Continuous Random Variables

Dr. Mine Dogucu

Discrete Random Variables Review

Discrete Random Variables

A discrete random variable has a countable number of possible numeric outcomes.

Probability mass function (pmf):

\(P(X = x) = f(x)\)

Cumulative distribution function (cdf):

\(P(X \leq x) = F(x)\)

\(E(X) = \sum_{S} x f(x)\)

\(Var(X)= E(X^2) - [E(X)]^2\)

Bernoulli Distribution

If X is a random variable that takes value 1 with probability of success \(\pi\) and 0 with probability \(1-\pi\), then X follows a Bernoulli distribution.

\(S = \{0, 1 \}\)

\(X \sim \text{Bernoulli} (\pi)\)

\(E(X) = \mu = \pi\)

\(Var(X)=\sigma^2 = \pi(1-\pi)\)

Geometric Distribution

Let X be the number of failures needed before the first success is observed in independent trials. \(X\) follows a geometric distribution

\(S = \{0, 1, 2, 3, 4, ...\}\)

\(X \sim \text{Geometric} (\pi)\)

\(f(x) = (1-\pi)^x(\pi)\)

\(E(X)=\frac{1-\pi}{\pi}\)

\(Var(X) = \frac{1-\pi}{\pi^2}\)

dgeom(x, prob) #pmf

pgeom(q, prob) #cdf

Binomial Distribution

The random variable X represents the number of successes in \(n\) trials where in independent trial the probability of success is \(\pi\).

\(S = \{0, 1, ..., n \}\)

\(X\sim \text{Binomial}(n, \pi)\)

\(P(X = x) = f(x) = {n \choose x}\pi^{x} (1-\pi)^{n-x}\)

\(E(X) = n\pi\)

\(Var(X) = n\pi(1-\pi)\)

dbinom(x, size, prob) #pmf

pbinom(q, size, prob) #cdf

Poisson Distribution

Let \(X\) represent the number of occurrences of an event within a fixed time or space.

\(S = \{0, 1, 2, 3, 4, ...\}\)

\(X \sim Poisson (\lambda)\)

\(P(X = x) = f(x) =\frac{\lambda^x}{x!} e^{-\lambda}\)

\(E(X) = Var(X) = \lambda\)

dpois(x, lambda) #pmf

ppois(q, lambda) #cdf

Continuous Random Variables

Continuous Random Variables

A continuous random variable \(X\) would have a sample space ( \(S_X\) ) that is uncountably infinite.

Let X be the the proportion of bike owners on campus. Then \(S_x = [0, 1]\).

Let Y be the the survival time after some surgery. Then \(S_Y = [0, \infty)\).

Probability Density Function (pdf) - \(f(x)\)

A probability density function gives the relative likelihood of the continuous random variable within the sample space.

\[f(x) \geq0 \text{ for all } x \ \epsilon \ S_X\]

\[\int_{x \ \epsilon \ S_X} f(x)dx = 1\]

\[P(a \leq X \leq b) = \int_a^b f(x)dx\]

Cumulative Distribution Function

Let \(X\) be a continuous random variable. Then the cumulative distribution function \(F(x) = P(x \leq X)\) defined as:

\[F(x) = P(x \leq X) = \int_{-\infty}^x f(t) dt\]

1. \(\frac{d}{dx}F = f\)
2. \(P(a \leq X \leq b) = F (b)− F(a)\)
3. \(P(X \geq a)= 1 − F (a) = \int_a^\infty f (x)dx\)

Example - pdf

\(x^2 \geq0\)

\((1-x) \geq0\)

\(f(x) \geq0 \text{ for all } x \ \epsilon \ S_X\)

Area Under the Curve = 1

\(\int_0^1 12(x^2)(1-x)dx\)

\(12\int_0^1 (x^2-x^3)dx\)

\(12\big[\frac{x^3}{3} -\frac{x^4}{4}\bigg\rvert_0^1\big] = 1\)

\(\int_{x \ \epsilon \ S_X} f(x)dx = 1\)

Probability is Area Under the Curve

\(P(0.25<X<0.50) = \int_{0.25}^{0.50} 12(x^2)(1-x)dx =12\int_{0.25}^{0.50} (x^2-x^3)dx\)

\(=12\big[\frac{x^3}{3} -\frac{x^4}{4}\bigg\rvert_{0.25}^{0.50}\big] = 0.2617188\)

\(P(X\ \epsilon \ B) = \int_{x \ \epsilon \ B} f(x)dx\)

\(P(X=x_i) = 0\)

\(P(X=0.40) =\)

\(\int_{0.40}^{0.40} 12(x^2)(1-x)dx\)

\(12\int_{0.40}^{0.40} (x^2-x^3)dx\)

\(12\big[\frac{x^3}{3} -\frac{x^4}{4}\bigg\rvert_{0.40}^{0.40}\big] = 0\)

cdf

\(P(X\leq x) = \int_{-\infty}^x f(t)dt\)

\(P(X\leq 0.70) =\)

\(\int_{0}^{0.70} 12(t^2)(1-t)dt\)

\(12\big[\frac{t^3}{3} -\frac{t^4}{4}\bigg\rvert_{0}^{0.70}\big] = 0.6516996\)

Note: \(f(x) = \frac{dF(x)}{dx}\)

Expected Value

\(E(X) = \int_{x \ \epsilon \ S_X} xf(x)dx\)

\(\int_0^1 x12(x^2)(1-x)dx\)

\(12\int_0^1 (x^3-x^4)dx\)

\(12\big[\frac{x^4}{4} -\frac{x^5}{5}\bigg\rvert_0^1\big] = 0.6\)

Variance

\(Var(X) = E(X^2)- [E(X)]^2\)

\(E(X^2) = ?\)

\(E(X^2) = \int_0^1 x^212(x^2)(1-x)dx\)

\(E(X^2) = 12\int_0^1 (x^4-x^5)dx\)

\(E(X^2) = 12\big[\frac{x^5}{5} -\frac{x^6}{6}\bigg\rvert_0^1\big] = 0.4\)

\(Var(X) = 0.4- 0.6^2 = 0.04\)