Continuous Random Variables [PDF]

Apr 21, 2017 - (eg height, weight, time). Continuous Random Variables. Probability Density Functions. X is a Continuous

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Idea Transcript


Chris Piech CS109

Lecture #9 April 21st , 2017

Continuous Random Variables So far, all random variables we have seen have been discrete. In all the cases we have seen in CS109 this meant that our RVs could only take on integer values. Now it’s time for continuous random variables which can take on values in the real number domain. They usually represent measurements with arbitrary precision (eg height, weight, time).

Continuous Random Variables Probability Density Functions X is a Continuous Random Variable if there is a Probability Density Function (PDF) f (x) for −∞ ≤ x ≤ ∞ such that: P(a ≤ X ≤ b) =

Z b

f (x)dx a

The following properties must also hold. These preserve the axiom that P(a ≤ X ≤ b) is a probability: 0 ≤ P(a ≤ X ≤ b) ≤ 1 P(−∞ < X < ∞) = 1 A common misconception is to think of f (x) as a probability. It is instead what we call a probability density. It represents probability/unit of X. Generally this is not particularly meaningful without either taking the interval over X or comparing it to another probability density. Of special note, the probability that a continuous random variable takes on a specific value (to infinite precision) is 0. Z a

P(X = a) =

f (x)dx = 0 a

That is pretty different than in the discrete world where we often talked about the probability of a random variable taking on a particular value.

Cumulative Distribution Function For a continuous random variable X the Cumulative Distribution Function, written F(a) or as (CDF) is: F(a) = P(X ≤ a) =

Z a

f (x)dx −∞

Example 1 Let X be a continuous random variable (CRV) with PDF: ( C(4x − 2x2 ) when 0 < x < 2 f (x) = 0 otherwise In this function, C is a constant. What value is C? Since we know that the PDF must sum to 1: Z 2

C(4x − 2x2 )dx = 1   2x3 2 2 C 2x − =1 3 0    16 C 8− −0 = 1 3 0

And if you solve the equation for C you find that C = 3/8. What is P(X > 1) Z ∞

f (x)dx =

      3 2x3 2 3 16 2 1 2 = (4x − 2x )dx = 2x − 8− − 2− = 8 8 3 1 8 3 3 2

Z 2 3

1

1

2

Example 2 Let X be a random variable which represents the number of days of use before your disk crashes with PDF: ( λ ex/100 when x ≥ 0 f (x) = 0 otherwise First, determine λ . Recall that eu du = eu : R

−1 x/100 λ ex/100 dx = 1 ⇒ −100λ e dx = 1 100 ∞ 1 − 100λ −x/100 = 1 ⇒ 100λ = 1 ⇒ λ = 100 0

Z

Z

What is the P(X < 10)? F(10) =

Z 10 1

100

0

e

−x/100

10 −x/100

dx = −e

= −e−1/10 + 1 ≈ 0.095



0

Expectation and Variance For continuous RV X: Z ∞

E[X] =

x f (x)dx −∞

Z ∞

E[g(X)] =

g(x) f (x)dx −∞

E[X n ] =

Z ∞

xn f (x)dx

−∞

For both continuous and discrete RVs: E[aX + b] = aE[X] + b Var(X) = E[(X − µ)2 ] = E[X 2 ] − (E[X])2

Uniform Random Variable X is a Uniform Random Variable X ∼ Uni(α, β ) if: ( 1 when α ≤ x ≤ β f (x) = β −α 0 otherwise The key properties of this RV are: P(a ≤ X ≤ b) =

Z b

Z ∞

E[X] =

Z β

x f (x)dx = −∞

Var(X) =



b−a (for α ≤ a ≤ b ≤ β ) β −α β x x2 = α +β dx = β −α 2(β − α) 2

f (x)dx = a

α

α

− α)2 12

2

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