ma 320-001: introductory probability - mathematicsdmu228/ma320/lectures/sec8.1.pdf · sum of...

MA 320-001: Introductory Probability

David Murrugarra

Department of Mathematics,University of Kentucky

http://www.math.uky.edu/~dmu228/ma320/

Spring 2017

David Murrugarra (University of Kentucky) MA 320: Section 8.1 Spring 2017 1 / 11

mailto:[email protected]

http://www.math.uky.edu/

http://www.uky.edu/

http://www.math.uky.edu/~dmu228/ma320/


Sum of Independent Normal Random Variables

Example (Homework12: Problem 5)Let Xk be independent and normally distributed with common mean 2and standard deviation 1 (so their common variance is 1.)Compute

P

(−∞ ≤

25∑k=1

Xk ≤ 56.95

)

Solution:

Let Y =25∑

k=1

Xk . Then Y is N(50,25). We want

P(−∞ ≤ Y ≤ 56.95) = 0.5 + P(0 ≤ Z ≤ 1.39) = 0.9177



Sum of Two Independent Poisson Random Variables

ExampleSuppose X and Y are two independent random variables, each withpoisson density.

LetfX (x) =

(λr)x

x!e−λr and fY (x) =

(λs)x

x!e−λs

Let Z = X + Y . Then we have

fZ (z) =(λ(r + s))x

x!e−λ(r+s)



Sum of Two Independent Poisson Random Variables

Example (Homework12: Problem 2)Let X1 and X2 have Poisson distributions with the same average rateλ = 0.9 on independent time intervals of length 4 and 1 respectively.Find

Prob (X1 + X2 = 3)

to at least 6 decimal places.

Solution:

Let λ = 0.9, r = 4, and s = 1. Then

fZ (3) =(λ(r + s))x

x!e−λ(r+s) =

(0.9)3(5)3

6e−5(0.9)



Section 8.1 Law of Large Numbers for DiscreteRandom Variables

Theorem (Chebyshev Inequality)Let X be a discrete random variable with expected value µ = E(X ),and let ε > 0 be any positive real number. Then

P(|X − µ| ≥ ε) ≤ V [X ]

ε2




Example (Example 8.1)

Let X by any random variable with E(X ) = µ and V (X ) = σ2. Then, ifε = kσ, Chebyshev’s Inequality states that

P(|X − µ| ≥ kσ) ≤ σ2

k2σ2 =1k2

Thus, for any random variable, the probability of a deviation from themean of more than k standard deviations is 1

k2 .




Example (Homework13: Problem 1)Let X be a random variable with expected value 4 and variance 9.According to the Chebyshev inequality,

P(|X − 4| ≥ 0.63) ≤

Solution:

P(|X − 4| ≥ 0.63) ≤ 9(0.63)2




Theorem (Law of Large Numbers)Let X1,X2, . . . ,Xn be an independent trial process, with finite expectedvalue µ = E(Xi) and finite variance σ2 = V (Xi). Let

Sn = X1 + X2 + · · ·+ Xn.

Then for any positive real number ε > 0,

P(∣∣∣∣Sn

n− µ

∣∣∣∣ ≥ ε)→ 0 as n→∞.

Equivalently,

P(∣∣∣∣Sn

n− µ

∣∣∣∣ < ε

)→ 1 as n→∞.



Example (Homework13: Problem 4)Let X1, . . . ,Xn be an independent trials process with common expectedvalue 3 and common variance 17.

What is the exact value of the expected value of the average An?

What is the variance of the average An?

According to the Chebyshev inequality, P(|An − 3| ≥ 0.1) ≤

Using your bound from the Chebyshev inequality, how many trialsare needed so that P(|An − 3| ≥ 0.1) ≤ 0.31?

Solution:

P(|An − 3| ≥ 0.1) ≤ 17n(0.1)2



Section 8.2 Law of Large Numbers for ContinuousRandom Variables

Theorem (Chebyshev Inequality)Let X be a continuous random variable with expected value µ = E(X ),and let ε > 0 be any positive real number. Then

P(|X − µ| ≥ ε) ≤ V [X ]

ε2



Section 8.2 Law of Large Numbers for ContinuousRandom Variables

Theorem (Law of Large Numbers)Let X1,X2, . . . ,Xn be an independent trial process with a continuousdensity function f , finite expected value µ = E(Xi), and finite varianceσ2 = V (Xi). Let

Sn = X1 + X2 + · · ·+ Xn.

Then for any positive real number ε > 0,

P(∣∣∣∣Sn

n− µ

∣∣∣∣ ≥ ε)→ 0 as n→∞.

Equivalently,

P(∣∣∣∣Sn

n− µ

∣∣∣∣ < ε

)→ 1 as n→∞.



ma 320-001: introductory probability - mathematicsdmu228/ma320/lectures/sec8.1.pdf · sum of...

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