space of probability for an infinite sequence of coin flips

Question

Let $(\Omega,\mathcal{F},P)$ be the usual space of probability for an infinite sequence of coin flips. Let $X$ be the function from $\Omega$ to $(\mathbb{R}^2,\mathcal{B})$ defined by $$ X((w_1,w_2,\ldots))=(0.w_1w_3w_5\dots two,0.w_2w_4w_6\dots two ) $$

($two = binary$)

Here I understand that the distribution of $X$ is a generalization of the area, restricted to the unit square $[0,1] \times [0,1]$, i.e, $$ Q([a,b]\times[c,d]) = (b-a)(d-c) = area([a,b]\times[c,d]) $$

For the random variable $X=(X_1,X_2)$: How can I calculate the probability that $X_1+X_2 \geq 1/2$, well I first try to calculate $$ P(X_1+X_2\geq 1/2) = P (\{ w: X(w_1)+X(w_2) \geq 1/2\}) $$

but now I don't know how to continue calculating this probability if I have that $$ X((w_1,w_2,\ldots))=(0.w_1w_3w_5\dots two,0.w_2w_4w_6\dots two ) $$ because we have only $w_1$ and $w_2$ then Is it true that

$$ X((w_1,w_2))=(0.w_1,0.w_2) ? $$ Someone could help me calculating this probability in this space please...

Thanks for your time and help.

Answer 1

$X_1 = 0.\omega_1\omega_3\omega_5\ldots_{\mathsf{two}}$ which is to say a binary fraction generated by the digits $\omega_1,\omega_3,\omega_5$, and so on.   Each of which is either $0$ or $1$ as assigned by an independent coin flip.

$X_1$ is thus a continuous uniformly distributed random variable over the support $[0;1]\cap \Bbb Q$.

$X_2$ is likewise defined, and independently so from $X_1$. (Since it it defined by the infinite set of 'even' indexed coin flips.)

Hence $X(\omega) ~{= (X_1(\omega_1,\omega_3,\omega_5,\ldots),X_2(\omega_2,\omega_4,\omega_6,\ldots)) \\ = (0.\omega_1\omega_3\omega_5\ldots_{\mathsf{two}},0.\omega_2\omega_4\omega_6\ldots_{\mathsf{two}})}$

The probability you want is $\mathsf P(\{(\omega_1,\omega_2,\ldots): 0.\omega_1\omega_3\ldots_{\mathsf {two}}+0.\omega_2\omega_4\ldots_{\mathsf {two}}\geq {0.1}_{\mathsf{two}}\}) \\ = \mathsf P(0.\omega_1\omega_3\ldots_{\mathsf {two}}+0.\omega_2\omega_4\ldots_{\mathsf {two}}\geq {0.1}_{\mathsf{two}}) \\ = \mathsf P(\{\omega: X_1(\omega_{\sf odd})+X_2(\omega_{\sf even})\geq 1/2\})\\=\mathsf P(X_1+X_2\geq 1/2)\\\ldots\text{ and such}$

This is easy to find by using what is know of uniform continuous distributions, but you could do it knowing that the probability equals:

$${\quad{\mathsf P(\omega_1=1)+\mathsf P(\omega_1=0)\mathsf P(\omega_2=1)\\+\mathsf P(\omega_1=0)\mathsf P(\omega_2=0)\,\mathsf P(0.\omega_3\omega_5\ldots_{\textsf{two}}+0.\omega_4\omega_6\ldots_{\textsf{two}}\geq 1_{\textsf{two}})} \\ ~\vdots\\ = \tfrac 12+\tfrac 14+\tfrac 18\\ = \tfrac 78}$$

You just have to fill in the missing steps.

Answer 2

SInce each of $X_1,X_2$ can be considered a uniformly distributed random variate on $[0,1]$, and they are not correlated, what you are looking for is the probability that the sum of two such variates exceeds $\frac12$.

Geometrically, that corresponds to the area of the unit square lying above the line $x+y=\frac12$. Since the triangle defined by $(0,0),(\frac12,0),(0\frac12)$ has area $\frac18$ the probability you are looking for is $$ 1-\frac18 = \frac78 $$