Exam P — Discrete Distributions Flashcards

(1) Non-negativity: $p(x) \geq 0$ for every $x$ in the support. (2) Total mass one: $\sum_{x} p(x) = 1$, summed over all support points. Values outside the support have $p(x)=0$.

The cdf accumulates mass: $F(x)=\Pr(X \leq x)=\sum_{k \leq x} p(k)$. It is a right-continuous non-decreasing step function with $F(-\infty)=0$ and $F(\infty)=1$. For a discrete variable, $\Pr(a < X \leq b)=F(b)-F(a)$.

The jump in the cdf gives the point mass: $\Pr(X=x)=F(x)-F(x^{-})$, the size of the step at $x$. The survival function is $S(x)=\Pr(X>x)=1-F(x)$, so for integer support $\Pr(X \geq x)=1-F(x-1)$.

$E[X]=\sum_{k=1}^{\infty}\Pr(X \geq k)=\sum_{k=0}^{\infty}\Pr(X>k)$. This sums the survival function and is often faster than $\sum_{k} k\,p(k)$ — it is the discrete analogue of $E[X]=\int_{0}^{\infty}S(x)\,dx$.

**Binomial:** fixed number of independent trials $n$; count the successes; replacement (constant $p$). **Geometric / negative binomial:** fixed number of successes ($1$ or $r$); count the trials (or failures) until you reach them. **Hypergeometric:** fixed sample size $n$ drawn **without replacement**; count successes from a finite population — trials are dependent.

pmf: $p(x)=\dfrac{1}{n}$ for $x=1,\dots,n$. Mean: $E[X]=\dfrac{n+1}{2}$. Variance: $\mathrm{Var}(X)=\dfrac{n^{2}-1}{12}$. Each outcome is equally likely (e.g. a fair die has $n=6$).

This is discrete uniform on $\{1,\dots,6\}$, so $n=6$. Mean: $E[X]=\dfrac{n+1}{2}=\dfrac{7}{2}=3.5$. Variance: $\mathrm{Var}(X)=\dfrac{n^{2}-1}{12}=\dfrac{35}{12}\approx 2.917$.

It takes value $1$ ('success') with probability $p$ and $0$ ('failure') with probability $q=1-p$. Mean: $E[X]=p$. Variance: $\mathrm{Var}(X)=pq=p(1-p)$. MGF: $M_X(t)=q+pe^{t}$.

The variance $\mathrm{Var}(X)=p(1-p)$ is a downward parabola in $p$, maximized at $p=\tfrac{1}{2}$. Maximum value: $\mathrm{Var}(X)=\tfrac{1}{2}\cdot\tfrac{1}{2}=0.25$.

pmf: $p(x)=\binom{n}{x}p^{x}q^{\,n-x}$ for $x=0,1,\dots,n$, where $q=1-p$. Mean: $E[X]=np$. Variance: $\mathrm{Var}(X)=npq$. MGF: $M_X(t)=(q+pe^{t})^{n}$. It counts successes in $n$ independent identical trials.

Let $X\sim\text{Binomial}(10,0.3)$. $\Pr(X=4)=\binom{10}{4}(0.3)^{4}(0.7)^{6}$. $\binom{10}{4}=210$, so $\Pr(X=4)=210\,(0.0081)(0.117649)\approx 0.2001$.

Use the complement of 'no successes': $\Pr(X \geq 1)=1-\Pr(X=0)=1-(0.75)^{8}$. $(0.75)^{8}\approx 0.1001$, so $\Pr(X \geq 1)\approx 0.8999$.

Subtract the $0$ and $1$ terms from one: $\Pr(X \geq 2)=1-\Pr(X=0)-\Pr(X=1)$. $\Pr(X=0)=(0.8)^{5}=0.32768$; $\Pr(X=1)=\binom{5}{1}(0.2)(0.8)^{4}=5(0.2)(0.4096)=0.4096$. $\Pr(X \geq 2)=1-0.32768-0.4096\approx 0.2627$.

Support $\{1,2,3,\dots\}$ with $X=$ trial of first success. pmf: $p(x)=q^{x-1}p$, where $q=1-p$. Mean: $E[X]=\dfrac{1}{p}$. Variance: $\mathrm{Var}(X)=\dfrac{q}{p^{2}}$. MGF: $M_X(t)=\dfrac{pe^{t}}{1-qe^{t}}$ for $t<-\ln q$.

Support $\{0,1,2,\dots\}$ with $Y=$ failures before the first success. pmf: $p(y)=q^{y}p$. Mean: $E[Y]=\dfrac{q}{p}$. Variance: $\mathrm{Var}(Y)=\dfrac{q}{p^{2}}$ (same as convention A). Note $Y=X-1$, so the means differ by exactly $1$: $\tfrac{1}{p}$ vs $\tfrac{q}{p}$. Always read the problem to pick the support.

Here $p=\tfrac{1}{6}$, $q=\tfrac{5}{6}$. Mean: $E[X]=\dfrac{1}{p}=6$. $\Pr(X=4)=q^{3}p=\left(\tfrac{5}{6}\right)^{3}\tfrac{1}{6}=\dfrac{125}{1296}\approx 0.0965$.

Variance: $\mathrm{Var}(X)=\dfrac{q}{p^{2}}=\dfrac{0.8}{0.04}=20$. Survival: $\Pr(X>k)=q^{k}$, so $\Pr(X>5)=(0.8)^{5}=0.32768$.

Past failures do not change future waiting time: for $s,t \geq 0$, $\Pr(X>s+t \mid X>s)=\Pr(X>t)$. The geometric is the only discrete distribution with this property (the exponential is its continuous analogue). It follows from $\Pr(X>k)=q^{k}$.

By memorylessness, $\Pr(X>5 \mid X>2)=\Pr(X>3)$. Using $\Pr(X>k)=q^{k}$ with $q=0.75$: $\Pr(X>3)=(0.75)^{3}=0.421875\approx 0.4219$.

Support $\{r,r+1,\dots\}$ with $X=$ trial of the $r$-th success. pmf: $p(x)=\binom{x-1}{r-1}p^{r}q^{\,x-r}$, $q=1-p$. Mean: $E[X]=\dfrac{r}{p}$. Variance: $\mathrm{Var}(X)=\dfrac{rq}{p^{2}}$. MGF: $M_X(t)=\left(\dfrac{pe^{t}}{1-qe^{t}}\right)^{r}$. It reduces to the geometric when $r=1$.

Support $\{0,1,2,\dots\}$. pmf: $p(w)=\binom{w+r-1}{w}p^{r}q^{w}$. Mean: $E[W]=\dfrac{rq}{p}$. Variance: $\mathrm{Var}(W)=\dfrac{rq}{p^{2}}$ (same as the trials form). Note $W=X-r$, so the means differ by $r$.

Negative binomial, trials form, $r=3$, $p=0.4$, $q=0.6$. $\Pr(X=5)=\binom{5-1}{3-1}p^{3}q^{2}=\binom{4}{2}(0.4)^{3}(0.6)^{2}$. $\binom{4}{2}=6$, so $\Pr(X=5)=6\,(0.064)(0.36)=0.13824\approx 0.1382$.

Mean: $E[X]=\dfrac{r}{p}=\dfrac{3}{0.4}=7.5$. Variance: $\mathrm{Var}(X)=\dfrac{rq}{p^{2}}=\dfrac{3(0.6)}{0.16}=\dfrac{1.8}{0.16}=11.25$.

$\Pr(W=3)=\binom{w+r-1}{w}p^{r}q^{w}=\binom{3+2-1}{3}(0.3)^{2}(0.7)^{3}$. $\binom{4}{3}=4$, so $\Pr(W=3)=4\,(0.09)(0.343)=0.12348\approx 0.1235$.

Draw $n$ items without replacement from $N$ items of which $K$ are 'successes'; $X=$ successes drawn. pmf: $p(x)=\dfrac{\binom{K}{x}\binom{N-K}{\,n-x}}{\binom{N}{n}}$. Mean: $E[X]=n\dfrac{K}{N}$. Variance: $\mathrm{Var}(X)=n\dfrac{K}{N}\left(1-\dfrac{K}{N}\right)\dfrac{N-n}{N-1}$. The last factor $\tfrac{N-n}{N-1}$ is the finite-population correction.

Use the hypergeometric for sampling **without replacement** from a finite population (the trials are dependent because each draw changes the remaining composition). Use the binomial for independent trials with constant success probability — i.e. sampling **with replacement**, or drawing from a population so large the depletion is negligible.

Hypergeometric with $N=20$, $K=8$, $n=5$. $\Pr(X=2)=\dfrac{\binom{8}{2}\binom{12}{3}}{\binom{20}{5}}=\dfrac{28\cdot 220}{15504}=\dfrac{6160}{15504}\approx 0.3973$.

Hypergeometric: $N=12$, $K=7$ (women), $n=4$. $\Pr(X=2)=\dfrac{\binom{7}{2}\binom{5}{2}}{\binom{12}{4}}=\dfrac{21\cdot 10}{495}=\dfrac{210}{495}\approx 0.4242$.

Hypergeometric: $N=50$, $K=5$, $n=4$. $\Pr(X=0)=\dfrac{\binom{5}{0}\binom{45}{4}}{\binom{50}{4}}=\dfrac{1\cdot 148995}{230300}\approx 0.6470$.

Mean: $E[X]=n\dfrac{K}{N}=5\cdot\dfrac{8}{20}=2$. Variance: $\mathrm{Var}(X)=n\dfrac{K}{N}\left(1-\dfrac{K}{N}\right)\dfrac{N-n}{N-1}=5(0.4)(0.6)\dfrac{15}{19}=1.2\cdot\dfrac{15}{19}\approx 0.9474$.

pmf: $p(x)=\dfrac{e^{-\lambda}\lambda^{x}}{x!}$ for $x=0,1,2,\dots$. Mean: $E[X]=\lambda$. Variance: $\mathrm{Var}(X)=\lambda$ (mean equals variance). MGF: $M_X(t)=\exp[\lambda(e^{t}-1)]$.

$\Pr(X \geq 2)=1-\Pr(X=0)-\Pr(X=1)$. $\Pr(X=0)=e^{-3}\approx 0.049787$; $\Pr(X=1)=3e^{-3}\approx 0.149361$. $\Pr(X \geq 2)=1-0.049787-0.149361\approx 0.8009$.

$\Pr(X=3)=\dfrac{e^{-2.5}(2.5)^{3}}{3!}=\dfrac{e^{-2.5}\cdot 15.625}{6}$. $e^{-2.5}\approx 0.082085$, so $\Pr(X=3)\approx \dfrac{0.082085\cdot 15.625}{6}\approx 0.2138$.

Sum the first three terms: $\Pr(X \leq 2)=e^{-1.5}\left(1+1.5+\dfrac{1.5^{2}}{2}\right)=e^{-1.5}(1+1.5+1.125)=e^{-1.5}(3.625)$. $e^{-1.5}\approx 0.223130$, so $\Pr(X \leq 2)\approx 0.8088$.

Over two weeks the rate scales: $\lambda'=4\cdot 2=8$. $\Pr(X=6)=\dfrac{e^{-8}8^{6}}{6!}=\dfrac{e^{-8}\cdot 262144}{720}$. $e^{-8}\approx 0.000335463$, so $\Pr(X=6)\approx 0.1221$.

The Poisson factorial moment is $E[X(X-1)]=\lambda^{2}$. So $E[X^{2}]=E[X(X-1)]+E[X]=\lambda^{2}+\lambda$. With $\lambda=4$: $E[X^{2}]=16+4=20$ (consistent with $\mathrm{Var}=E[X^{2}]-(E[X])^{2}=20-16=4=\lambda$).

If $n\to\infty$ and $p\to 0$ with the product $np\to\lambda$ held fixed, then $\text{Binomial}(n,p)\to\text{Poisson}(\lambda)$. In practice it is accurate when $n$ is large and $p$ small (rare events), setting $\lambda=np$.

Set $\lambda=np=200(0.01)=2$. $\Pr(X=0)\approx e^{-\lambda}=e^{-2}\approx 0.1353$. (The exact binomial value $(0.99)^{200}\approx 0.1340$ is very close, confirming the approximation.)

For independent $X_i\sim\text{Poisson}(\lambda_i)$, MGFs multiply: $M_{\sum X_i}(t)=\prod_i \exp[\lambda_i(e^{t}-1)]=\exp\!\Big[\big(\sum_i\lambda_i\big)(e^{t}-1)\Big]$. This is the MGF of $\text{Poisson}\big(\sum_i\lambda_i\big)$, so the sum is Poisson with the rates added.

If events occur Poisson$(\lambda)$ and each is independently classified type-1 with probability $p$ (type-2 with $1-p$), then the type-1 count is $\text{Poisson}(\lambda p)$ and the type-2 count is $\text{Poisson}(\lambda(1-p))$. Moreover these two counts are **independent** of each other.

By thinning, sales $\sim\text{Poisson}(\lambda p)=\text{Poisson}(20\cdot 0.25)=\text{Poisson}(5)$. Mean sales per hour $=5$ (variance also $5$).

Expectation is linear: $E[aX+b]=aE[X]+b$. Variance ignores the shift and squares the scale: $\mathrm{Var}(aX+b)=a^{2}\mathrm{Var}(X)$, so $\mathrm{SD}(aX+b)=|a|\,\sigma$. These hold for any distribution, discrete or continuous.

Differentiate and evaluate at $0$: $E[X^{n}]=M_X^{(n)}(0)$, the $n$-th derivative at $t=0$. In particular $E[X]=M_X'(0)$ and $E[X^{2}]=M_X''(0)$. Every MGF satisfies $M_X(0)=1$ (since $e^{0\cdot X}=1$). The MGF, when it exists, uniquely determines the distribution.

This matches the Poisson MGF $\exp[\lambda(e^{t}-1)]$ with $\lambda=5$, so $X\sim\text{Poisson}(5)$. Therefore $E[X]=\lambda=5$ and $\mathrm{Var}(X)=\lambda=5$.

This matches the binomial MGF $(q+pe^{t})^{n}$ with $n=6$, $p=0.3$, $q=0.7$, so $X\sim\text{Binomial}(6,0.3)$. Mean: $E[X]=np=1.8$. Variance: $\mathrm{Var}(X)=npq=6(0.3)(0.7)=1.26$.

This matches the geometric MGF (convention A) $\dfrac{pe^{t}}{1-qe^{t}}$ with $p=0.4$, $q=0.6$. So $X\sim\text{Geometric}(0.4)$ on $\{1,2,\dots\}$, giving $E[X]=\dfrac{1}{p}=2.5$.