2 Fixed typo and added more steps to the derivation.

This problem has delighted me since I first encountered it before college. I wrote up a generalization for a less mathematical audience in a poker forum.

One way to look at the original is that if f(x) = E(N(X)), then $f(x) = 1 + \int_{x-1}^x f(t) dt$, satisfying the initial condition that $f(x) = 0$ on $(-1,0)$. $f$ is 1 more than the average of $f$ on the previous interval.

We can get rid of the constant by using $g(x) = f(x) - 2x$ which satisfies $g(x) = \int_{x-1}^x g(t) dt$. $g$ is equal to its average on the previous interval. Now the initial condition becomes nontrivial: $g(x) = -2x$ on $(-1,0)$, and it is from this $-2x$ that the asymptotic value of $2/3$ comes.

$g$ is asymptotically constant, and we can guess from geometry and verify that the weighted average $\int_0^1 2t~g(x+t)dt$ is independent of $x$, hence equal to the asymptotic value of $g$. The value at $x=-1$ is $\int_0^1 4t(1-t)dt = \frac23$, so that is the asymptotic value of $g$.

The same technique works for more general continuous distributions supported on $\mathbb R^+$ than the uniform distribution, and the answer turns out to be remarkably simple. Let's find an analogous conserved quantity.

Let $\alpha$ be the density function. Let the $n$th moment be $\alpha_n$. We'll want $\alpha_2$ to exist.

Let $f(x)$ be the average index of the first partial sum which is at least $x$ of IID random variables with density $\alpha$. $f(x) = 1 + \int_{-\infty}^x f(t) \alpha(x-t) dt$, and $f(x) = 0$ for $x\lt0$. (We could let the lower limit be 0, too.)

Let $g(x) = f(x) - x/\alpha_1$. Then $g(x) = \int_{-\infty}^xg(t)\alpha(x-t)dt$, with $g(x) = -x/\alpha_1$ on $\mathbb R^-$.

We'll choose $h$ so that $H(x) = \int_{-\infty}^x g(xg(t) h(x-t) dt$ is constant.

$0 = H'(x) = g(x)h(0) + \int_{\infty}^x g(t) h'(x-t)dt.$

This is satisfied when integral looks like the integral equation for $g$ if we choose $h'(x) = c\alpha(x)$. $c=-1$ satisfies the equation. So, if $h(x) = \int_x^\infty \alpha(t)dt$. alpha(t)dt$then$H(x)$is constant. Let the asymptotic value of$g$be$v$. Then the value of$H(x)$is both$H(\infty) = v~\alpha_1$and$H(0) = \int_{-\infty}^0 g(t) h(0-t)dtH(0) = \alpha_1^{-1} 1/\alpha_1 \int_0^\infty y~h(y) dyH(0) = 1/(2\alpha_1) \int_0^\infty y^2 ~\alpha(y)dy~~$(by parts)$H(0) = \alpha_2 / (2 \alpha_1)v~ \alpha_1 = \alpha_2 / (2 \alpha_1)v = \alpha_2 / (2 \alpha_1^2)$. So,$f(x)$is asymptotic to$x/\alpha_1 + \alpha_2/(2 \alpha_1^2)$. For the original interval, uniform distribution on [0,1],$\alpha_1 = \frac12$and$\alpha_2 = \frac13$. frac13$, so $f(x) = 2x + \frac23 + o(1)$.

As a check, an exponential distribution with mean 1 has second moment 2, and we get that $f(x)$ is asymptotic to $x+1$. In fact, in that case, $f(x) = x+1$ on $\mathbb R^+$. If you have memoryless light bulbs with average life 1, then at time $x$, an average of $x$ bulbs have burned out, and you are on the $x+1$st bulb.

1

This problem has delighted me since I first encountered it before college. I wrote up a generalization for a less mathematical audience in a poker forum.

One way to look at the original is that if f(x) = E(N(X)), then $f(x) = 1 + \int_{x-1}^x f(t) dt$, satisfying the initial condition that $f(x) = 0$ on $(-1,0)$. $f$ is 1 more than the average of $f$ on the previous interval.

We can get rid of the constant by using $g(x) = f(x) - 2x$ which satisfies $g(x) = \int_{x-1}^x g(t) dt$. $g$ is equal to its average on the previous interval. Now the initial condition becomes nontrivial: $g(x) = -2x$ on $(-1,0)$, and it is from this $-2x$ that the asymptotic value of $2/3$ comes.

$g$ is asymptotically constant, and we can guess from geometry and verify that the weighted average $\int_0^1 2t~g(x+t)dt$ is independent of $x$, hence equal to the asymptotic value of $g$. The value at $x=-1$ is $\int_0^1 4t(1-t)dt = \frac23$, so that is the asymptotic value of $g$.

The same technique works for more general continuous distributions supported on $\mathbb R^+$ than the uniform distribution, and the answer turns out to be remarkably simple. Let's find an analogous conserved quantity.

Let $\alpha$ be the density function. Let the $n$th moment be $\alpha_n$. We'll want $\alpha_2$ to exist.

Let $f(x)$ be the average index of the first partial sum which is at least $x$ of IID random variables with density $\alpha$. $f(x) = 1 + \int_{-\infty}^x f(t) \alpha(x-t) dt$, and $f(x) = 0$ for $x\lt0$. (We could let the lower limit be 0, too.)

Let $g(x) = f(x) - x/\alpha_1$. Then $g(x) = \int_{-\infty}^xg(t)\alpha(x-t)dt$, with $g(x) = -x/\alpha_1$ on $\mathbb R^-$.

We'll choose $h$ so that $H(x) = \int_{-\infty}^x g(x) h(x-t) dt$ is constant. This is satisfied when $h(x) = \int_x^\infty \alpha(t)dt$.

Let the asymptotic value of $g$ be $v$. Then the value of $H(x)$ is both $H(\infty) = v~\alpha_1$ and

$H(0) = \int_{-\infty}^0 g(t) h(0-t)dt$

$H(0) = \alpha_1^{-1} \int_0^\infty y~h(y) dy$

$H(0) = \alpha_2 / (2 \alpha_1)$

$v~ \alpha_1 = \alpha_2 / (2 \alpha_1)$

$v = \alpha_2 / (2 \alpha_1^2)$.

So, $f(x)$ is asymptotic to $x/\alpha_1 + \alpha_2/(2 \alpha_1^2)$.

For the original interval, $\alpha_1 = \frac12$ and $\alpha_2 = \frac13$.

As a check, an exponential distribution with mean 1 has second moment 2, and we get that $f(x)$ is asymptotic to $x+1$. In fact, in that case, $f(x) = x+1$ on $\mathbb R^+$. If you have memoryless light bulbs with average life 1, then at time $x$, an average of $x$ bulbs have burned out, and you are on the $x+1$st bulb.