Statistics Graduate Level Quiz: Random Variables And Distributions

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Random Variables And DistributionsQuestion 1 of 10

A random variable $X$ has density $f_X(x)=(1+x)/2$ for $-1<x<1$ and zero otherwise. If $Y=X^2$ , which expression is the CDF of $Y$ for $0<y<1$ ?

F_Y(y)=2sqrt(y)-y

F_Y(y)=y

F_Y(y)=[sqrt(y)+y]/2

F_Y(y)=sqrt(y)

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Random Variables And Distributions

Practice Random Variables And Distributions 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 Random Variables And Distributions, 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

A random variable $X$ has density $f_X(x)=(1+x)/2$ for $-1<x<1$ and zero otherwise. If $Y=X^2$ , which expression is the CDF of $Y$ for $0<y<1$ ?

$F_Y(y)=2sqrt(y)-y$
$F_Y(y)=y$
$F_Y(y)=[sqrt(y)+y]/2$
$F_Y(y)=sqrt(y)$ (correct answer)

Explanation: When a transformation like

Y = X^2

maps both negative and positive values of

X

to the same value of

Y

, you must account for both branches of the inverse transformation when computing the CDF. For

0 < y < 1

, the event

\{Y \leq y\} = \{X^2 \leq y\} = \{-\sqrt{y} \leq X \leq \sqrt{y}\}

. So the CDF is:

F_Y(y) = \int_{-\sqrt{y}}^{\sqrt{y}} \frac{1+x}{2}\, dx

Split the integral:

= \int_{-\sqrt{y}}^{\sqrt{y}} \frac{1}{2}\, dx + \int_{-\sqrt{y}}^{\sqrt{y}} \frac{x}{2}\, dx

The first integral gives

\frac{1}{2}(2\sqrt{y}) = \sqrt{y}

. The second integral vanishes because

x/2

is an odd function integrated over a symmetric interval. Therefore

F_Y(y) = \sqrt{y}

, confirming answer D. Now consider why the distractors fail. Answer A,

2\sqrt{y} - y

, likely comes from incorrectly integrating only over

[0, \sqrt{y}]

and doubling, while mishandling the

x

term. Answer B,

F_Y(y) = y

, is the CDF of a Uniform(0,1) — it ignores the non-uniform density entirely and is a classic distractor. Answer C,

[\sqrt{y} + y]/2

, retains the

y

term as if the odd-function integral contributed, reflecting a failure to recognize the symmetry cancellation. As a strategy: whenever you integrate an asymmetric density over a symmetric interval, always check whether pieces of the integrand are odd functions — they cancel and simplify your work considerably. This symmetry argument is a frequent shortcut on graduate-level problems.

Question 2

A device lifetime $X$ is zero with probability $0.3$ . Conditional on being positive, it has an exponential distribution with rate $1$ . The recorded lifetime is censored at time $1$ , so the observed random variable is $Y=min(X,1)$ .

Which description gives the complete distribution of $Y$ ?

An atom of size $0.3$ at $0$ , density $0.7e^{-y}$ on $0<y<1$ , and an atom of size $0.7e^{-1}$ at $1$ . (correct answer)
An atom of size $0.3$ at $0$ , density $e^{-y}$ on $0<y<1$ , and an atom of size $e^{-1}$ at $1$ .
No atom at $0$ , density $0.7e^{-y}$ on $0<y<1$ , and an atom of size $0.3+0.7e^{-1}$ at $1$ .
An atom of size $0.3$ at $0$ , density $0.7e^{-y}$ on $0<y<1$ , and no atom at $1$ .

Explanation: When you encounter a mixed distribution (one that has both discrete mass points and a continuous density), your job is to carefully track every place probability accumulates. Here,

X

is a mixture: point mass

0.3

at zero, and with probability

0.7

it's

\text{Exp}(1)

. Censoring at

1

means

Y = \min(X, 1)

, so you need to find probability at

y=0

, density on

(0,1)

, and probability at

y=1

. At $y=0$ :

Y=0

only when

X=0

, which happens with probability

0.3

. So there's an atom of size

0.3

at zero. On $(0,1)$ :

Y=y

when

X=y\in(0,1)

, which requires

X

to be positive (probability

0.7

) and exponentially distributed at that value. The density is

0.7 \cdot e^{-y}

for

0 < y < 1

. At $y=1$ :

Y=1

when

X \geq 1

, which happens with probability

0.7 \cdot P(\text{Exp}(1)\geq 1) = 0.7e^{-1}

. So there's an atom of size

0.7e^{-1}

at one. You can verify:

0.3 + \int_0^1 0.7e^{-y}\,dy + 0.7e^{-1} = 0.3 + 0.7(1-e^{-1}) + 0.7e^{-1} = 1

. ✓ This confirms A is correct. B omits the

0.7

mixing weight, giving an unnormalized distribution. C incorrectly removes the atom at zero and bundles it into the atom at one, misidentifying when

Y=0

occurs. D correctly handles zero and the density but ignores the censoring atom at

1

, losing probability mass. Always verify a mixed distribution by checking that atoms plus integrated density sum to exactly

1

— this instantly catches missing or misplaced mass.

Question 3

Random variables $X$ and $Y$ both have uniform marginal distributions on $[0,1]$ . Their joint CDF on $[0,1]^2$ is $F_{X,Y}(x,y)=min(x,y)$ . Let $W=X+Y$ .

Which expression gives $F_W(w)$ for $0<=w<=2$ ?

$F_W(w)=sqrt(w/2)$ for $0<=w<=2$
$F_W(w)=w^2/2$ for $0<=w<=1$ and $F_W(w)=1-(2-w)^2/2$ for $1<w<=2$
$F_W(w)=w$ for $0<=w<=1$ and $F_W(w)=1$ for $1<w<=2$
$F_W(w)=w/2$ (correct answer)

Explanation: When you see a joint distribution defined by

F_{X,Y}(x,y) = \min(x,y)

, recognize immediately that this is the Fréchet upper bound copula, which forces

X = Y

almost surely. That is, this joint CDF describes perfect positive dependence —

X

and

Y

are not independent; they are literally the same random variable. To confirm this, notice that the joint density would be concentrated on the diagonal

x = y

, meaning

P(X = Y) = 1

. Therefore

W = X + Y = 2X

almost surely. Since

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

, we have

W = 2X \sim \text{Uniform}(0,2)

. The CDF of a

\text{Uniform}(0,2)

random variable is simply

F_W(w) = w/2

for

0 \leq w \leq 2

, confirming answer D. Answer B is the classic trap — it gives the correct CDF for

W = X + Y

when

X

and

Y

are independent uniforms. This is the convolution result students often memorize, but it completely ignores the dependence structure encoded in the joint CDF. Answer A introduces a square root form with no probabilistic justification; it doesn't correspond to any standard distribution arising from this setup. Answer C implies

W

is uniform on

[0,1]

, which would require

W

to stay bounded by 1 — impossible when both

X

and

Y

range up to 1. The key study tip: always extract the dependence structure before computing a sum's distribution. If

X

and

Y

are not independent, convolution does not apply — analyze the joint behavior first.

Question 4

Suppose two random variables have CDFs satisfying $F_X(t) \geq F_Y(t)$ for every real $t$ . No assumption is made about their joint distribution. Which conclusion is guaranteed solely by this condition?

Under every possible coupling of $X$ and $Y$ , we have $P(X \leq Y)=1$ .
There exists a coupling under which $P(X \leq Y)=1$ , but an arbitrary coupling need not have this property. (correct answer)
There exists a coupling under which $P(X \geq Y)=1$ , because $F_X(t) \geq F_Y(t)$ places $X$ stochastically to the right of $Y$ .
For every bounded increasing function $g$ , the inequality $E[g(X)] \geq E[g(Y)]$ holds.

Explanation: When you see a condition like

F_X(t) \geq F_Y(t)

for all

t

, you're looking at first-order stochastic dominance — a concept that connects marginal CDFs to the existence of special joint distributions called couplings. The key insight is that a statement about marginal CDFs says nothing about how

X

and

Y

are jointly distributed; you must reason carefully about what is and isn't guaranteed across all possible joint distributions. The correct answer is B. When

F_X(t) \geq F_Y(t)

for all

t

X

is stochastically smaller than

Y

— its distribution is shifted left. The Strassen/quantile coupling theorem guarantees that you can always construct a joint distribution where

P(X \leq Y) = 1

by pairing quantiles: set

X = F_X^{-1}(U)

and

Y = F_Y^{-1}(U)

for the same uniform

U

. Since

F_X \geq F_Y

implies

F_X^{-1} \leq F_Y^{-1}

, this coupling delivers

X \leq Y

almost surely. However, if you choose an arbitrary joint distribution — say, making

X

and

Y

independent — there is no such guarantee. A is wrong because "every coupling" is far too strong; you can construct couplings where

X > Y

with positive probability. C reverses the direction.

F_X(t) \geq F_Y(t)

means

X

has more probability mass to the left, so

X

is stochastically smaller, not larger. The coupling

P(X \geq Y) = 1

would require

X

to dominate

Y

, which is the opposite condition. D is also backwards. For a bounded increasing

g

, stochastic dominance of

Y

over

X

gives

E[g(Y)] \geq E[g(X)]

, not the other way around. Study tip: Always check the direction of stochastic dominance carefully —

F_X \geq F_Y

means

X

is smaller in distribution. Pair this with the quantile coupling construction as your go-to proof technique for existence results.

Question 5

For each positive integer $n$ , let $X_n$ equal $1/n$ with probability $1/2$ and $1+1/n$ with probability $1/2$ . Let $X$ equal $0$ with probability $1/2$ and $1$ with probability $1/2$ .

Which statement about the distribution functions $F_n$ and $F$ is correct?

$X_n$ does not converge in distribution to $X$ because $F_n(0)$ does not converge to $F(0)$ .
$X_n$ converges in distribution to $X$ even though convergence fails at the discontinuity $x=0$ of $F$ . (correct answer)
$X_n$ converges in distribution to the constant $1/2$ because each variable has two equally likely values.
$X_n$ converges in distribution to $X$ only if $F_n(x)$ converges to $F(x)$ at every real $x$ .

Explanation: Convergence in distribution has a precise definition that trips up many students:

X_n \xrightarrow{d} X

if and only if

F_n(x) \to F(x)

at every continuity point of

F

. This exemption for discontinuity points is not a loophole — it is fundamental to the definition. Here,

F_n(x) = 0

for

x < 1/n

, jumps to

1/2

x = 1/n

, and jumps to

1

x = 1 + 1/n

. Meanwhile,

F(x) = 0

for

x < 0

, jumps to

1/2

x = 0

, and jumps to

1

x = 1

. The continuity points of

F

are all

x \notin \{0, 1\}

. At any such point, you can verify

F_n(x) \to F(x)

: for example, at

x = 0.5

F_n(0.5) = 1/2 \to 1/2 = F(0.5)

. So B is correct — convergence in distribution holds even though

F_n(0) = 0 \not\to 1/2 = F(0)

. A misidentifies what the definition requires. Failure at a discontinuity of

F

is explicitly permitted and does not block convergence in distribution. C is nonsense — having two equally likely values does not pin down a limiting distribution; the actual values matter. D states a stricter condition than necessary and is flatly false as a characterization of convergence in distribution. Study tip: Whenever convergence in distribution is tested, immediately ask yourself: where is $F$ continuous? Pointwise convergence at discontinuities of

F

is neither required nor expected — only continuity points count.

Question 6

For a constant $c$ , consider the function $F$ defined by $F(x)=0$ for $x<0$ , $F(x)=c$ for $0<=x<1$ , and $F(x)=1-(1/2)e^{-(x-1)}$ for $x>=1$ . Which statement correctly characterizes when $F$ is a cumulative distribution function and the resulting probability at $x=1$ ?

$F$ is a CDF only when $c=1/2$ , in which case $P(X=1)=0$ .
$F$ is a CDF for $0<=c<=1/2$ , and then $P(X=1)=1/2-c$ . (correct answer)
$F$ is a CDF for $0<=c<1/2$ , and then $P(X=1)=1/2$ .
$F$ is a CDF for $1/2<=c<=1$ , and then $P(X=1)=c-1/2$ .

Explanation: When analyzing whether a piecewise function qualifies as a CDF, you need to verify three properties: right-continuity, non-decreasing behavior, and boundary conditions (limits of 0 and 1 at

-\infty

and

+\infty

). The trickiest requirement here is ensuring no decrease occurs at the junction point

x = 1

. Evaluate the given pieces at the junction. The left-hand limit as

x \to 1^-

c

(from the middle piece), and the function's value at

x = 1

from the right piece is

F(1) = 1 - \frac{1}{2}e^0 = 1 - \frac{1}{2} = \frac{1}{2}

. For

F

to be non-decreasing, you need

c \leq \frac{1}{2}

. Also, since

F(x) = 0

for

x < 0

and

c

for

0 \leq x < 1

, you need

c \geq 0

for non-decrease there. So

F

is a valid CDF precisely when

0 \leq c \leq \frac{1}{2}

. The atom (point mass) at

x = 1

equals the jump:

P(X = 1) = F(1) - F(1^-) = \frac{1}{2} - c

. This confirms B is correct. Choice A is too restrictive — it claims only

c = \frac{1}{2}

works, but any

c \in [0, \frac{1}{2}]

satisfies the CDF requirements. Choice C wrongly excludes

c = \frac{1}{2}

(which is valid) and incorrectly states

P(X=1) = \frac{1}{2}

regardless of

c

. Choice D gets the inequality backwards; values

c > \frac{1}{2}

would cause

F

to decrease at

x = 1

, violating the non-decreasing requirement. Your strategy: always check junction points in piecewise CDFs by comparing left-hand limits with the function's value — that difference directly gives the point mass at that location.

Question 7

Let $X_1,X_2,X_3$ be independent and identically distributed. Each variable equals $0$ with probability $0.25$ ; with probability $0.75$ , it is uniformly distributed on $[0,1]$ . Let $M=max(X_1,X_2,X_3)$ .

Which pair correctly gives the point mass $P(M=0)$ and the density of $M$ for $0<m<1$ ?

$P(M=0)=1/64$ and $f_M(m)=2.25(0.25+0.75m)^2$ . (correct answer)
$P(M=0)=0.25$ and $f_M(m)=3(0.75m)^2$ .
$P(M=0)=1/64$ and $f_M(m)=3(0.75m)^2$ .
$P(M=0)=3/64$ and $f_M(m)=2.25(0.25+0.75m)^2$ .

Explanation: When you see a maximum of mixed discrete-continuous random variables, recognize that the CDF of each

X_i

has a jump at zero — this makes the distribution of

M

a mixed distribution with both a point mass and a continuous density. Start by finding the CDF of a single

X_i

. For

0 \leq x \leq 1

P(X_i \leq x) = P(X_i = 0) + P(0 < X_i \leq x) = 0.25 + 0.75x

. This CDF has a jump of 0.25 at zero, then rises continuously. Since

M = \max(X_1, X_2, X_3)

and the variables are i.i.d., the CDF of

M

F_M(m) = [F_X(m)]^3 = (0.25 + 0.75m)^3

. The point mass at zero is

P(M = 0) = F_M(0) = (0.25)^3 = 1/64

. The density for

0 < m < 1

f_M(m) = F_M'(m) = 3(0.25 + 0.75m)^2 \cdot 0.75 = 2.25(0.25 + 0.75m)^2

. This confirms answer A is correct. Answer B incorrectly sets

P(M=0) = 0.25

, confusing the marginal point mass of one variable with the joint event that all three equal zero. Its density also ignores the 0.25 jump in the CDF entirely. Answer C gets the point mass right (

1/64

) but uses the wrong density — dropping the 0.25 offset means it treats each

X_i

as purely uniform. Answer D uses the correct density form but sets

P(M=0) = 3/64

, which would come from mistakenly computing

3 \times (0.25)^3

instead of

(0.25)^3

. Your key takeaway: always build the full CDF of each variable — including any point masses — before raising it to the

n

-th power for order statistics. Differentiating an incomplete CDF is the most common trap here.

Question 8

A random variable $X$ has probability $0.2$ at $-1$ , density $0.4$ on $[0,1]$ , probability $0.4$ at $2$ , and no probability elsewhere. Define the generalized quantile function by $Q(p)=\inf\{x:F(x)\geq p\}$ .

Which ordered quadruple equals $[Q(0.2),\, Q(0.5),\, Q(0.6),\, Q(0.61)]$ ?

$[-1,\; 0.75,\; 1,\; 2]$ (correct answer)
$[0,\; 0.75,\; 2,\; 2]$
$[-1,\; 0.50,\; 1,\; 2]$
$[-1,\; 0.75,\; 2,\; 2]$

Explanation: Whenever you see a mixed distribution (point masses plus a continuous density), your first move should be to carefully construct the CDF

F(x)

before touching the quantile function. Here, the distribution places mass

0.2

x=-1

, uniform density

0.4

over

[0,1]

, and mass

0.4

x=2

. This gives:

$F(x) = 0$ for $x < -1$
$F(-1) = 0.2$ (jump to 0.2)
$F(x) = 0.2 + 0.4x$ for $x \in [0,1]$ , so $F(0)=0.2$ , $F(1)=0.6$
$F(x) = 0.6$ for $x \in (1,2)$
$F(2) = 1.0$ (jump to 1.0)

Now apply

Q(p) = \inf\{x : F(x) \geq p\}

$Q(0.2)$ : We need the smallest $x$ with $F(x) \geq 0.2$ . Since $F(-1) = 0.2$ , that's $x = -1$ .
$Q(0.5)$ : Solve $0.2 + 0.4x = 0.5 \Rightarrow x = 0.75$ .
$Q(0.6)$ : Solve $0.2 + 0.4x = 0.6 \Rightarrow x = 1$ . Since $F(1) = 0.6 \geq 0.6$ , the infimum is $x = 1$ .
$Q(0.61)$ : $F(x) = 0.6$ for all $x \in [1,2)$ , which never reaches $0.61$ . The first point where $F(x) \geq 0.61$ is $x = 2$ .

This confirms answer A:

[-1,\; 0.75,\; 1,\; 2]

. Choice B incorrectly gives

Q(0.2) = 0

, missing that the jump at

-1

already achieves

p = 0.2

. Choice C gives

Q(0.5) = 0.50

, confusing the probability level with the quantile value. Choice D correctly finds

Q(0.2)

and

Q(0.61)

but wrongly sets

Q(0.6) = 2

, overlooking that

F(1) = 0.6

exactly satisfies the condition. The key trap: at jump points and flat regions, always check whether the CDF reaches

p

before you jump to the next mass.

Question 9

Let $X$ have an arbitrary cumulative distribution function $F$ , possibly with atoms. Let $V$ be independent of $X$ and uniformly distributed on $[0,1]$ . Which random variable is guaranteed to be uniformly distributed on $[0,1]$ ?

$F(X)$
$F(X-)$
$F(X-)+V[F(X)-F(X-)]$ (correct answer)
$F(X)+V[1-F(X)]$

Explanation: When a CDF has jump discontinuities (atoms), the standard probability integral transform breaks down. The key concept here is the generalized inverse transform, which produces a uniform random variable even when

F

is not continuous. For a continuous CDF,

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

— but atoms create point masses that pile probability at specific values, so

F(X)

is no longer uniform. The fix is to "smear out" each atom uniformly across the jump it creates. Why C works: At a point

x

where

F

jumps, the atom has mass

F(x) - F(x-)

. The expression

F(X-) + V[F(X) - F(X-)]

takes the left-limit

F(X-)

and adds a uniform fraction

V

of the jump size. This randomization distributes probability mass uniformly across the interval

[F(X-), F(X)]

, effectively "filling in" each atom. One can verify that

P(C \leq t) = t

for all

t \in [0,1]

, confirming uniformity. Why A fails:

F(X)

assigns positive probability to individual points wherever

F

has jumps — it cannot be uniform on

[0,1]

when atoms exist. Why B fails:

F(X-)

uses left-limits, but this has the same atom problem in reverse — point masses still survive, just shifted to the left endpoint of each jump interval. Why D fails:

F(X) + V[1 - F(X)]

smears mass above

F(X)

rather than across the atom's jump, so it does not correctly handle the discontinuities and fails to produce a uniform distribution. Study tip: Whenever a problem involves a CDF "possibly with atoms," immediately think about what happens at jump discontinuities — the correct transformation must randomize within each jump interval, which is exactly what option C accomplishes.

Question 10

Let $X$ be uniformly distributed on $[0,1]$ . Conditional on $X=x$ , a binary random variable $Y$ satisfies $P(Y=1|X=x)=x$ . Let $Z$ have the conditional distribution of $X$ given $Y=1$ .

Which expression is the CDF of $Z$ for $0<=z<=1$ ?

$F_Z(z)=z$
$F_Z(z)=2z-z^2$
$F_Z(z)=z^2$ (correct answer)
$F_Z(z)=sqrt(z)$

Explanation: When you see a question about conditioning on an observed outcome, reach for Bayes' theorem for continuous distributions. The key tool here is finding the conditional density of

X

given

Y=1

, then integrating to get the CDF. Start with the marginal probability

P(Y=1)

. Since

X \sim \text{Uniform}[0,1]

and

P(Y=1|X=x)=x

, the law of total probability gives

P(Y=1) = \int_0^1 x \cdot 1\, dx = \frac{1}{2}

. Now apply Bayes' theorem to find the conditional density of

X

given

Y=1

f_{X|Y=1}(x) = \frac{P(Y=1|X=x) \cdot f_X(x)}{P(Y=1)} = \frac{x \cdot 1}{1/2} = 2x

. This is a Beta(2,1) density. Integrating to get the CDF of

Z

F_Z(z) = \int_0^z 2x\, dx = z^2

, confirming answer C. Choice A,

F_Z(z) = z

, corresponds to a Uniform[0,1] distribution — this is the CDF of

X

itself, ignoring the conditioning on

Y=1

entirely. Choice B,

F_Z(z) = 2z - z^2

, is the CDF of a Beta(1,2) distribution, which would arise if you conditioned on

Y=0

instead (where

P(Y=0|X=x) = 1-x

), giving conditional density

2(1-x)

. Choice D,

F_Z(z) = \sqrt{z}

, has density

\frac{1}{2\sqrt{z}}

, which decreases on [0,1] — opposite to the intuition that higher

x

makes

Y=1

more likely. The study tip: always compute the normalizing constant (the marginal probability) before writing the conditional density. Forgetting to divide by

P(Y=1)

is the most common error in Bayesian conditioning problems.