$\left(\frac{3}{4}\right)^3$

Explanation: When you see a Poisson rate drawn from a Gamma prior, think Negative Binomial / Gamma-Poisson conjugacy. The key insight is that observing data updates your Gamma prior, and you must condition on that updated posterior before predicting new observations. Start with the prior:

\lambda \sim \text{Gamma}(\alpha=3, \beta=2)

(rate parameterization). After observing

X=2

counts in one interval, the posterior is

\lambda \mid X=2 \sim \text{Gamma}(3+2,\ 2+1) = \text{Gamma}(5, 3)

, since the Gamma is conjugate to the Poisson with posterior shape

\alpha + x

and posterior rate

\beta + 1

. Now find

P(Y=0 \mid X=2)

by averaging over this posterior:

P(Y=0 \mid X=2) = E\!\left[e^{-\lambda} \mid X=2\right] = M_\lambda(-1)

where

M_\lambda

is the moment-generating function of

\text{Gamma}(5,3)

. For a Gamma with rate

\beta

, the MGF evaluated at

-1

\left(\frac{\beta}{\beta+1}\right)^{\alpha} = \left(\frac{3}{4}\right)^5

. This confirms answer A. Answer B,

\left(\frac{2}{3}\right)^3

, uses the original prior parameters

(3, 2)

without updating — it ignores the observation entirely. Answer C,

\exp(-5/3)

, treats the posterior mean

5/3

as a fixed rate and plugs it into the Poisson formula, confusing

E[e^{-\lambda}]

with

e^{-E[\lambda]}

(Jensen's inequality violation). Answer D,

\left(\frac{3}{4}\right)^3

, correctly uses the updated rate

\beta=3

but forgets to update the shape, leaving it at the prior value of

3

instead of

5

. Study tip: Always update both the shape and rate of your Gamma posterior before computing predictive probabilities — forgetting either update is the most common trap in Bayesian Poisson problems.

Question 5

Let $(P_1,P_2,P_3)$ have a Dirichlet distribution with parameters $(2,3,5)$ , and define $S=P_1+P_2$ . For a fixed $s\in(0,1)$ , what is $P(P_1>s/2\mid S=s)$ ?

$\frac{1}{2}$
$\frac{2}{5}$
$\frac{5}{16}$ (correct answer)
$\frac{11}{16}$

Explanation: Whenever you see a conditional probability question involving a Dirichlet distribution, your first instinct should be to exploit the aggregation and conditional properties of the Dirichlet. Here's the key theorem: if

(P_1, P_2, P_3) \sim \text{Dir}(2,3,5)

, then the sum

S = P_1 + P_2 \sim \text{Beta}(5, 5)

, and given

S = s

, the conditional distribution of

P_1/S

\text{Beta}(2, 3)

. This is the Dirichlet's "neutrality" property — conditioned on the sum of a subset, the relative split within that subset follows its own Beta distribution with the original parameters. So you need

P(P_1 > s/2 \mid S = s) = P\!\left(\frac{P_1}{S} > \frac{1}{2} \,\Big|\, S = s\right)

, which equals

P(V > 1/2)

where

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

. Computing this: the Beta(2,3) density is

f(v) = 12v(1-v)^2

, so

P(V > 1/2) = 12\int_{1/2}^{1} v(1-v)^2\,dv = 12 \cdot \frac{5}{192} = \frac{5}{16}.

That confirms the answer is C. Choice A,

1/2

, would only be correct if

V

were uniform or symmetric — Beta(2,3) is skewed left, so the median is below

1/2

, making the upper-tail probability less than

1/2

. Choice B,

2/5

, is simply the mean of Beta(2,3), which students sometimes confuse with a tail probability. Choice D,

11/16

, is actually

P(V < 1/2) = 1 - 5/16

, the complementary event — a classic sign-flip error. Study tip: Always check whether a Dirichlet conditional reduces to a Beta ratio problem — this transforms a multivariate integral into a one-dimensional Beta tail computation you can directly evaluate.

Question 6

Let $X_1,X_2,X_3$ be independent uniform random variables on $(0,\theta)$ , and define $L=\min_i X_i$ and $M=\max_i X_i$ . For $0<t<m<\theta$ , what is the conditional density of $L$ given $M=m$ ?

$f_{L\mid M}(t\mid m)=\dfrac{3(m-t)^2}{m^3}$
$f_{L\mid M}(t\mid m)=\dfrac{2(m-t)}{m^2}$ (correct answer)
$f_{L\mid M}(t\mid m)=\dfrac{1}{m}$
$f_{L\mid M}(t\mid m)=\dfrac{2t}{m^2}$

Explanation: When you encounter a question about conditional distributions of order statistics, your first instinct should be to work from the joint density of the min and max, then apply the definition of conditional density. For

X_1, X_2, X_3 \overset{iid}{\sim} \text{Uniform}(0,\theta)

, the joint density of

(L, M)

— the minimum and maximum — is derived from order statistic theory. With

n=3

, the joint density of the minimum

X_{(1)}=L

and maximum

X_{(3)}=M

is:

f_{L,M}(t,m) = \frac{n!}{(1-1)!(n-2)!(1-1)!} \cdot \frac{1}{\theta^n}(m-t)^{n-2} = \frac{3!}{1} \cdot \frac{1}{\theta^3}(m-t)^1 = \frac{6(m-t)}{\theta^3}

for

0 < t < m < \theta

. The marginal density of

M

is:

f_M(m) = \int_0^m \frac{6(m-t)}{\theta^3}\,dt = \frac{6}{\theta^3} \cdot \frac{m^2}{2} = \frac{3m^2}{\theta^3}.

The conditional density is then:

f_{L\mid M}(t\mid m) = \frac{f_{L,M}(t,m)}{f_M(m)} = \frac{6(m-t)/\theta^3}{3m^2/\theta^3} = \frac{2(m-t)}{m^2},

confirming answer B. Answer A uses

(m-t)^2

in the numerator, which would arise if

n=4

, not

n=3

. Answer C gives a uniform distribution on

(0,m)

, ignoring the ordering constraint imposed by the joint density. Answer D has

t

in the numerator instead of

(m-t)

, reflecting a confusion about which endpoint the middle observation clusters near. Your strategy: always write out the joint density of the relevant order statistics explicitly, then divide by the appropriate marginal. The

\theta

cancels cleanly, which is a good self-check that your answer is correct.

Question 7

Let $X$ be uniform on $(0,1)$ . Given $X=x$ , the random variable $Y$ equals zero with probability $x$ ; with the remaining probability $1-x$ , it is uniformly distributed on $(0,1)$ .

Which statement correctly compares the conditional means at the atom $Y=0$ and at a fixed continuous value $y\in(0,1)$ ?

$E(X\mid Y=0)=\frac{2}{3}$ and $E(X\mid Y=y)=\frac{1}{3}$ (correct answer)
$E(X\mid Y=0)=\frac{1}{3}$ and $E(X\mid Y=y)=\frac{2}{3}$
$E(X\mid Y=0)=\frac{1}{2}$ and $E(X\mid Y=y)=\frac{1}{2}$
$E(X\mid Y=0)=\frac{1}{3}$ and $E(X\mid Y=y)=\frac{1}{2}$

Explanation: This problem tests your ability to work with mixed distributions — where a random variable has both a discrete atom and a continuous component — and to apply Bayes' theorem to find conditional expectations. The key insight is that

Y=0

is an atom (positive probability mass), while

Y=y

for

y\in(0,1)

requires conditioning on a continuous event. You must use different tools for each. For the atom $Y=0$ : The probability that

Y=0

given

X=x

x

. So the conditional density of

X

given

Y=0

is proportional to

x \cdot 1 = x

(since

X\sim\text{Uniform}(0,1)

). Normalizing:

f_{X|Y=0}(x) = 2x

, a Beta(2,1) distribution. Thus

E(X\mid Y=0) = \frac{2}{3}

. Intuitively, larger

x

makes

Y=0

more likely, so observing

Y=0

shifts the posterior toward larger

x

. For $Y=y\in(0,1)$ : Given

X=x

, the density of

Y

y

(1-x)\cdot 1 = 1-x

. So

f_{X|Y=y}(x) \propto (1-x)

, a Beta(1,2) distribution. Thus

E(X\mid Y=y) = \frac{1}{3}

. Observing a continuous value means

Y$ wasn't zero, favoring smaller

x$$. This confirms answer A is correct. B swaps the two values — a tempting error if you confuse which event (atom vs. continuous) favors large or small

x

. C incorrectly assigns

\frac{1}{2}

to both, ignoring how the observation updates the prior. D mixes up the continuous case with the marginal mean of

X

. When you see mixed distributions, always apply Bayes by weighting the prior by the likelihood of the observed event — and remember that atoms and densities require separate likelihood calculations.

Question 8

Let $X$ and $Y$ be binary. The conditional probabilities are $P(X=1\mid Y=0)=0.2$ and $P(X=1\mid Y=1)=0.8$ . It is also known that $P(Y=1\mid X=1)=2/3$ .

What is $P(Y=1\mid X=0)$ ?

$\frac{1}{9}$ (correct answer)
$\frac{1}{6}$
$\frac{1}{3}$
$\frac{2}{9}$

Explanation: When you see a problem mixing conditional probabilities in both directions, your instinct should be Bayes' theorem combined with the law of total probability — these two tools together let you recover any marginal or conditional probability from the given information. Start by recovering

P(Y=1)

. You know

P(X=1 \mid Y=0) = 0.2

P(X=1 \mid Y=1) = 0.8

, and

P(Y=1 \mid X=1) = 2/3

. By Bayes' theorem:

P(Y=1 \mid X=1) = \frac{P(X=1 \mid Y=1)\,P(Y=1)}{P(X=1)}

The denominator expands via total probability:

P(X=1) = 0.8p + 0.2(1-p)

where

p = P(Y=1)

. Setting up the equation:

\frac{2}{3} = \frac{0.8p}{0.8p + 0.2(1-p)} = \frac{0.8p}{0.2 + 0.6p}

Cross-multiplying:

0.4 + 1.2p = 2.4p

, so

p = 1/3

. Now apply the same logic to find

P(Y=1 \mid X=0)

. With

P(X=0 \mid Y=1) = 0.2

P(X=0 \mid Y=0) = 0.8

, and

P(Y=1) = 1/3

P(Y=1 \mid X=0) = \frac{0.2 \cdot (1/3)}{0.2\cdot(1/3) + 0.8\cdot(2/3)} = \frac{1/15}{1/15 + 8/15} = \frac{1}{9}

This confirms A is correct. Choice B (

1/6

) likely comes from arithmetic errors in the Bayes calculation. Choice C (

1/3

) is simply

P(Y=1)

— a classic trap of confusing the marginal with the conditional. Choice D (

2/9

) may result from incorrectly using

P(Y=1 \mid X=1)

symmetrically without recomputing. As a study habit: always distinguish whether you have

P(A \mid B)

P(B \mid A)

— flipping them without Bayes is one of the most common errors in probability problems.

Question 9

Let $X$ and $Y$ be independent exponential random variables with respective rates $2$ and $1$ . Define $U=X/(X+Y)$ and $S=X+Y$ . Which pair gives both $P(U>1/2)$ and $E(S\mid U=u)$ ?

$P(U>1/2)=\frac{1}{3}$ and $E(S\mid U=u)=\frac{1}{1+u}$
$P(U>1/2)=\frac{1}{2}$ and $E(S\mid U=u)=\frac{1}{1+u}$
$P(U>1/2)=\frac{2}{3}$ and $E(S\mid U=u)=\frac{2}{1+u}$
$P(U>1/2)=\frac{1}{3}$ and $E(S\mid U=u)=\frac{2}{1+u}$ (correct answer)

Explanation: When you see a transformation like

U = X/(X+Y)

with

S = X+Y

, your first instinct should be to find the joint distribution via the Jacobian, then check for independence between

U

and

S

— a beautiful result that appears frequently with exponentials. With

X \sim \text{Exp}(2)

and

Y \sim \text{Exp}(1)

, the joint density is

f(x,y) = 2e^{-2x}e^{-y}

. Substituting

x = us

and

y = s(1-u)

with Jacobian

s

, the joint density of

(U,S)

becomes

f(u,s) = 2e^{-2us}e^{-s(1-u)} \cdot s = 2s\,e^{-s(1+u)}

, for

u \in (0,1)

s > 0

. This factors into a marginal in

s

(a Gamma) and a marginal in

u

f_U(u) = \frac{2}{(1+u)^2}

, confirming

U

and

S

are independent. For

P(U > 1/2)

: integrate

\int_{1/2}^{1} \frac{2}{(1+u)^2}\,du = \left[-\frac{2}{1+u}\right]_{1/2}^{1} = \frac{4}{3} - 1 = \frac{1}{3}

. So

P(U>1/2) = 1/3

. For

E(S \mid U = u)

: because

U

and

S

are independent, this equals

E(S)

. The marginal of

S

is Gamma(2,

1+u

)... wait — actually integrating out

u

, the conditional

S \mid U=u

has density

\propto s\,e^{-s(1+u)}

, a Gamma(2,

1+u

) with mean

\frac{2}{1+u}

. This confirms answer D. Choice A gets

E(S\mid U=u)

wrong by missing the factor of 2 from the Gamma shape parameter. Choice B compounds that error with the wrong probability. Choice C correctly computes

E(S\mid U=u) = \frac{2}{1+u}

but miscalculates the probability as

2/3

— likely by forgetting to normalize or reversing the integration bounds. Study tip: Always check if a transformation yields independence — when

U

and

S

factor cleanly, conditional expectations collapse to marginal ones, and the Gamma shape parameter directly controls the mean.

Question 10

Suppose $X\sim N(0,1)$ and, conditionally on $X=x$ , $Y$ is normal with mean $x^2$ and variance $1+x^2$ . Which statement about the marginal relationship between $X$ and $Y$ is correct?

$E(Y)=1$ , $\operatorname{Var}(Y)=3$ , and $X,Y$ are uncorrelated and independent
$E(Y)=1$ , $\operatorname{Var}(Y)=2$ , and $X,Y$ are uncorrelated and independent
$E(Y)=0$ , $\operatorname{Var}(Y)=4$ , and $\operatorname{Cov}(X,Y)=1$
$E(Y)=1$ , $\operatorname{Var}(Y)=4$ , and $X,Y$ are uncorrelated but dependent (correct answer)

Explanation: When you see a conditional distribution problem like this, your first instinct should be to reach for the Law of Total Expectation and Law of Total Variance — these are your workhorses for computing marginal moments. Finding E(Y): By the tower property,

E(Y) = E[E(Y|X)] = E[X^2] = \operatorname{Var}(X) + (E[X])^2 = 1 + 0 = 1

. Finding Var(Y): The Law of Total Variance gives

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

. The first term is

E[1+X^2] = 1 + E[X^2] = 2

. The second term is

\operatorname{Var}(X^2)

. Since

X \sim N(0,1)

X^2 \sim \chi^2(1)

, so

\operatorname{Var}(X^2) = 2

. Thus

\operatorname{Var}(Y) = 2 + 2 = 4

. Finding Cov(X,Y):

\operatorname{Cov}(X,Y) = E[XY] - E[X]E[Y] = E[E(XY|X)] - 0 = E[X \cdot E(Y|X)] = E[X \cdot X^2] = E[X^3] = 0

, since odd moments of a symmetric distribution vanish. So

X

and

Y

are uncorrelated, but are they independent? Since

E(Y|X) = X^2 \neq E(Y) = 1

, the conditional mean depends on

X

, confirming dependence. This is answer D. A and B both claim independence, which fails because the conditional distribution of

Y

clearly depends on

X

. B also gets the variance wrong. C incorrectly computes

E(Y) = 0

and claims a nonzero covariance. The key study tip: zero covariance never implies independence — this question is a perfect illustration. Always check whether the conditional distribution itself depends on the conditioning variable.