Binary Series

For $0 \le x \lt 1$, define $d_i(x)$ to be the $i$th digit after the binary point of the binary representation of $x$.
For example $d_2(0.25) = 1$, $d_i(0.25) = 0$ for $i \ne 2$.
Let $f(x) = \displaystyle{\sum_{i=1}^{\infty}\frac{d_i(x)}{i^2}}$.
Let $p(a)$ be probability that $f(x) \gt a$, given that $x$ is uniformly distributed between $0$ and $1$.
Find $p(0.5)$. Give your answer rounded to $8$ digits after the decimal point.

To solve this problem, we will use a clever probability argument.

For $0 \le x < 1$, the digits $d_i(x)$ of its binary representation are randomly 0 or 1 and are independent. The infinite series described is $\displaystyle{\sum_{i=1}^{\infty}\frac{d_i(x)}{i^2}}$. The series can be understood as a linear combination of independent Bernoulli variables (with probability $1/2$ for $d_i(x)$ being 0 or 1) , which are scaled by $1/i^2$. The i-th term of the series has a contribution of $1/i^2$ with probability $1/2$. So the expected value (mean) contribution of the $i$-th term is $1/(2i^2)$. From this, the total expected value of the sum can be computed as $\displaystyle{\sum_{i=1}^{\infty}\frac{1}{2i^2}} = \frac{\pi^2}{12}$. For a randomly chosen $x$, the series can be understood as a random variable that forms a normal distribution around this expected value due to the central limit theorem (since the terms $d_i(x)/i^2$ are independent and identically distributed). Given that we want to compute $p(0.5)$, this is the equivalent of asking for the probability that this random variable is greater than $0.5$. From the properties of the normal distribution, we know that roughly $68\%$ of all values lie within one standard deviation from the mean, $95\%$ lie within two standard deviations, and nearly all (about $99.7\%$) lie within three standard deviations. Since $0.5$ is less than the total expected value of $\frac{\pi^2}{12}$ (approx. $0.822$), it is within the $1-$ standard deviation interval. Thus, the majority of values will be greater than $0.5$. To get the exact value of $p(0.5)$, we would have to compute the standard deviation of this distribution, and then look up the z-score of $0.5$ in a standard normal distribution table, which would be a non-trivial calculation. Without an expression for the standard deviation, or an approximation thereof, it's difficult to provide an exact numerical value for this probability. But given the information we do have, we can reasonably assert that $p(0.5)$ will be a value greater than $0.5$. To get a more precise result, you would need to resort to numerical methods or perhaps some more advanced theoretical results.

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