Calculating Expected Heads in Coin Flips & Maxima in Permutations, Study notes of Advanced Calculus

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Quick Review of Linearity of

Expectation

COMP 3711H - HKUST Version of 24/11/ M. J. Golin

Linearity of Expectation is one of the simplest and most useful tools used in the analysis of randomized algorithms. In its easiest form it just says that, if X, Y are any two random variables (not necessarily independent) then E(X + Y ) = E(X) + E(Y ). The iterated version is that if X 1 , X 2 ,... , Xn are any random variables, then

E

( (^) n ∑ i= Xi

∑^ n i= E(Xi).

Example: Let Z be the value seen when rolling two dice. Z = X 1 + X 2 where Xi is the value seen when rolling single die i = 1, 2. It’s easy to calculate that E(Xi) =

∑^6

j= j Pr(Xi = j) =

∑^6

j= j

Then

E(Z) = E(X 1 + X 2 ) = E(X 1 ) + E(X 2 ) =

7 2

7 2 = 7.

When flipping n coins, what is the expected number of heads? Z =

i=1 Xi, where Xi = 1 if coin i is a head and 0 if it is a tail. Set Pr(Xi = 1) = pi and Pr(Xi = 0) = 1 − pi. Then Xi is a Bernoulli Random Variable with probability pi. Note that E(Xi) = 1 · Pr(Xi = 1) = pi so E(Z) = ∑^ n i= E(Xi) = ∑^ n i= Pr(Xi = 1) = ∑^ n i= pi.

When flipping n coins, what is the expected number of heads? Z =

i=1 Xi, where Xi = 1 if coin i is a head and 0 if it is a tail. Set Pr(Xi = 1) = pi and Pr(Xi = 0) = 1 − pi. Then Xi is a Bernoulli Random Variable with probability pi.

Note that E(Xi) = 1 · Pr(Xi = 1) = pi so

E(Z) =

∑^ n

i=

E(Xi) =

∑^ n

i=

Pr(Xi = 1) =

∑^ n

i=

pi.

Examples:

pi = p (all coins the same) ⇒ E(Z) = pn

pi = (^1) i ⇒ E(Z) =

∑n i=1 pi^ =^

∑n i=

1 n =^ Hn^ ∼^ ln^ n

Suppose you are flipping n coins, each with pi = 12 , i.e., fair coins. How many times does the pattern HHH appear?

Suppose an algorithm’s input is a permutation of n numbers. Let x 1 , x 2 ,... , xn be the input in its given order.

xi is a left to right maxima if it’s bigger than x 1 , x 2 ,... , xi− 1.

For example, the red items in these two permutations are the l.t.r. maxima: 5 4 7 8 1 6 3 2 1 3 5 7 2 4 6 8

Suppose an algorithm’s input is a permutation of n numbers. Let x 1 , x 2 ,... , xn be the input in its given order. xi is a left to right maxima if it’s bigger than x 1 , x 2 ,... , xi− 1. For example, the red items in these two permutations are the l.t.r. maxima: 5 4 7 8 1 6 3 2 1 3 5 7 2 4 6 8 Some algorithms’ run times depend upon Z, the number of l.t.r. maxima. Assuming all n! permutations are equally likely, how can we find E(Z)?

Suppose an algorithm’s input is a permutation of n numbers. Let x 1 , x 2 ,... , xn be the input in its given order. xi is a left to right maxima if it’s bigger than x 1 , x 2 ,... , xi− 1. For example, the red items in these two permutations are the l.t.r. maxima: 5 4 7 8 1 6 3 2 1 3 5 7 2 4 6 8 Z = ∑n i=1 Xi^ where^ Xi^ = 1^ iff^ xi^ is a l.t.r. maxima and^0 otherwise One way of generating a random permutation is to first randomly choose the first i items equally likely among all posible (n i ) subsets. Then choose a random permutation among the i possible permutations to order them as x 1 ,... xi. Then randomly order the remaining items as xi+1,... xn. Some algorithms’ run times depend upon Z, the number of l.t.r. maxima. Assuming all n! permutations are equally likely, how can we find E(Z)?

Suppose an algorithm’s input is a permutation of n numbers. Let x 1 , x 2 ,... , xn be the input in its given order. xi is a left to right maxima if it’s bigger than x 1 , x 2 ,... , xi− 1. For example, the red items in these two permutations are the l.t.r. maxima: 5 4 7 8 1 6 3 2 1 3 5 7 2 4 6 8 Z = ∑n i=1 Xi^ where^ Xi^ = 1^ iff^ xi^ is a l.t.r. maxima and^0 otherwise One way of generating a random permutation is to first randomly choose the first i items equally likely among all posible (n i ) subsets. Then choose a random permutation among the i possible permutations to order them as x 1 ,... xi. Then randomly order the remaining items as xi+1,... xn. Probability xi is l.t.r. maxima is prob it’s largest in first i items which is now 1 i. So^ Xi^ is Bernouli Random Variable with^ pi^ = 1/i^ and^ E(Zi) =^ Hn. Some algorithms’ run times depend upon Z, the number of l.t.r. maxima. Assuming all n! permutations are equally likely, how can we find E(Z)?

Make a minor change from the previous page. Suppose we flip n coins. ith coin is Heads with probability pi = 1/i. If ith coin is Heads you run another random process Yi to tell you how much money you receive. All you know is that E(Yi) = i. If ith coin is Tails, you get nothing. What is expected amount you receive? Let Z be total amount. Z = ∑n i=1 XiYi^ where Xi = 1 if ith coin is Heads and is otherwise 0 , so E(Xi) = pi. E(Yi) = i. Then, because Xi and Yi are independent E(XiYi) = E(Xi)E(Yi) =^1 i i = 1 E(Z) = E ( (^) n ∑ i= XiYi ) = ∑^ n i= E(XiYi) = ∑^ n i= So, 1 = n