Exercises — Discrete random variables — PMF, CDF

WHAT we check: the two-and-only-two rules.

1. Every entry $\ge 0$? Yes — $0.1,0.4,0.3,0.2$ are all non-negative. ✓
2. Do they sum to 1? $0.1+0.4+0.3+0.2=1.0$. ✓

Both rules hold, so yes, it is a legal PMF. Since $5$ is not a listed value, no mass sits there: $p_X(5)=0$.

WHAT we do: sum all mass at values $\le x$.

• $F_X(1)=p_X(0)+p_X(1)=0.1+0.4=0.5$. (Values $0,1$ are $\le 1$.)
• $F_X(2.5)=p_X(0)+p_X(1)+p_X(2)=0.1+0.4+0.3=0.8$. (Values $0,1,2$ are $\le 2.5$; the value $3$ is not.)

Notice $F$ is flat between $2$ and $3$: nothing new to collect until you hit $3$.

WHY normalization: the total mass rule $\sum p_X=1$ is the equation that pins down the unknown $c$. $c(1^2+2^2+3^2)=c(1+4+9)=14c=1\Rightarrow c=\frac{1}{14}.$ So $p_X(1)=\frac{1}{14},p_X(2)=\frac{4}{14},p_X(3)=\frac{9}{14}$.

$P(X\ge 2)$: add the mass at values $2$ and $3$: $P(X\ge 2)=\frac{4}{14}+\frac{9}{14}=\frac{13}{14}.$ Cross-check with the complement: $1-p_X(1)=1-\frac{1}{14}=\frac{13}{14}$. ✓

WHY two answers: the bracket type is not decoration — points carry positive mass, so including or excluding an endpoint changes the count.

$P(2<X\le 5)$ excludes $2$, includes $5$ → values $\{3,4,5\}$: $F_X(5)-F_X(2)=\frac{5}{6}-\frac{2}{6}=\frac{3}{6}=\frac{1}{2}.$

$P(2\le X\le 5)$ includes $2$ → values $\{2,3,4,5\}$: $F_X(5)-F_X(2^{-})=F_X(5)-F_X(1)=\frac{5}{6}-\frac{1}{6}=\frac{4}{6}=\frac{2}{3}.$ The extra $\frac{1}{6}$ is exactly $p_X(2)$, the mass we chose to keep.

WHY jumps = mass: by the parent note, $p_X(x_i)=F_X(x_i)-F_X(x_i^{-})$ — the height of each vertical step is the probability sitting exactly there. Flat pieces mean no mass.

The staircase jumps at $x=0,1,2,4$:

• at $0$: $0.2-0=0.2$
• at $1$: $0.5-0.2=0.3$
• at $2$: $0.9-0.5=0.4$
• at $4$: $1-0.9=0.1$

So the RV lives on $\{0,1,2,4\}$ (not $3$ — $F$ is flat across $[2,4)$, so no mass there) with $p_X(0)=0.2,p_X(1)=0.3,p_X(2)=0.4,p_X(4)=\mathrm{0.1.}$ Sanity check: $0.2+0.3+0.4+0.1=1$. ✓

• $P(X=2)$ = jump at $2$ = $0.4$.
• $P(1\le X<2)$: values $\ge 1$ and $<2$ → just $\{1\}$, so $=p_X(1)=0.3$. (Via CDF: $F_X(2^{-})-F_X(1^{-})=0.5-0.2=0.3$.)
• $P(X>1)$ = $1-P(X\le 1)=1-F_X(1)=1-0.5=0.5$. (Values $2$ and $4$: $0.4+0.1=0.5$. ✓)

WHY binomial: independent trials, each Head with the same $p=0.6$; $X$ counts successes. The number of ways to get exactly $k$ Heads in $3$ flips is $\left(\genfrac{}{}{0ex}{}{3}{k}\right)$. $p_X(k)=\left(\genfrac{}{}{0ex}{}{3}{k}\right)(0.6{)}^k(0.4{)}^{3-k}.$

• $p_X(0)=\left(\genfrac{}{}{0ex}{}{3}{0}\right)(0.6{)}^0(0.4{)}^3=1\cdot 1\cdot 0.064=0.064$
• $p_X(1)=\left(\genfrac{}{}{0ex}{}{3}{1}\right)(0.6{)}^1(0.4{)}^2=3\cdot 0.6\cdot 0.16=0.288$
• $p_X(2)=\left(\genfrac{}{}{0ex}{}{3}{2}\right)(0.6{)}^2(0.4{)}^1=3\cdot 0.36\cdot 0.4=0.432$
• $p_X(3)=\left(\genfrac{}{}{0ex}{}{3}{3}\right)(0.6{)}^3(0.4{)}^0=1\cdot 0.216\cdot 1=0.216$

Check total: $0.064+0.288+0.432+0.216=1.000$. ✓

$F_X(1)=p_X(0)+p_X(1)=0.064+0.288=\mathrm{0.352.}$

WHY Poisson: it is the standard PMF for "how many independent rare events land in a fixed window," governed by one parameter $\lambda =$ mean count. $P(X\le 2)=p_X(0)+p_X(1)+p_X(2)=e^{-2}\left(\frac{2^0}{0!}+\frac{2^1}{1!}+\frac{2^2}{2!}\right)=e^{-2}(1+2+2)=5e^{-2}.$ Numerically $5e^{-2}\approx 5\times 0.13534=0.6767$.

(a) Normalization with an infinite tail. The values are countably infinite, so we sum a geometric series: ${\sum }_{k=1}^{\infty }c{\left(\frac{1}{2}\right)}^k=c\cdot \frac{1/2}{1-1/2}=c\cdot 1=c=1.$ (Used ${\sum }_{k=1}^{\infty }{r}^k=\frac{r}{1-r}$ for $|r|<1$, here $r=\frac{1}{2}$.) So $c=1$ and $p_X(k)=(\frac{1}{2}{)}^k$.

(b) $P(X\le 2)=p_X(1)+p_X(2)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$.

(c) By definition $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$.

• $A=\{X\le 2\}$, $B=\{X\ge 2\}$. Their intersection is exactly $\{X=2\}$, so $P(A\cap B)=p_X(2)=\frac{1}{4}$.
• $P(B)=P(X\ge 2)=1-p_X(1)=1-\frac{1}{2}=\frac{1}{2}$. $P(X\le 2\mid X\ge 2)=\frac{1/4}{1/2}=\frac{1}{2}.$

Proof. The events $\{X\le a\}$ and $\{X>a\}$ are disjoint (no value can be both) and their union is everything $X$ can do. By the probability axioms (see Probability Axioms), disjoint events that cover $\Omega$ have probabilities summing to $1$: $P(X\le a)+P(X>a)=1\Rightarrow P(X>a)=1-P(X\le a)=1-F_X(a).$

Apply. Die, uniform on $\{1,\dots ,6\}$: $F_X(4)=\frac{4}{6}=\frac{2}{3}$, so $P(X>4)=1-\frac{2}{3}=\frac{1}{3}.$ (Check directly: values $5,6$ give $\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$. ✓)

Read/verify a PMF (rules) → sum to a CDF (staircase) → read intervals with correct endpoints → invert jumps to recover mass → build PMFs from named models (Binomial, Poisson, geometric) → prove complement/conditional identities from the axioms.