Statistics Graduate Level Quiz: Computing Via Conditioning

10 questions · exam conditions

0:00

Computing Via ConditioningQuestion 1 of 10

Let $X\sim N(0,1)$ . Conditional on $X$ , a random variable $Y$ has conditional mean $E[Y|X]=X^2$ and conditional variance $Var(Y|X)=1+X^2$ . What is $Var(Y)$ ?

2

3

4

5

← Back to quizzes

Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Computing Via Conditioning

Practice Computing Via Conditioning 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 Computing Via Conditioning, 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

Let $X\sim N(0,1)$ . Conditional on $X$ , a random variable $Y$ has conditional mean $E[Y|X]=X^2$ and conditional variance $Var(Y|X)=1+X^2$ . What is $Var(Y)$ ?

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

Explanation: Whenever you see a question involving a conditional mean and conditional variance, your instinct should be to reach for the Law of Total Variance:

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

. This decomposition splits total variance into two pieces — the average within-group variance and the variance of the group means. Here, you're given

E[Y|X] = X^2

and

\text{Var}(Y|X) = 1 + X^2

, with

X \sim N(0,1)

. Compute each piece separately. For the first term:

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

, since

E[X^2] = \text{Var}(X) = 1

for a standard normal. For the second term:

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

. Since

X \sim N(0,1)

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

, which has variance

2

. So

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

. Adding both pieces:

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

, confirming the answer is C. Choice A (

2

) comes from computing only the first term

E[\text{Var}(Y|X)]

and forgetting the second term entirely. Choice B (

3

) likely arises from correctly computing

E[\text{Var}(Y|X)] = 2

but mistakenly using

\text{Var}(X^2) = 1

instead of

2

. Choice D (

5

) may result from using

E[X^4] = 3

directly as

\text{Var}(X^2)

without subtracting

(E[X^2])^2 = 1

, giving

3

for the second term instead of

2

. Study tip: Always memorize that for

X \sim N(0,1)

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

(not 1!), and never skip the second term in the Law of Total Variance — it's the most common trap on questions like this.

Question 2

Let $P$ have a beta distribution with parameters $\alpha=2$ and $\beta=3$ . Conditional on $P$ , the random variables $X$ and $Y$ are independent and each has a binomial distribution with parameters $6$ and $P$ . What is $Var(X-Y)$ ?

$1.20$
$2.40$ (correct answer)
$2.88$
$5.28$

Explanation: When you see a problem mixing a Beta-distributed random variable with conditional Binomials, your framework should be the Law of Total Variance:

Var(X-Y) = E[Var(X-Y|P)] + Var(E[X-Y|P])

. Since

X

and

Y

are conditionally independent given

P

, we have

Var(X-Y|P) = Var(X|P) + Var(Y|P) = 2 \cdot 6P(1-P) = 12P(1-P)

, and

E[X-Y|P] = 6P - 6P = 0

. For the Beta(2,3) distribution, the key moments are:

E[P] = \frac{\alpha}{\alpha+\beta} = \frac{2}{5} = 0.4

E[P^2] = \frac{\alpha(\alpha+1)}{(\alpha+\beta)(\alpha+\beta+1)} = \frac{6}{30} = 0.2

Var(P) = E[P^2] - (E[P])^2 = 0.2 - 0.16 = 0.04

Now apply total variance:

E[Var(X-Y|P)] = 12E[P(1-P)] = 12(E[P] - E[P^2]) = 12(0.4 - 0.2) = 2.40

Var(E[X-Y|P]) = Var(0) = 0

Var(X-Y) = 2.40 + 0 = \mathbf{2.40}

, confirming answer B. Choice A (1.20) likely comes from forgetting that

Var(X-Y|P) = Var(X|P) + Var(Y|P)

— using only one binomial's variance instead of summing both. Choice C (2.88) results from incorrectly including a

Var(E[X-Y|P])

term using

Var(6P)

rather than recognizing the conditional mean is identically zero. Choice D (5.28) reflects adding

Var(6P)

twice, double-counting the second term of total variance. Your study tip: always check whether the conditional mean of the quantity of interest is a constant — if so, the second term of total variance vanishes, simplifying your calculation dramatically.

Question 3

An urn contains $20$ positions. Before sampling, each position is independently designated a success with probability $0.3$ , so the total number of successes $K$ is binomial with parameters $20$ and $0.3$ . After the designations are made, $5$ positions are sampled without replacement, and $X$ is the number of sampled successes.

Using conditioning on $K$ , what is $Var(X)$ ?

$0.2625$
$0.7875$
$0.8290$
$1.0500$ (correct answer)

Explanation: When a problem involves a random variable whose distribution depends on another random variable, the Law of Total Variance is your essential tool:

Var(X) = E[Var(X \mid K)] + Var(E[X \mid K])

. Given

K

successes in the urn,

X

follows a Hypergeometric distribution with parameters

N=20

K

, and

n=5

. Its conditional moments are:

E[X \mid K] = \frac{5K}{20} = \frac{K}{4}, \quad Var(X \mid K) = \frac{5 \cdot K \cdot (20-K)}{20^2} \cdot \frac{20-5}{19} = \frac{5K(20-K)}{400} \cdot \frac{15}{19}

Since

K \sim Bin(20, 0.3)

, we have

E[K]=6

Var(K)=4.2

, and

E[K^2]=Var(K)+[E(K)]^2=4.2+36=40.2

. First term:

E[Var(X\mid K)] = \frac{75}{400 \cdot 19}E[K(20-K)] = \frac{75}{7600}(20E[K]-E[K^2]) = \frac{75}{7600}(120-40.2) = \frac{75 \times 79.8}{7600} \approx 0.7875

Second term:

Var(E[X\mid K]) = Var\!\left(\frac{K}{4}\right) = \frac{Var(K)}{16} = \frac{4.2}{16} = 0.2625

Adding both:

Var(X) = 0.7875 + 0.2625 = 1.0500

, confirming D. Choice A (0.2625) is only the second term — forgetting

E[Var(X\mid K)]

. Choice B (0.7875) is only the first term — forgetting

Var(E[X\mid K])

. Choice C (0.8290) likely arises from misapplying the hypergeometric variance formula, perhaps omitting the finite-population correction. Always decompose variance using both components of the Law of Total Variance — dropping either piece is the most common trap on conditioning problems.

Question 4

A device is tested repeatedly until its first success. Its success probability $P$ is selected once and remains fixed across trials. With equal probability, $P=1/2$ or $P=1/4$ . Conditional on $P$ , the waiting time $T$ has a geometric distribution on $1,2,\ldots$ .

What is $Var(T)$ ?

$40/9$
$6$
$7$
$8$ (correct answer)

Explanation: When a parameter is itself random, you need the Law of Total Variance:

\text{Var}(T) = E[\text{Var}(T\mid P)] + \text{Var}(E[T\mid P])

. This decomposition separates within-group variance from between-group variance, and forgetting either piece is exactly the trap this question sets. For a Geometric

(p)

distribution, recall

E[T\mid P=p] = 1/p

and

\text{Var}(T\mid P=p) = (1-p)/p^2

. Since

P = 1/2

P = 1/4

each with probability

1/2

First term:

E[\text{Var}(T\mid P)] = \frac{1}{2}(2) + \frac{1}{2}(12) = 7

Second term:

E[T\mid P]

takes values

2

and

4

with equal probability, so

E[E[T\mid P]] = 3

and

\text{Var}(E[T\mid P]) = \frac{1}{2}(2-3)^2 + \frac{1}{2}(4-3)^2 = 1

Therefore

\text{Var}(T) = 7 + 1 = \mathbf{8}

, confirming D. Choice C (7) is the most seductive trap — it equals only

E[\text{Var}(T\mid P)]

, forgetting the between-group variance term entirely. Choice B (6) likely arises from averaging the conditional variances incorrectly or using wrong geometric formulas. Choice A (40/9) may come from computing

\text{Var}(T)

as if

P

were its mean value

3/8

, plugging a single averaged

p

into the formula instead of applying total variance. Your study tip: whenever a parameter is randomly selected, never plug in its average value directly — always apply the Law of Total Variance, since the

\text{Var}(E[\cdot\mid\theta])

term accounts for uncertainty in the parameter itself.

Question 5

In a Galton–Watson branching process, each individual independently produces a number of offspring having mean $m=1.5$ and variance $v=2$ . The process begins with one individual, so $Z_0=1$ , and $Z_t$ denotes the population in generation $t$ .

What is $Var(Z_2)$ ?

$3.0$
$4.5$
$6.0$
$7.5$ (correct answer)

Explanation: Galton–Watson branching processes require a specific variance formula that accounts for how randomness compounds across generations. When you see a question asking for

\text{Var}(Z_t)

, your first instinct should be to recall the recursive variance formula rather than trying to reason from scratch. For a supercritical process with

m > 1

, the variance of

Z_t

is given by:

\text{Var}(Z_t) = v \cdot m^{t-1} \cdot \frac{m^t - 1}{m - 1}

where

v

is the offspring variance and

m

is the offspring mean. For

t = 2

m = 1.5

, and

v = 2

\text{Var}(Z_2) = 2 \cdot m^{1} \cdot \frac{m^2 - 1}{m - 1} = 2 \cdot 1.5 \cdot \frac{2.25 - 1}{0.5} = 3 \cdot \frac{1.25}{0.5} = 3 \cdot 2.5 = 7.5

This confirms D is correct. Choice A (3.0) likely comes from computing

v \cdot m = 2 \cdot 1.5

, which only captures one layer of variability and ignores the compounding from generation 1 to 2. Choice B (4.5) may reflect multiplying

v \cdot m^2 = 2 \cdot 2.25

, misapplying the exponent without the correct summation structure. Choice C (6.0) could arise from using

v \cdot (m + m^2) = 2(1.5 + 2.25) = 7.5

but making an arithmetic error, or from an incorrect simplified formula altogether. A reliable study tip: memorize the Galton–Watson variance formula and practice applying it at

t = 1, 2, 3

to build fluency — exam questions almost always test

t = 2

because that's where compounding first becomes nontrivial.

Question 6

An insurance portfolio has a random claim intensity. Conditional on $\Lambda$ , the number of claims $N$ is Poisson with mean $\Lambda$ . The mixing variable satisfies $E[\Lambda]=3$ and $Var(\Lambda)=2$ . Claim amounts $X_1,X_2,\ldots$ are independent of $N$ and $\Lambda$ , with $E[X_i]=2$ and $Var(X_i)=5$ .

If the aggregate loss is $S=\sum_{i=1}^{N}X_i$ , what is $Var(S)$ ?

$23$
$27$
$35$ (correct answer)
$45$

Explanation: When you see aggregate loss variance with a random claim intensity, reach for the law of total variance combined with the compound Poisson variance formula. The key insight is that

\Lambda

itself is random, so you need an extra layer of conditioning. Start by conditioning on

\Lambda

. Given

\Lambda

S

is a compound Poisson sum, so:

Var(S|\Lambda) = \Lambda \cdot E[X_i^2] = \Lambda(Var(X_i) + E[X_i]^2) = \Lambda(5+4) = 9\Lambda

E[S|\Lambda] = \Lambda \cdot E[X_i] = 2\Lambda

Now apply the law of total variance:

Var(S) = E[Var(S|\Lambda)] + Var(E[S|\Lambda])

= E[9\Lambda] + Var(2\Lambda)

= 9E[\Lambda] + 4\,Var(\Lambda)

= 9(3) + 4(2) = 27 + 8 = 35

So C) 35 is correct. A) 23 ignores the extra variance from the random intensity entirely — it's roughly what you'd get treating

\Lambda

as fixed at its mean, computing

9(3) = 27

minus an arithmetic error. B) 27 is a common trap: it computes

9E[\Lambda] = 27

but forgets the second term

Var(E[S|\Lambda]) = 4\,Var(\Lambda) = 8

. This is the mistake of omitting the between-group variance component. D) 45 likely confuses

Var(\Lambda)

E[X_i^2]

with incorrect values, overstating one of the components. Study tip: Memorize the two-step structure — condition on the mixing variable, then apply total variance — because this exact framework appears frequently in actuarial and graduate-level insurance problems whenever the Poisson mean is itself random.

Question 7

After observing current data, a Poisson rate $\Lambda$ has posterior mean $4$ and posterior variance $2$ . Conditional on $\Lambda$ , two future counts $Y_1$ and $Y_2$ are independent Poisson random variables, each with mean $\Lambda$ .

What is the posterior predictive variance of $Y_1+Y_2$ ?

$8$
$12$
$16$ (correct answer)
$20$

Explanation: When you need to find the variance of a posterior predictive quantity, reach for the law of total variance:

\text{Var}(Y) = E[\text{Var}(Y|\Lambda)] + \text{Var}(E[Y|\Lambda])

. This decomposes uncertainty into two sources — randomness in the outcome given the parameter, and uncertainty about the parameter itself. Here, let

S = Y_1 + Y_2

. Conditional on

\Lambda

, both

Y_1

and

Y_2

are independent Poisson

(\Lambda)

, so

S|\Lambda \sim \text{Poisson}(2\Lambda)

, meaning

E[S|\Lambda] = 2\Lambda

and

\text{Var}(S|\Lambda) = 2\Lambda

. Applying the law of total variance:

\text{Var}(S) = E[\text{Var}(S|\Lambda)] + \text{Var}(E[S|\Lambda])

= E[2\Lambda] + \text{Var}(2\Lambda)

= 2E[\Lambda] + 4\text{Var}(\Lambda)

= 2(4) + 4(2) = 8 + 8 = 16

So C) 16 is correct. Choice A) 8 captures only the first term — the expected within-

\Lambda

variance — ignoring the additional uncertainty from not knowing

\Lambda

exactly. Choice B) 12 likely comes from adding

E[\Lambda]

terms incorrectly or misapplying the variance of a sum. Choice D) 20 may result from using

\text{Var}(2\Lambda) = 2\cdot\text{Var}(\Lambda) = 4

instead of

4\cdot\text{Var}(\Lambda) = 8

, forgetting to square the constant when computing variance. Your key takeaway: whenever a predictive variance problem involves an unknown parameter, always apply the law of total variance — ignoring the

\text{Var}(E[Y|\Lambda])

term is the single most common mistake on Bayesian prediction problems.

Question 8

In a heteroskedastic random-design regression, the predictor satisfies $E[X]=1$ and $Var(X)=2$ . The response satisfies $E[Y|X]=3X$ and $Var(Y|X)=1+X^2$ .

What is the marginal variance $Var(Y)$ ?

$22$ (correct answer)
$20$
$18$
$31$

Explanation: Whenever you see a question asking for a marginal variance involving a conditional mean and conditional variance, your instinct should be the Law of Total Variance:

Var(Y) = E[Var(Y|X)] + Var(E[Y|X])

. This decomposition is essential in random-design regression, where both the conditional mean and conditional variance depend on a random predictor. Start with the two pieces. For the first term,

E[Var(Y|X)] = E[1 + X^2] = 1 + E[X^2]

. Since

Var(X) = E[X^2] - (E[X])^2

, we get

E[X^2] = Var(X) + (E[X])^2 = 2 + 1 = 3

. So

E[Var(Y|X)] = 1 + 3 = 4

. For the second term,

Var(E[Y|X]) = Var(3X) = 9 \cdot Var(X) = 9 \cdot 2 = 18

. Adding them:

Var(Y) = 4 + 18 = 22

, confirming that A) 22 is correct. Choice B) 20 likely comes from forgetting to include the 1 in

E[1 + X^2]

, computing

E[Var(Y|X)] = E[X^2] = 3

instead of 4, giving

3 + 18 = 21

, or from a similar arithmetic slip. Choice C) 18 reflects only computing

Var(3X)

and ignoring

E[Var(Y|X)]

entirely — a common mistake of confusing marginal variance with just the variance of the conditional mean. Choice D) 31 likely results from incorrectly using

E[X]

instead of

E[X^2]

somewhere, inflating one of the components. Your study tip: always write out the Law of Total Variance formula explicitly before computing — students most often lose points by dropping one of the two terms entirely.

Question 9

A random total count $N$ has mean $10$ and variance $14$ . Conditional on $N$ , each of the $N$ items is independently retained with probability $0.3$ , and $X$ denotes the number retained.

Which pair gives $E[X]$ and $Var(X)$ , respectively?

$(3,3.36)$ (correct answer)
$(3,3.00)$
$(3,2.10)$
$(4.2,3.36)$

Explanation: When you see a random count

N

feeding into a binomial-like selection process, you're in the territory of compound distributions — specifically, the law of total expectation and the law of total variance. For the mean, the law of total expectation gives

E[X] = E[E[X \mid N]]

. Since conditional on

N

X \sim \text{Binomial}(N, 0.3)

, we have

E[X \mid N] = 0.3N

. Therefore,

E[X] = 0.3 \cdot E[N] = 0.3 \times 10 = 3

. For the variance, the law of total variance states

\text{Var}(X) = E[\text{Var}(X \mid N)] + \text{Var}(E[X \mid N])

. The two components are:

E[\text{Var}(X \mid N)] = E[N \cdot 0.3 \cdot 0.7] = 0.21 \cdot E[N] = 0.21 \times 10 = 2.10

, and

\text{Var}(E[X \mid N]) = \text{Var}(0.3N) = 0.09 \cdot \text{Var}(N) = 0.09 \times 14 = 1.26

. Adding these:

\text{Var}(X) = 2.10 + 1.26 = 3.36

. This confirms answer A,

(3, 3.36)

. Answer B gives variance

3.00

, which likely comes from treating

N

as fixed at its mean (ignoring the extra variability from

N

being random). Answer C reports only the first component

E[\text{Var}(X \mid N)] = 2.10

, forgetting to add the

\text{Var}(E[X \mid N])

term entirely. Answer D uses an incorrect mean of

4.2

, perhaps confusing the retention probability with something else. The key study tip: whenever

N

is random, never plug

E[N]

directly into the binomial variance formula — always apply the full law of total variance, which adds an extra term proportional to

\text{Var}(N)

Question 10

Suppose $\Theta\sim N(0,4)$ and, conditional on $\Theta$ , an observation satisfies $Y|\Theta\sim N(\Theta,5)$ . What is $Var(E[\Theta|Y])$ ?

$4/9$
$16/9$ (correct answer)
$20/9$
$4$

Explanation: When you see a question involving a normal prior and normal likelihood, your first instinct should be Bayesian updating and the Law of Total Variance:

Var(Y) = E[Var(Y|X)] + Var(E[Y|X])

. Here, the question asks for

Var(E[\Theta|Y])

, which is precisely one piece of that decomposition — but applied to

\Theta

given

Y

. Start by finding the posterior

E[\Theta|Y]

. With

\Theta \sim N(0,4)

(prior variance

\tau^2 = 4

) and

Y|\Theta \sim N(\Theta, 5)

(likelihood variance

\sigma^2 = 5

), the posterior mean in a normal-normal model is:

E[\Theta|Y] = \frac{\tau^2}{\tau^2 + \sigma^2} Y = \frac{4}{4+5}Y = \frac{4}{9}Y

Since

Y \sim N(0, \tau^2 + \sigma^2) = N(0, 9)

marginally, we get:

Var(E[\Theta|Y]) = \left(\frac{4}{9}\right)^2 Var(Y) = \frac{16}{81} \cdot 9 = \frac{16}{9}

That confirms B is correct. Now for the distractors: A (

4/9

) is the shrinkage weight itself, not squared and multiplied through — a partial-computation trap. C (

20/9

) likely comes from incorrectly using

\sigma^2 = 5

in the numerator instead of

\tau^2 = 4

, confusing prior and likelihood variances. D (

4

) is simply the prior variance

Var(\Theta)

, which would equal

Var(E[\Theta|Y])

only if

Y

were a perfect signal — forgetting that shrinkage reduces variance. A reliable strategy: in normal-normal Bayesian problems, memorize that

E[\Theta|Y] = \frac{\tau^2}{\tau^2+\sigma^2}Y

and always remember to compute the variance of this linear function of

Y

using the marginal variance of

Y