2.1 Definition and Distribution Functions
Definition. A random variable is a measurable function X:Ω→R. The cumulative distribution function (CDF) of X is
FX(x)=P(X≤x)
Proposition 2.1 (Properties of the CDF).
- F is non-decreasing: if a≤bThen F(a)≤F(b).
- limx→−∞F(x)=0 and limx→+∞F(x)=1.
- F is right-continuous: limx→a+F(x)=F(a).
Proof. (1) If a≤bThen {X≤a}⊆{X≤b}So F(a)=P(X≤a)≤P(X≤b)=F(b) by Proposition 1.1(3).
(2) As x→−∞The events {X≤x} decrease to ∅So by continuity from above of probability measures, F(x)→0. As x→+∞The events increase to ΩSo F(x)→1.
(3) As x→a+The events {X≤x} decrease to {X≤a}Giving right-continuity. ■
2.2 Discrete Random Variables
A random variable is discrete if its range is countable. The probability mass function (PMF) is pX(x)=P(X=x).
Definition (Expected Value). For a discrete random variable:
E[X]=∑xxpX(x)
Provided the sum converges absolutely.
Definition (Variance). Var(X)=E[(X−μ)2]=E[X2]−(E[X])2 where μ=E[X].
Proposition 2.2 (Linearity of Expectation). E[aX+bY]=aE[X]+bE[Y] for any random variables X, Y and constants a, b.
Proof. Direct computation from the definition of expected value. For the discrete case:
E[aX+bY]=∑x,y(ax+by)pX,Y(x,y)=a∑xxpX(x)+b∑yypY(y)=aE[X]+bE[Y]
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2.3 Continuous Random Variables
A random variable is continuous if its CDF is absolutely continuous, i.e., there exists a probability density function (PDF) fX such that
FX(x)=∫−∞xfX(t)dt
Key properties:
- fX(x)≥0 for all x.
- ∫−∞∞fX(x)dx=1.
- P(a≤X≤b)=∫abfX(x)dx.
- P(X=a)=0 for any single point a.
2.4 Common Distributions
Discrete distributions:
| Distribution | PMF | E[X] | Var(X) |
|---|
| Bernoulli(p) | px(1−p)1−x, x∈{0,1} | p | p(1−p) |
| Binomial(n,p) | (xn)px(1−p)n−x | np | np(1−p) |
| Poisson(λ) | e−λλx/x! | λ | λ |
| Geometric(p) | (1−p)x−1p, x≥1 | 1/p | (1−p)/p2 |
Continuous distributions:
| Distribution | PDF | E[X] | Var(X) |
|---|
| Uniform(a,b) | 1/(b−a) on [a,b] | (a+b)/2 | (b−a)2/12 |
| Exponential(λ) | λe−λx, x≥0 | 1/λ | 1/λ2 |
| N(μ,σ2) | σ2π1e−(x−μ)2/(2σ2) | μ | σ2 |
2.5 The Normal Distribution
Definition. X∼N(μ,σ2) if X has PDF f(x)=σ2π1exp(−2σ2(x−μ)2).
Theorem 2.3 (Standardisation). If X∼N(μ,σ2)Then Z=(X−μ)/σ∼N(0,1).
Proof. The CDF of Z: P(Z≤z)=P(X≤μ+σz)=∫−∞μ+σzσ2π1e−t2/2dt. Substituting u=(t−μ)/σ: =∫−∞z2π1e−u2/2duWhich is the CDF of N(0,1). ■
Theorem 2.4 (Moment Generating Function). If X∼N(μ,σ2)Then
MX(t)=E[etX]=exp(μt+2σ2t2)
Proof. MX(t)=∫−∞∞etxσ2π1e−(x−μ)2/(2σ2)dx. Completing the square in the exponent and evaluating the Gaussian integral gives the result. ■
2.6 Moment Generating Functions
Definition. The moment generating function (MGF) of X is MX(t)=E[etX] (when it exists in a neighbourhood of t=0).
Theorem 2.5. If the MGF exists in a neighbourhood of 0, it uniquely determines the distribution. Furthermore, E[Xn]=MX(n)(0).
2.7 Worked Examples
Problem. Let X∼Poisson(3) and Y∼Poisson(5) be independent. Find the distribution of X+Y.
Solution
The MGF of X∼Poisson(λ) is MX(t)=eλ(et−1).
MX+Y(t)=MX(t)⋅MY(t)=e3(et−1)⋅e5(et−1)=e8(et−1).
This is the MGF of Poisson(8). Since the MGF uniquely determines the distribution, X+Y∼Poisson(8).
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Worked Example: Minimum of Exponential Random Variables
Solution. Let X1,…,Xn be independent with Xi∼Exp(λi). Find the distribution of M=min(X1,…,Xn).
P(M>t)=P(X1>t,…,Xn>t)=∏i=1nP(Xi>t)=∏i=1ne−λit=e−(λ1+⋯+λn)t
So P(M≤t)=1−e−λt where λ=∑i=1nλi. This means M∼Exp(λ). ■