The book

Standard discrete and continuous distributions

Why this is asked: Know the mean/variance table cold and the standard relationships (sum of exponentials = gamma, min of exponentials = exponential, memorylessness).

In one sentence

Almost every named distribution is a sum, a limit, a square or a ratio of the others, and knowing which relationship connects two of them answers most questions without any integration.

Why the exam asks it

Because the relationships are exactly what makes the standard results computable — the $t$ statistic, the $F$ test, the chi-square goodness of fit — and because memorylessness characterises precisely two distributions, which is a clean fact to test.

The idea

The standard distributions and how they arise from one another. Almost every named distribution in the syllabus is a sum, a limit, a square or a ratio of the others.

Discrete

distribution	mean	variance	arises as
Bernoulli $(p)$	$p$	$p(1-p)$	one trial
Binomial $(n,p)$	$np$	$np(1-p)$	sum of $n$ Bernoullis
Poisson $(\lambda)$	$\lambda$	$\lambda$	limit of Binomial, $np\to\lambda$
Geometric $(p)$	$1/p$	$(1-p)/p^2$	trials until first success
Negative binomial	$r/p$	$r(1-p)/p^2$	trials until $r$ -th success
Hypergeometric	$n\frac{K}{N}$	—	sampling without replacement

Poisson has mean equal to variance — a fact used constantly to check whether count data is Poisson, and the reason overdispersion is a recognised phenomenon.

Hypergeometric against binomial is sampling without against with replacement; the hypergeometric variance is smaller by the finite-population correction, and it tends to binomial as $N\to\infty$ .

Continuous

distribution	mean	variance	arises as
Uniform $(a,b)$	$\frac{a+b}2$	$\frac{(b-a)^2}{12}$	—
Exponential $(\lambda)$	$1/\lambda$	$1/\lambda^2$	gaps in a Poisson process
Gamma $(n,\lambda)$	$n/\lambda$	$n/\lambda^2$	sum of $n$ exponentials
Normal $(\mu,\sigma^2)$	$\mu$	$\sigma^2$	CLT limit
$\chi^2(n)$	$n$	$2n$	sum of $n$ squared standard normals
$t(n)$	$0$ ( $n>1$ )	$\frac{n}{n-2}$ ( $n>2$ )	$Z/\sqrt{\chi^2(n)/n}$
$F(m,n)$	$\frac{n}{n-2}$	—	ratio of two $\chi^2$ over their dfs
Beta $(\alpha,\beta)$	$\frac{\alpha}{\alpha+\beta}$	—	conjugate to the binomial

$\chi^2(n) = \text{Gamma}(n/2, 1/2)$ , so the gamma family contains the exponential and the chi-square as special cases. That single observation collapses several formulas into one.

Memorylessness

P(X>s+t \mid X>s) = P(X>t).

The exponential is the only continuous distribution with this property, and the geometric the only discrete one. Nothing else forgets. This characterisation is examined directly, and it is also why exponential inter-arrival times are what make a Poisson process Poisson.

Tails and limits

$t(1)$ is Cauchy — no mean. $t(n)\to$ Normal as $n\to\infty$ , and the $t$ has heavier tails at every finite $n$ , which is why its critical values exceed the normal's.
Sums of independent normals are normal; sums of independent Poissons are Poisson; sums of independent chi-squares are chi-square. These families are closed under convolution, and most are not.
Binomial $\to$ Poisson when $n\to\infty$ with $np\to\lambda$ : $(1-\lambda/n)^n\to e^{-\lambda}$ .
Binomial $\to$ Normal by the CLT, which is the better approximation when $p$ is not small.

Where intuition breaks

"The $t$ distribution always has a mean and variance." $t(1)$ is Cauchy with no mean; $t(2)$ has a mean and infinite variance. The variance $n/(n-2)$ needs $n>2$ .

"Poisson mean and variance are different parameters." Both are $\lambda$ . There is only one parameter.

"Any distribution can be memoryless." Exponential and geometric only.

"The sum of two independent uniforms is uniform." It is triangular. Closure under convolution is special — normal, Poisson, gamma with common rate, chi-square.

"Sampling with and without replacement give the same variance." Without replacement is hypergeometric, with a smaller variance by the finite-population correction.

" $\chi^2$ and gamma are unrelated." $\chi^2(n)=\text{Gamma}(n/2,1/2)$ .

"The normal approximation to the binomial always works." It needs $np$ and $n(1-p)$ both reasonably large. For small $np$ the Poisson approximation is the right one.

The exam's angle

Identify the mechanism, not the formula — counting successes, waiting for the first, summing squares, taking a ratio. The distribution follows.
Use the relationships to avoid integration. Mean of a gamma is $n$ times the mean of an exponential, and so on.
Check the degrees of freedom against existence for $t$ and $F$ moments.
Recognise memorylessness as a characterisation, so "which distribution has this property" has exactly one answer in each of the discrete and continuous cases.
Match the approximation to the regime: Poisson for rare events, normal for large $np$ .

The night before

Binomial $=$ sum of Bernoullis; Poisson $=$ limit with $np\to\lambda$ ; Poisson mean $=$ variance $=\lambda$ .
Geometric and exponential are the only memoryless distributions.
Gamma $=$ sum of exponentials; $\chi^2(n)=$ Gamma $(n/2,1/2)=$ sum of $n$ squared normals, mean $n$ , variance $2n$ .
$t(n)=Z/\sqrt{\chi^2(n)/n}$ : $t(1)$ is Cauchy, variance $n/(n-2)$ for $n>2$ , tends to normal.
$F(m,n)$ is a ratio of chi-squares over their degrees of freedom.
Closed under convolution: normal, Poisson, gamma (same rate), chi-square. Not uniform.
Hypergeometric is sampling without replacement, with smaller variance than the binomial.

See it move

A distribution's own shape, as its parameters moveinteractivefree

The trap here

“The maximum of independent exponentials is exponential” — false

$\max(X_{1}, X_{2})$ with $X_{i}$ ~ Exp(1)

The minimum is exponential (rate $\lambda_{1}+\lambda_{2})$ ; the maximum has CDF (1 − $e^{-x})^{2}$ , which is not exponential.