STAT 400

Wed. February 12th, 2020

Parameters (cont.)

Bernoulli Distribution

Ex. For a Bernoulli random variable, introducing a parameter α\alpha gives us the Bernoulli distribution: p(x;α)={αx=11−αx=0p(x;\alpha)=\begin{cases} \alpha & x=1\\1-\alpha & x=0 \end{cases} To say that a random variable xx follows this distribution, we write x→Bernoulli(α).x\rightarrow\text{Bernoulli}(\alpha). For example, if xx is the number of heads that occur from a coin toss, then x→Bernoulli(0.5).x\rightarrow \text{Bernoulli}(0.5).

Geometric Distribution

Ex. Geometric distribution with parameter pp:
Suppose we’re testing the battery life of batteries. Let p=P({S})p=P(\{S\}). xx is the number of tests we must do before finding a battery with acceptable battery life (an SS). Then: p(1)=P(x=1)=pp(2)=P(x=2)=(1−p)pp(3)=P(x=3)=(1−p)2pp(k)=P(x=k)=(1−p)k−1p\begin{aligned} p(1)&=P(x=1)=p\\ p(2)&=P(x=2)=(1-p)p\\ p(3)&=P(x=3)=(1-p)^2p\\ p(k)&=P(x=k)=(1-p)^{k-1}p \end{aligned} In general, p(k;p)={(1−p)k−1pk=1,2,3...0otherwisep(k;p)=\begin{cases} (1-p)^{k-1}p & k=1,2,3...\\ 0 & \text{otherwise} \end{cases}
We say that this xx follows a geometric distribution with parameter pp: x→Geo(p).x\rightarrow \text{Geo}(p).

Cumulative Distribution Function (cdf)

Definition. The cumulative distribution function (cdf) F(x)F(x) of a discrete random variable xx with probability mass function p(x)p(x) is defined for every xx by F(x)=P(X≤x)=∑y: y≤xp(y).F(x)=P(X\le x)=\sum_{y:\ y\le x}p(y).

Ex. Converting a cdf to a geometric distribution with pp: In general, F(x)=∑y: y≤xp(y;p)=∑y=1⌊x⌋p(y;p)=∑y=1⌊x⌋(1−p)y−1p.\begin{aligned} F(x)&=\sum_{y:\ y\le x}p(y;p)\\&=\sum_{y=1}^{\lfloor x\rfloor}p(y;p)\\&=\sum_{y=1}^{\lfloor x\rfloor}(1-p)^{y-1}p. \end{aligned}
Now we can apply the formula for the sum of a geometric series, which is: ∑y=0kay=1−ak+11−a if 0<a<1.\sum_{y=0}^k a^y=\frac{1-a^{k+1}}{1-a}\text{ if } 0<a<1.
F(x)=p∑y=1⌊x⌋(1−p)y−1=p∑y=0⌊x⌋−1(1−p)y=p⋅1−(1−p)⌊x⌋1−(1−p)=1−(1−p)⌊x⌋\begin{aligned} F(x)&=p\sum_{y=1}^{\lfloor x\rfloor}(1-p)^{y-1}\\&=p\sum_{y=0}^{\lfloor x\rfloor - 1}(1-p)^y\\&=p\cdot \frac{1-(1-p)^{\lfloor x\rfloor}}{1-(1-p)}\\&=1-(1-p)^{\lfloor x\rfloor} \end{aligned}