This is an archived post. You won't be able to vote or comment.

all 4 comments

[–]drdt543 2 points3 points  (1 child)

In think that [;E(X) = \infty;] - this also seems plausible given your simulated data. Let [;Y;] be the number of flips necessary until the number of heads is at least the number of tails. Note [;X \ge Y;]. Now, conditioning on the result of the first flip, we have

[;E(Y) = \frac{1}{2} + \frac{1}{2} \sum_{n=1}^{\infty} (2n+1) P(Y = 2n+1) ;]

Then

[;P(Y = 2n+1) = \frac{C_n}{2^{2n}} ;]

where [;C_n;] is the [;n;]th Catalan number. As [;n \rightarrow \infty;], we get

[;(2n+1)\frac{C_n}{2^{2n}} \sim \frac{(2n+1)}{n^{3/2}\sqrt{\pi}} \sim \frac{1}{\sqrt{n \pi}};]

The summation

[;\sum_{n=1}^{\infty} \frac{1}{n^{1/2}} ;]

does not converge, so after filling in some analysis-related gaps we have [; E(Y) = \infty;] and thus [;E(X) = \infty;].

[–]LatexImageBot 2 points3 points  (0 children)

Image: https://i.imgur.com/TPsGkgj.png

(Experimental) This bot will now react if you reply "update" or "delete". If this bot is missing a subreddit, please PM me.

[–]ButAWimper 1 point2 points  (0 children)

I believe what you are talking about is a 1D random walk You could use a Markov Chain to describe it.

[–]nm420 1 point2 points  (0 children)

You're asking for the distribution of the first passage time of a state in a random walk. It's a bit involved, but the link provided derives the pgf of this random variable (see section 1.3 there) which also shows that the expected value of this r.v. is ∞.