Uniform Fraction

Let \(X,Y \sim\) \(U\left( 0,1 \right)\),

\[ A = \left\{ \text{The closest interget to } \frac{X}{Y} \text{is even.} \right\} \]

What is \(\mathbb{P}\left( A \right)\)?

Let's condition on \(\left\{ Y=y \right\}\).

Let \(A_k = \left\{ \frac{X}{Y} \text{ closest to } k \right\}\), \(k \in \mathbb{N}_0\).

\(\frac{X}{Y}\) closest to \(k\) is \(\iff\) \(X \in \left[(k-\frac{1}{2})y, (k+\frac{1}{2})y \right] \coloneqq I_k\)

Let \(q\) be the largest even integer such that the \(q\)th \(y\) interval is in \(\left[ 0,1 \right]\).

\[ \left( q-\frac{1}{2} \right)y<1 \]

and

\[ \left( q+2-\frac{1}{2} \right)y > 1 \iff \left( q+\frac{3}{2} \right)y > 1 \]

If \(q=0\) then \(1<y<\frac{2}{3}\). The \(y=\frac{2}{3}\) case is when \(I_2\) moves into \(\left[ 0,1 \right]\).

\[ \begin{array}{cccc} I_0 & &I_1 \\ | & )(& &) \\ 0 &\frac{1}{3}=\frac{y}{2} & & 1 \end{array} \]

In this case \(\mathbb{P}\left( A | Y=y\right) = \frac{y}{2}\)

If even \(q >0\) then \(\frac{1}{\left( q+\frac{3}{2} \right)}<y<\frac{1}{\left( q-\frac{1}{2} \right)}\).

This interval is fully in \(\left[ 0,1 \right] \iff \left( q+\frac{1}{2} \right)y<1\).

In this case there is \(\frac{y}{2} + \frac{q}{2}y\) even area in \(\left[ 0,1 \right]\).

If the interval is not fully in \(\left[ 0,1 \right] \iff \left( q+\frac{1}{2} \right)y>1\).

In this case the even area is

\[ \frac{y}{2} + \left( \frac{q}{2} - 1 \right)y + 1-\left( q-\frac{1}{2} \right)y \\ = 1 + y\left( 1/2+q/2-1-q+1/2 \right) \\ = 1 - y \frac{q}{2} \]

So

\[ \begin{align*} \frac{1}{ q+\frac{3}{2} }<y<\frac{1}{ q+\frac{1}{2} } &\iff \mathbb{P}\left( A | Y=y \right) = \left( \frac{q+1}{2} \right)y \\ \\ \frac{1}{ q+\frac{1}{2} }<y<\frac{1}{ q-\frac{1}{2} } &\iff \mathbb{P}\left( A | Y=y \right) = 1-y\frac{q}{2} \end{align*} \]

Therefore,

\[ \begin{align*} \mathbb{P} \left( A \right) &= \int_{2/3}^{1} \frac{y}{2} \, dy + \sum_{n=1}^{\infty} \left[ \int_{\frac{1}{2n+1/2}}^{\frac{1}{2n-1/2}} 1-ny \, dy + \int_{\frac{1}{2n+3/2}}^{\frac{1}{2n+1/2}} \frac{2n+1}{2}y \, dy \right] \\ &= \frac{5}{36} + \sum_{n=1}^{\infty} \left[ y - n \frac{y^2}{2} \right]_{\frac{1}{2n+1/2}}^{\frac{1}{2n-1/2}} + \left[ \frac{2n+1}{4}y^2 \right]_{\frac{1}{2n+3/2}}^{\frac{1}{2n+1/2}} \\ &= \frac{5}{36} + \sum_{n=1}^{\infty} \left[ \begin{aligned} & \frac{1}{2n-1/2} - \frac{1}{2n+1/2} -\frac{n}{2}\left( \frac{1}{(2n-1/2)^2} - \frac{1}{(2n+1/2)^2} \right) \\ & + \frac{2n+1}{4} \left( \frac{1}{(2n+1/2)^2} - \frac{1}{(2n+3/2)^2} \right) \end{aligned} \right] \\ &= \frac{5}{36} + \sum_{n=1}^{\infty} \left[ \begin{aligned} & \frac{1}{2n-1/2} - \frac{1}{2n+1/2} -\frac{n}{2}\left( \frac{4n}{(2n-1/2)^2 (2n+1/2)^2} \right) \\ & + \frac{2n+1}{4} \left( \frac{6n + 9/4 - 2n - 1/4}{(2n+1/2)^2 (2n+3/2)^2} \right) \end{aligned} \right] \\ &= \frac{5}{36} + \sum_{n=1}^{\infty} \left[ \begin{aligned} & \frac{1}{2n-1/2} - \frac{1}{2n+1/2} - \frac{2n^2}{(2n-1/2)^2 (2n+1/2)^2} \\ & + \frac{(2n+1)^2}{2} \frac{1}{(2n+1/2)^2 (2n+3/2)^2} \end{aligned} \right] \\ &= \frac{5}{36} + \sum_{n=1}^{\infty} \left[ \begin{aligned} & \frac{1}{(2n-1/2)^2 (2n+1/2)^2} - \frac{2n^2}{(2n-1/2)^2 (2n+1/2)^2} \\ & + \frac{(2n+1)^2}{2} \frac{1}{(2n+1/2)^2 (2n+3/2)^2} \end{aligned} \right] \\ &= \frac{5}{36} + \sum_{n=1}^{\infty} \left[ \begin{aligned} & \frac{16}{(4n-1)^2 (4n+1)^2} - \frac{32n^2}{(4n-1)^2 (4n+1)^2} \\ & + (2n+1)^2 \frac{8}{(4n+1)^2 (4n+3)^2} \end{aligned} \right] \\ &= \frac{5}{36} + \sum_{n=1}^{\infty} \left[ \begin{aligned} & \frac{-9}{2} \frac{1}{4n-1} + \frac{7}{2} \frac{1}{(4n-1)^2} + \frac{9}{2} \frac{1}{4n+1} + \frac{7}{2} \frac{1}{(4n+1)^2} \\ & + (2n+1)^2 \frac{8}{(4n+1)^2 (4n+3)^2} \end{aligned} \right] \end{align*} \]

Alt Way

\[ \begin{align*} \mathbb{P} \left( A \right) &= \int_{2/3}^{1} \frac{y}{2} \, dy + \sum_{n=1}^{\infty} \left[ \int_{\frac{1}{2n+1/2}}^{\frac{1}{2n-1/2}} 1-ny \, dy + \int_{\frac{1}{2n+3/2}}^{\frac{1}{2n+1/2}} \frac{2n+1}{2}y \, dy \right] \\ &= \frac{5}{36} + \sum_{n=1}^{\infty} \left[ y - n \frac{y^2}{2} \right]_{\frac{1}{2n+1/2}}^{\frac{1}{2n-1/2}} + \left[ \frac{2n+1}{4}y^2 \right]_{\frac{1}{2n+3/2}}^{\frac{1}{2n+1/2}} \\ &= \frac{5}{36} + \sum_{n=1}^{\infty} \left[ \begin{aligned} & \frac{1}{2n-1/2} - \frac{1}{2n+1/2} -\frac{n}{2}\left( \frac{1}{(2n-1/2)^2} - \frac{1}{(2n+1/2)^2} \right) \\ & + \frac{2n+1}{4} \left( \frac{1}{(2n+1/2)^2} - \frac{1}{(2n+3/2)^2} \right) \end{aligned} \right] \\ &= \frac{5}{36} + \sum_{n=1}^{\infty} \left[ \begin{aligned} & \frac{2}{4n-1} - \frac{2}{4n+1} - \frac{2n}{(4n-1)^2} + \frac{2n}{(4n+1)^2} \\ & + \frac{2n+1}{(4n+1)^2} - \frac{2n+1}{(4n+3)^2} \end{aligned} \right] \\ &= \frac{5}{36} + \sum_{n=1}^{\infty} \left[ \begin{aligned} & \frac{2}{4n-1} - \frac{2}{4n+1} \\ & - 1/2 \frac{1}{(4n-1)^2} -1/2 \frac{1}{4n-1} + -1/2 \frac{1}{(4n+1)^2} +1/2 \frac{1}{4n+1} \\ & + 1/2 \frac{1}{(4n+1)^2} + 1/2 \frac{1}{4n+1} + 1/2 \frac{1}{(4n+3)^2} - 1/2 \frac{1}{4n+3} \end{aligned} \right] \\ &= \frac{5}{36} + \sum_{n=1}^{\infty} \left[ \begin{aligned} & 3/2 \frac{1}{4n-1} -1/2 \frac{1}{(4n-1)^2} - \frac{1}{4n+1} + 1/2 \frac{1}{(4n+3)^2} - 1/2 \frac{1}{4n+3} \end{aligned} \right] \\ &= \frac{5}{36} + \sum_{n=1}^{\infty} \left[ \begin{aligned} & 1/2 \frac{1}{4n-1} - 1/2 \frac{1}{4n+3} -1/2 \frac{1}{(4n-1)^2} + 1/2 \frac{1}{(4n+3)^2} \\ & + \frac{1}{4n-1} - \frac{1}{4n+1} \end{aligned} \right] \end{align*} \]