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The bookUnit 4 · Probability72 / 83

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).

Reference table

Distribution Mean Variance
Bernoulli(p) p pq
Binomial(n,p) np npq
Poisson(λ)(\lambda) λ\lambda λ\lambda
Geometric (trials, p) 1/p q/p2q/p^{2}
Negative binomial(r,p) r/p rq/p2/p^{2}
Hypergeometric nK/N nK(NK)(Nn)/(N2(N1))(N-K)(N-n)/(N^{2}(N-1))
Uniform(a,b) (a+b)/2 (ba)2/12(b-a)^{2}/12
Exponential(λ)(\lambda) 1/λ1/\lambda 1/λ21/\lambda^{2}
Gamma(k,λ)(k,\lambda) k/λk/\lambda k/λ2k/\lambda^{2}
Beta(a,b) a/(a+b) ab/((a+b)2(a+b+1))/((a+b)^{2}(a+b+1))
N(μ,σ2)N(\mu,\sigma^{2}) μ\mu σ2\sigma^{2}
χn2\chi^{2}_{n} n 2n
tνt_\nu 0(ν>1)0 (\nu>1) ν/(ν2)(ν>2)\nu/(\nu-2) (\nu>2)
Cauchy

Relationships to quote

  • Sum of n i.i.d. Exp(λ)=(\lambda) = Gamma(n,λ)(n, \lambda); Gamma(n/2, ½)=χn2) = \chi^{2}_{n}.
  • **min of independent Exp(λi)(\lambda_{i}) ~ Exp(λi)(\sum\lambda_{i})** — the series-system fact.
  • Sum of independent Poissons is Poisson; of independent normals is normal; of independent χ2\chi^{2} is χ2\chi^{2}.
  • Z ~ N(0,1)Z2N(0,1) \Rightarrow Z^{2} ~ χ12.t=Z/χn2/n;F=(χm2/m)/(χn2/n)\chi^{2}_{1}. t = Z/\sqrt{\chi^{2}_{n}/n}; F = (\chi^{2}_{m}/m)/(\chi^{2}_{n}/n).
  • Binomial → Poisson (n large, np fixed) → Normal (CLT).
  • Memorylessness characterises the exponential (continuous) and geometric (discrete): P(X > s+t | X > s) = P(X > t).

Order statistics

For i.i.d. with CDF F:P(X(n)x)=F(x)nF: P(X_{(n)} \le x) = F(x)^{n} and P(X(1)>x)=(1F(x))nP(X_{(1)} > x) = (1 - F(x))^{n}. For Uniform(0,1),X(k)(0,1), X_{(k)} ~ Beta(k, n−k+1).

Key takeaways

  • Exponential: memoryless, minimum stays exponential, sum becomes gamma. χ2,t- \chi^{2}, t and F are all built from normals — know the constructions.
  • Uniform order statistics are Beta.

See it move

The signatures that identify a distribution instantlyinteractive

One characterising property each for Poisson, exponential and normal.

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The trap here

“The maximum of independent exponentials is exponential” — false

max(X1,X2)\max(X_{1}, X_{2}) with XiX_{i} ~ Exp(1)

The minimum is exponential (rate λ1+λ2)\lambda_{1}+\lambda_{2}); the maximum has CDF (1 − ex)2e^{-x})^{2}, which is not exponential.

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Check yourself — select all that apply

Let X1X_{1} and X2X_{2} be independent, X1X_{1} gamma with mean 10 and variance 10, and X2X_{2} ~ N(3, 4). Let f1,f2f_{1}, f_{2} be their densities. Define Y with density f(y)=0.4f1(y)+qf2(y)f(y) = 0.4 f_{1}(y) + q f_{2}(y). Which of the following are true?

Next: Joint distributions, transformations, order statistics

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Open this in the full syllabus view · Unit 4