Stopping time using a sequences of i.i.d Uniform(0,1) random variables

Question

Let $(U_1,U_2,...) , (V_1,V_2,...)$ be two independent sequences of i.i.d. Uniform (0, 1) random variables. Define the stopping time $N = \min\left(n\geqslant 1\mid U_n \leqslant V^2_n\right)$.

Obtain $P(N = n)$ and $P(V_N \leqslant v)$ for $n = 1,2,...,1\geqslant v \geqslant$0.

I know that I should use conditioning in order to get the probability.

I also know that I have to check if $U_1 \leqslant V_1$ then $N=1$

Answer 1

Here are some hints.

1. The event $\{N=n\}$ can be written as $A_n^c\cap\bigcap_{i=1}^{n-1}A_i$, where $A_i=\{U_i\leqslant V_i^2\}$.
2. Show that the collection of events $A_n^c,A_1,\dots,A_{n-1}$ is independent.
3. Compute the probability of $A_i$.

For the second part, start from $$ P\left(V_N\leqslant v\right)=\sum_{n\geqslant 1}P\left(V_n\leqslant v,N=n\right). $$ Then use the previous decomposition of $\{N=n\}$ and use independence to get the value of $P\left(V_n\leqslant v,N=n\right)$ for all $n$.

Answer 2

For $0<v<1$ we have $$\mathbb P(V_1^2\leqslant v) = \mathbb P(V_1\leqslant \sqrt v) = \sqrt v$$ and hence $V_1$ has density $f_{V_1}(v)=\frac12 v^{-\frac12}\mathsf 1_{(0,1)}(v)$.

For positive integers $n$ we have $$ \{N=n\} = \{U_n\leqslant V_n^2\}\cap\bigcap_{i=1}^{n-1}\{U_i>V_i^2\}. $$ We compute \begin{align} \mathbb P(U_1\leqslant V_1^2) &= \iint_{\mathbb R^2} f_{U_1,V_1}(u,v)\ \mathsf d(u\times v)\\ &= \int_0^1\int_0^v\frac12 v^{-\frac12}\ \mathsf du\ \mathsf dv\\ &= \frac13. \end{align} it follows that $$ \mathbb P(N=n) = \left(\frac23\right)^{n-1}\frac13,\ n=1,2,\ldots, $$ so that $N$ has geometric distribution with parameter $\frac13$. Finally, $V_N$ just has the same distribution as $V_1$.

Answer 3

To solve with minimal calculation, and focusing on your comment "I know that I should use conditioning in order to get the probability".

It is common to try to do "first step analysis" in these sorts of problems. Letting $A$ be the event that $\{V_1^2 \gt U_1\}$ (ignoring the zero probability set where $V_1 = U_1$), introduce the indicator (Bernouli) random variable $\mathbb I_A$:

$p = E\Big[\mathbb I_A\Big] = E\Big[E\big[\mathbb I_A\big \vert U_1\big]\Big]$
and in particular, for $x \in[0,1]$
$E\big[\mathbb I_A\big \vert U_1 = x\big] = Pr(V_1^2 \gt U_1 =x) = Pr(V_1 \gt \sqrt{x}) = 1 - \sqrt{x}$
which is given by the complementary CDF of $V_1$. Each $x \in [0,1]$ has density 1, which gives

$p = E\Big[\mathbb I_A\Big] = E\Big[E\big[\mathbb I_A\big \vert U_1\big]\Big] = \int_{0}^{1} (1-\sqrt{x})dx = 1-\int_{0}^{1} \sqrt{x}dx =\frac{1}{3}$

This is a Bernouli process so it is immediate that $N$ has a geometric distribution with success parameter $p$ and $P(V_N <=v) = P(V_1 <=v) = p$