A coin is equally likely to be either $B_{1/3}$ (Bernoulli distributed with $p=1/3$) or $B_{2/3}$. To figure out the bias, we toss the coin 99 times and declare $B_{1/3}$ if the number of heads is less than 49.5 and $B_{2/3}$ otherwise. Bound the error probability using the Chernoff bound.
The Chernoff inequality or bound states: $P(X\geq a) = P(e^{tX}\geq e^{ta}) \leq \frac{E[e^{tX}]}{e^{ta}}$
In the special case of a binomial distribrution the lower bound of the Chernoff inequality is given by: $P(X\leq(1-\delta)\mu) = e^{-\frac{\delta^2}{2}\mu}$
From the question I understand that we are looking for the bound prob that
- (1) choosing $B_{1/3}$ is wrong if we see 49.5 or less heads after the 99 tosses.
- (2) Or we choose $B_{2/3}$ if we see 49.5 or more heads after the 99 tosses.
To calculate the first part (1): I tried using the binomial case of the Chernoff inequality. First, calculating $\delta$
$$ P(X\leq(1-\delta)\mu) = P(49.5 \leq (1-\delta)E(X_{2/3}))\\ E(X_{2/3})=\frac{2}{3}99 = 66\\ \delta = 1-\frac{49.5}{66} = 0.25\\ $$ Then I plugged delta into the inequality to get the lower bound probability: $$ P(X\leq(1-0.25)66)\leq e^{-\frac{0.25^2}{2}66} = 0.1271 $$
for the second part (2): first we need to use the binomial Chernoff upper bound inequality, which is: $P(X\geq(1+\delta)\mu) = e^{-\frac{\delta^2}{2+\delta}\mu}$
Again I calculte delta: $$ P(X\geq(1+\delta)\mu) = P(49.5 \geq (1-\delta)E(X_{1/3}))\\ E(X_{1/3})=\frac{1}{3}99 = 33\\ \delta = \frac{49.5}{33}-1 = 0.5\\ $$ Then I plugged delta into the inequality to get the upper bound probability: $$ P(X\geq(1+0.5)33)\leq e^{-\frac{0.5^2}{2.5}33} = 0.03688 $$
Finally, I add both upper and lower bound $P(wrong) = P(X\geq49.5)+P(X\leq49.5) = 0.1640189 $
However, this answer is wrong, and I am not sure what I am doing wrong. Maybe is the interpretation of the question? Any help or hint would be very appreciated!