1
$\begingroup$

There are $n$ flips of a fair coin, $X$ is a random variable which counts the number of $HH$:

for example the event '$HHH$' yields, $X=2$.

I'm trying to bound $P(X\le\frac{n}{8})$ from above.

What I tried to do is to define $X=\sum\limits _{i=1}^{n-1}Y_{i}$ where $Y_{i}\sim Ber\left(\frac{1}{4}\right)$ represent a H in the $i$-th and $(i+1)$-th flips. And to apply Chebyshev's inequality. But im having trouble to calculate $\mathbb{E}[X^{2}]$

1 Answers 1

1

You can bound it by Markov inequality.

Since the maximum of $X$ is $n-1$ you can define $Z=n-X$ and use Markov. So you need not to calculate Variance...

$$P(X\le\frac{n}8)\le P(Z>\frac{7n}{8})<\frac{EZ}{\frac{7n}8}\le\frac{\frac{3n}4}{\frac{7n}8}=\frac{6}{7}$$

  • 1
    Thank you ! but do you think it was possible to know this by Chebyshev's inequality ?because that was the guideness in the question – 2017-01-14
  • 0
    Your Welcome... You need to expand $X^2$ by binomial expansion and use linearity of Expectation. It's not too hard because most of $Y_i$s are independent. – 2017-01-14