Random Variables & Distributions - Statistics Graduate Level

1. Which expression is the CDF of $Z$ for $0<=z<=1$?
   Answer: $F_Z(z)=z^2$ 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.
2. Which description gives the complete distribution of $Y$?
   Answer: 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 $1$. 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.
3. 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 $x=1$?
   Answer: $F$ is a CDF for $0<=c<=1/2$, and then $P(X=1)=1/2-c$. 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 $x \to 1^-$ is $c$ (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.
4. Which expression gives $F_W(w)$ for $0<=w<=2$?
   Answer: $F_W(w)=w/2$ 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.
5. 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 $0<y<1$?
   Answer: $F_Y(y)=sqrt(y)$ 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.
6. Which ordered quadruple equals $[Q(0.2),\, Q(0.5),\, Q(0.6),\, Q(0.61)]$?
   Answer: $[-1,\; 0.75,\; 1,\; 2]$ 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 $0.2$ at $x=-1$, uniform density $0.4$ over $[0,1]$, and mass $0.4$ at $x=2$. 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.
7. 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 $[0,1]$?
   Answer: $F(X-)+V[F(X)-F(X-)]$ 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.
8. Which statement about the distribution functions $F_n$ and $F$ is correct?
   Answer: $X_n$ converges in distribution to $X$ even though convergence fails at the discontinuity $x=0$ of $F$. 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 $1/2$ at $x = 1/n$, and jumps to $1$ at $x = 1 + 1/n$. Meanwhile, $F(x) = 0$ for $x < 0$, jumps to $1/2$ at $x = 0$, and jumps to $1$ at $x = 1$. 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.
9. Which pair correctly gives the point mass $P(M=0)$ and the density of $M$ for $0<m<1$?
   Answer: $P(M=0)=1/64$ and $f_M(m)=2.25(0.25+0.75m)^2$. 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 $M$ is $F_M(m) = [F_X(m)]^3 = (0.25 + 0.75m)^3$. The point mass at zero is $P(M = 0) = F_M(0) = (0.25)^3 = 1/64$. The density for $0 < m < 1$ is $f_M(m) = F_M'(m) = 3(0.25 + 0.75m)^2 \cdot 0.75 = 2.25(0.25 + 0.75m)^2$. 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.
10. 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?
    Answer: There exists a coupling under which $P(X \leq Y)=1$, but an arbitrary coupling need not have this property. 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.

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