Question

jon flips a fair coin until a head is witnessed, and independently anna flips a fair coin (a different coin than jon’s) until a head is witnessed. let w be the minimum of the number of flips between john and anna. (a) derive the pmf for w. (b) compute e(w). (c) derive an expression for p(w ≥ k) for k

Answer 1

a) The pmf for w for certain value n is P(W = n) = 3 / 2²ⁿ⁺², b) The expected value on w, E(w) = 4 / 3, c) The expression for P(w ≥ k) =  $\frac{1}{2}^{2k}$, where k = 1, 2, 3...

Given: Jon flips a fair coin until a head is witnessed.

Anna flips a fair coin independently until a head is witnessed.

What is the probability density function?
The probability density function is a function that determines the probability for a discrete random variable X over a given sample space S for a certain value for X = x(some value).

It is also denoted as pmf and is written as P(X = x).

Let's solve the given question:

Let us assume the number of flips Jon did until he got a head to be X.

Also as Anna's flip independently so we need to consider a different random variable say Y.

So, the probability mass function for Jon (X) for a certain value of x is:

P(X = x) = $\frac{1}{2} * (1 - \frac{1}{2})^{x-1}$, where x = 1, 2, 3, ...........

P(X = x) = $\frac{1}{2}*\frac{1}{2}^{x - 1}$

P(X = x) = $\frac{1}{2}^{x - 1 + 1}$

P(X = x) = $\frac{1}{2}^x$ , where x = 1, 2, 3, ........

Similarly, for Anna, we will have the same probability mass function but with a different random variable Y for a certain value of y

P(Y = y) = $\frac{1}{2}^y$, where y = 1, 2, 3, ........

Now it is given that, w is the minimum number of flips between Jon and Anna.

w = minimum(P(X = x), P(Y = y))

Let us suppose the probability mass distribution over w is n for a certain value.

Then

P(w ≥ n) = P(minimum(P(X = x), P(Y = y)) ≥ n)

= $\frac{1}{2}^n * \frac{1}{2}^n$

= $\frac{1}{2}^{2n}$

Therefore, the probability mass function for w is

P(W = n) = P(W ≥ n) - P(W ≥ n + 1)

= $\frac{1}{2}^{2n}$ - $\frac{1}{2}^{2(n+1)}$

= $\frac{1}{2}^{2n}$ - $\frac{1}{2}^{2n + 2}$

= $\frac{1}{2}^{2n}$( 1 - $\frac{1}{2}^{2}$ )

= $\frac{1}{2}^{2n}$ ( 1 - 1 / 4)

= $\frac{1}{2}^{2n}$(4 - 1) / 4

= 3 / 4 ($\frac{1}{2}^{2n}$)

= 3 / 2² ($\frac{1}{2}^{2n}$)

= 3 / 2²ⁿ⁺²

Now the expected value of w, E(w) is:

E(w) = ∑P(w ≥ i) where i = 0 to ∞

= ∑$\frac{1}{2^{2i}}$ where i = 0 to ∞

= 1 + 1 / 4 + 1 / 16 + .......

This is infinite GP series. So the summation of infinite GP is

S = a / ( 1 - r )  

where a is the first term, r is the power and s is the summation.

Here a = 1, r = 1 / 4

S = 1 / (1 - 1 / 4)

S = 1 / 3 / 4

S = 4 / 3

Therefore, E(w) = 4 / 3

The expression for P(w ≥ k) is:

P(w ≥ k) =  $\frac{1}{2}^{2k}$, where k = 1, 2, 3...

Hence

a) The pmf for w for certain value n is P(W = n) = 3 / 2²ⁿ⁺²

b) The expected value on w, E(w) = 4 / 3

c) The expression for P(w ≥ k) =  $\frac{1}{2}^{2k}$, where k = 1, 2, 3...

Know more about "probability density function" here: brainly.com/question/14410995

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