Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $X_n$ is the number of coin tosses to get $n$ heads, then how can one show that there exists a constant $c>1$ such that $P(X_n \geq cn) \leq 1/n$? I am looking for a direct elementary proof.

Assume the coin tosses are fair and independent.

share|cite|improve this question
up vote 1 down vote accepted

Since $X_n$ is the sum of $n$ i.i.d. geometric random variables with parameter $\frac12$, $E[X_n]=2n$ and $\mathrm{var}(X_n)=2n$. Bienaymé-Chebychev inequality implies that, for every nonnegative $x$, $$ P[X_n\geqslant E[X_n]+x]\leqslant\mathrm{var}(X_n)/x^2. $$ If $x^2=2n^2$, the RHS is $1/n$ and $E[X_n]+x=(2+\sqrt2)n$ hence $c=2+\sqrt2$ answers the question. Or, $$ P[X_n\gt 4n]\leqslant1/(2n). $$

share|cite|improve this answer
Thank you. This would be a great answer but I was hoping, perhaps in vain, that there might be a proof that was elementary and entirely self contained. – felix Oct 6 '13 at 16:43
I suppose I could just incorporate a direct proof of… . – felix Oct 6 '13 at 17:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.