Math 20, Fall 2017 Edgar Costa Week 7 Dartmouth College Edgar - PowerPoint PPT Presentation

Math 20, Fall 2017 Edgar Costa Week 7 Dartmouth College Edgar Costa Math 20, Fall 2017 Week 7 1 / 32

Central Limit Theorem • Consider a Bernoulli trials process with probability p for success, i.e., a series Week 7 Math 20, Fall 2017 Edgar Costa j 2 / 32 and p . • We know that it is distributed as a binomial distribution with parameters n { X i } of i.i.d. Bernoulli trials. • X i = 1 or 0 if the i th outcome is a success or a failure, and let S n = X 1 + X 2 + · · · + X n . • Then S n is the number of successes in n trials. ( n ) p j ( 1 − p ) n − j P ( S n = j ) =

Edgar Costa Math 20, Fall 2017 Week 7 3 / 32

Edgar Costa Math 20, Fall 2017 Week 7 4 / 32

Standardized Sums • We can prevent the drifting of these spike graphs by subtracting the expected number of successes np from S n . • We obtain the new random variable S n np . • Now the maximum values of the distributions will always be near 0. • To prevent the spreading of these spike graphs, we can normalize S n np to have variance 1 by dividing by its standard deviation npq . Note: it does not spread as n Edgar Costa Math 20, Fall 2017 Week 7 5 / 32

Standardized Sums • We can prevent the drifting of these spike graphs by subtracting the expected number of successes np from S n . • Now the maximum values of the distributions will always be near 0. • To prevent the spreading of these spike graphs, we can normalize S n np to have variance 1 by dividing by its standard deviation npq . Note: it does not spread as n Edgar Costa Math 20, Fall 2017 Week 7 5 / 32 • We obtain the new random variable S n − np .

Standardized Sums • We can prevent the drifting of these spike graphs by subtracting the expected number of successes np from S n . • Now the maximum values of the distributions will always be near 0. Edgar Costa Math 20, Fall 2017 Week 7 5 / 32 • We obtain the new random variable S n − np . • To prevent the spreading of these spike graphs, we can normalize S n − np to have variance 1 by dividing by its standard deviation √ npq . Note: it does not spread as n → + ∞

Standardized Sum: Definition Edgar Costa Math 20, Fall 2017 Week 7 6 / 32 The Standardized sum of S n is given by n = S n − np S ∗ √ npq . Note: S ∗ n always has expected value 0 and variance 1.

• We make the height of the spikes at x j equal to the distribution value p j 1 p n n Week 7 Math 20, Fall 2017 Edgar Costa j j Standardized Sums 7 / 32 • We plot a spike graph with spikes placed at the possible values n = S n − np S ∗ √ npq . S ∗ n : x 0 , x 1 , . . . , x n , where x j = j − np √ npq

Standardized Sums j Week 7 Math 20, Fall 2017 Edgar Costa 7 / 32 • We plot a spike graph with spikes placed at the possible values n = S n − np S ∗ √ npq . S ∗ n : x 0 , x 1 , . . . , x n , where x j = j − np √ npq • We make the height of the spikes at x j equal to the distribution value ( n ) p j ( 1 − p ) n − j

Edgar Costa Can we make them match? Math 20, Fall 2017 Week 7 8 / 32 Standardized Sum n = 270 , p = 0 . 3 VS standard normal density 0.4 0.3 0.2 0.1 0 -4 -2 0 2 4

Edgar Costa Can we make them match? Math 20, Fall 2017 Week 7 8 / 32 Standardized Sum n = 270 , p = 0 . 3 VS standard normal density 0.4 0.3 0.2 0.1 0 -4 -2 0 2 4

g n x d x Can we make them match? j j 0 P S n j np npq n 0 j g n j np npq The last line is not an approximation for the integral! Why? Edgar Costa Math 20, Fall 2017 Week 7 n p j q n 9 / 32 j n 0 j n 1 x d x 1 ( n = j − np ) e − x 2 / 2 √ S ∗ φ ( x ) = g n ( x ) = P √ npq 2 π where j = round ( np + x √ npq ) In other words, x j = j − np √ npq is the closest point of that shape close to x .

g n x d x Can we make them match? j 0 P S n j np npq n g n 0 n j np npq The last line is not an approximation for the integral! Why? Edgar Costa Math 20, Fall 2017 Week 7 j 9 / 32 j 1 n ( n = j − np ) e − x 2 / 2 √ S ∗ φ ( x ) = g n ( x ) = P √ npq 2 π where j = round ( np + x √ npq ) In other words, x j = j − np √ npq is the closest point of that shape close to x . ( n ) ∫ ∑ φ ( x ) d x = 1 = p j q n − j R j = 0

g n x d x Can we make them match? g n n P n j 0 np j n npq The last line is not an approximation for the integral! Why? Edgar Costa Math 20, Fall 2017 Week 7 j 9 / 32 1 ( n = j − np ) e − x 2 / 2 √ S ∗ φ ( x ) = g n ( x ) = P √ npq 2 π where j = round ( np + x √ npq ) In other words, x j = j − np √ npq is the closest point of that shape close to x . ( n ) ( ) ∫ n = j − np p j q n − j = ∑ ∑ φ ( x ) d x = 1 = S ∗ √ npq R j = 0 j = 0

g n x d x Can we make them match? n Week 7 Math 20, Fall 2017 Edgar Costa The last line is not an approximation for the integral! Why? g n n P n j 9 / 32 1 ( n = j − np ) e − x 2 / 2 √ S ∗ φ ( x ) = g n ( x ) = P √ npq 2 π where j = round ( np + x √ npq ) In other words, x j = j − np √ npq is the closest point of that shape close to x . ( n ) ( ) ∫ n = j − np p j q n − j = ∑ ∑ φ ( x ) d x = 1 = S ∗ √ npq R j = 0 j = 0 ( j − np ) ∑ = √ npq j = 0

Can we make them match? n Week 7 Math 20, Fall 2017 Edgar Costa The last line is not an approximation for the integral! Why? g n n P j n 9 / 32 1 ( n = j − np ) e − x 2 / 2 √ S ∗ φ ( x ) = g n ( x ) = P √ npq 2 π where j = round ( np + x √ npq ) In other words, x j = j − np √ npq is the closest point of that shape close to x . ( n ) ( ) ∫ n = j − np p j q n − j = ∑ ∑ φ ( x ) d x = 1 = S ∗ √ npq R j = 0 j = 0 ( j − np ) ∫ ∑ = ̸ = g n ( x ) d x √ npq R j = 0

Math 20, Fall 2017 Edgar Costa Week 7 10 / 32 Standardized Sum n = 100 , p = 0 . 3 VS standard normal density 0.4 0.3 0.2 0.1 - 4 - 2 2 4

11 / 32 n Week 7 Math 20, Fall 2017 Edgar Costa npq 1 j n 1 1 j n 1 Integrating g n ( x ) ∫ ( j − np ) ∑ g n ( x ) d x = √ npqg n √ npq R j = 0 ( n ) ∑ = √ npq p j q n − j j = 0 ( n ) ∑ = p j q n − j √ npq j = 0

11 / 32 n Week 7 Math 20, Fall 2017 Edgar Costa 1 j n 1 1 j 1 n Integrating g n ( x ) ∫ ( j − np ) ∑ g n ( x ) d x = √ npqg n √ npq R j = 0 ( n ) ∑ = √ npq p j q n − j j = 0 ( n ) ∑ = p j q n − j √ npq j = 0 = √ npq

Math 20, Fall 2017 Edgar Costa Week 7 12 / 32 rescaled standardized Sum n = 100 , p = 0 . 3 VS standard normal density 0.4 0.3 0.2 0.1 - 4 - 2 2 4

Math 20, Fall 2017 Edgar Costa Week 7 13 / 32 rescaled standardized Sum n = 270 , p = 0 . 3 VS standard normal density 0.4 0.3 0.2 0.1 - 4 - 2 2 4

Central Limit Theorem for Binomial Distributions 1 Week 7 Math 20, Fall 2017 Edgar Costa 0 and assuming that np is an integer. Challenge: try to carry this out for x Theorem 14 / 32 j lim ( n ) Write b ( n , p , j ) := p j q n − j . We have √ npqb ( n , p , round ( np + x √ npq )) = φ ( x ) = e − x 2 / 2 √ 2 π n → + ∞ √ We can prove it directly using Stirling’s formula n ! ≈ 2 π nn n e − n as n → + ∞ .

Central Limit Theorem for Binomial Distributions 1 Week 7 Math 20, Fall 2017 Edgar Costa Theorem 14 / 32 lim j ( n ) Write b ( n , p , j ) := p j q n − j . We have √ npqb ( n , p , round ( np + x √ npq )) = φ ( x ) = e − x 2 / 2 √ 2 π n → + ∞ √ We can prove it directly using Stirling’s formula n ! ≈ 2 π nn n e − n as n → + ∞ . Challenge: try to carry this out for x = 0 and assuming that np is an integer.

Approximating Binomial Distributions 1 Week 7 Math 20, Fall 2017 Edgar Costa 15 / 32 • Solve for x • To find approximations for the values of b ( n , p , j ) , we set j = np + x √ npq x = j − np √ npq . φ ( x ) b ( n , p , j ) ≈ √ npq ( j − np ) = √ npq φ . √ npq

Example 2 1 and P S 100 55 1 5 1 5 1 e 50 1 2 0 0483941 • Indeed, P S 100 55 0 0484743 Edgar Costa Math 20, Fall 2017 Week 7 5 55 16 / 32 50 and 1 • Let us estimate the probability of exactly 55 heads in 100 tosses of a coin. • For this case np 100 1 • Thus x 2 npq 100 1 2 1 2 25 5. ( j − np ) b ( n , p , j ) ≈ √ npq φ √ npq

Example e 55 1 5 1 5 1 2 1 2 1 and 0 0483941 • Indeed, P S 100 55 0 0484743 Edgar Costa Math 20, Fall 2017 Week 7 P S 100 5 16 / 32 50 55 • Thus x 1 • Let us estimate the probability of exactly 55 heads in 100 tosses of a coin. ( j − np ) b ( n , p , j ) ≈ √ npq φ √ npq √ 2 = 50 and √ npq = √ • For this case np = 100 · 1 100 · 1 2 · 1 2 = 25 = 5.

Example 5 Week 7 Math 20, Fall 2017 Edgar Costa 0 0484743 55 • Indeed, P S 100 1 5 5 16 / 32 1 • Let us estimate the probability of exactly 55 heads in 100 tosses of a coin. ( j − np ) b ( n , p , j ) ≈ √ npq φ √ npq √ 2 = 50 and √ npq = √ • For this case np = 100 · 1 100 · 1 2 · 1 2 = 25 = 5. • Thus x = 55 − 50 = 1 and P ( S 100 = 55 ) ≈ φ ( 1 ) e − 1 / 2 = 1 √ 2 π = 0 . 0483941