Ожидание произведения зависимых случайных величин при

Пусть и , . Каково ожидание как ?$X_1 \sim U[0,1]$$X_i \sim U[X_{i - 1}, 1]$$i = 2, 3,...$$X_1 X_2 \cdots X_n$$n \rightarrow \infty$

Ответ действительно ,$1/e$ как и предполагалось в предыдущих ответах, основанных на моделировании и конечных приближениях.

Решение легко достигается введением последовательности функций . Хотя мы могли бы немедленно перейти к этому шагу, он может показаться довольно загадочным. Первая часть этого решения объясняет, как можно приготовить эти f n ( t ) . Вторая часть показывает, как они используются для нахождения функционального уравнения, удовлетворяемого предельной функцией f ( t ) = lim n → ∞ f n ( t $f_n:[0,1]\to[0,1]$$f_n(t)$$f(t) = \lim_{n\to\infty}f_n(t)$, Третья часть отображает (обычные) вычисления, необходимые для решения этого функционального уравнения.

1. Мотивация

Мы можем прийти к этому, применив некоторые стандартные математические методы решения проблем. В этом случае, когда какая-то операция повторяется до бесконечности, предел будет существовать как фиксированная точка этой операции. Ключ, поэтому, должен идентифицировать операцию.

Сложность в том, что переход от к E [ X 1 X 2 ⋯ X n - 1 X n ] выглядит сложным. Проще рассматривать этот шаг как результат присоединения X 1 к переменным ( X 2 , … , X n ), а не присоединения X n к переменным ( X 1$E[X_1X_2\cdots X_{n-1}]$$E[X_1X_2\cdots X_{n-1}X_n]$$X_1$$(X_2, \ldots, X_n)$$X_n$ . Если бы мы рассматривали ( X 2 , … , X n ) как конструируемые, как описано в вопросе - с X 2, равномерно распределенным на [ 0 , 1 ] , X 3, условно равномерно распределенным на [ X 2 , 1 ] , и так далее - то введения X 1 будет вызывать каждый из последующего X I к$(X_1, X_2, \ldots, X_{n-1})$$(X_2, \ldots, X_n)$$X_2$$[0,1]$$X_3$$[X_2, 1]$$X_1$$X_i$shrink by a factor of $1-X_1$ towards the upper limit $1$. This reasoning leads naturally to the following construction.

As a preliminary matter, since it's a little simpler to shrink numbers towards $0$ than towards $1$, let $Y_i = 1-X_i$. Thus, $Y_1$ is uniformly distributed in $[0,1]$ and $Y_{i+1}$ is uniformly distributed in $[0, Y_i]$ conditional on $(Y_1, Y_2, \ldots, Y_i)$ for all $i=1, 2, 3, \ldots.$ We are interested in two things:

1. The limiting value of $E[X_1X_2\cdots X_n]=E[(1-Y_1)(1-Y_2)\cdots(1-Y_n)]$.

2. How these values behave when shrinking all the $Y_i$ uniformly towards $0$: that is, by scaling them all by some common factor $t$, $0 \le t \le 1$.

To this end, define

Clearly each $f_n$ is defined and continuous (infinitely differentiable, actually) for all real $t$. We will focus on their behavior for $t\in[0,1]$.

2. The Key Step

The following are obvious:

1. Each $f_n(t)$ is a monotonically decreasing function from $[0,1]$ to $[0,1]$.

2. $f_n(t) \gt f_{n+1}(t)$ for all $n$.

3. $f_n(0) = 1$ for all $n$.

4. $E(X_1X_2\cdots X_n) = f_n(1).$

These imply that $f(t) = \lim_{n\to\infty} f_n(t)$ exists for all $t\in[0,1]$ and $f(0)=1$.

Observe that, conditional on $Y_1$, the variable $Y_2/Y_1$ is uniform in $[0,1]$ and variables $Y_{i+1}/Y_1$ (conditional on all preceding variables) are uniform in $[0, Y_i/Y_1]$: that is, $(Y_2/Y_1, Y_3/Y_1, \ldots, Y_n/Y_1)$ satisfy precisely the conditions satisfied by $(Y_1, \ldots, Y_{n-1})$. Consequently

This is the recursive relationship we were looking for.

In the limit as $n\to \infty$ it must therefore be the case that for $Y$ uniformly distributed in $[0,1]$ independently of all the $Y_i$,

That is, $f$ must be a fixed point of the functional $\mathcal{L}$ for which

3. Calculation of the Solution

Clear the fraction $1/t$ by multiplying both sides by $t$. Because the right hand side is an integral, we may differentiate it with respect to $t$, giving

Equivalently, upon subtracting $f(t)$ and dividing both sides by $t$,

for $0 \lt t \le 1$. We may extend this by continuity to include $t=0$. With the initial condition (3) $f(0)=1$, the unique solution is

Consequently, by (4), the limiting expectation of $X_1X_2\cdots X_n$ is $f(1)=e^{-1} = 1/e$, QED.

Because Mathematica appears to be a popular tool for studying this problem, here is Mathematica code to compute and plot $f_n$ for small $n$. The plot of $f_1, f_2, f_3, f_4$ displays rapid convergence to $e^{-t}$ (shown as the black graph).

a = 0 <= t <= 1;
l[g_] := Function[{t}, (1/t) Integrate[(1 - x) g[x], {x, 0, t}, Assumptions -> a]];
f = Evaluate@Through[NestList[l, 1 - #/2 &, 3][t]]
Plot[f, {t,0,1}]

Update

I think it's a safe bet that the answer is $1/e$. I ran the integrals for the expected value from $n=2$ to $n=100$ using Mathematica and with $n=100$ I got

0.367879441171442321595523770161567628159853507344458757185018968311538556667710938369307469618599737077005261635286940285462842065735614

(to 100 decimal places). The reciprocal of that value is

2.718281828459045235360287471351873636852026081893477137766637293458245150821149822195768231483133554

The difference with that reciprocal and $e$ is

-7.88860905221011806482437200330334265831479532397772375613947042032873*10^-31

I think that's too close, dare I say, to be a rational coincidence.

The Mathematica code follows:

Do[
 x = Table[ToExpression["x" <> ToString[i]], {i, n}];
 integrand = Expand[Simplify[(x[[n - 1]]/(1 - x[[n - 1]])) Integrate[x[[n]], {x[[n]], x[[n - 1]], 1}]]];
 Do[
   integrand = Expand[Simplify[x[[i - 1]] Integrate[integrand, {x[[i]], x[[i - 1]], 1}]/(1 - x[[i - 1]])]],
   {i, n - 1, 2, -1}]
  Print[{n, N[Integrate[integrand, {x1, 0, 1}], 100]}],
 {n, 2, 100}]

End of update

This is more of an extended comment than an answer.

If we go a brute force route by determining the expected value for several values of $n$, maybe someone will recognize a pattern and then be able to take a limit.

For $n=5$, we have the expected value of the product being

which is 96547/259200 or approximately 0.3724807098765432.

If we drop the integral from 0 to 1, we have a polynomial in $x_1$ with the following results for $n=1$ to $n=6$ (and I've dropped the subscript to make things a bit easier to read):

$x$

$(x + x^2)/2$

$(5x + 5x^2 + 2x^3)/12$

$(28x + 28x^2 + 13x^3 + 3x^4)/72$

$(1631x + 1631x^2 + 791x^3 + 231x^4 + 36x^5)/4320$

$(96547x + 96547x^2 + 47617x^3 + 14997x^4 + 3132x^5 + 360x^6)/259200$

If someone recognizes the form of the integer coefficients, then maybe a limit as $n\rightarrow\infty$ can be determined (after performing the integration from 0 to 1 that was removed to show the underlying polynomial).

Nice question. Just as a quick comment, I would note that:

• $X_n$ will converge to 1 rapidly, so for Monte Carlo checking, setting $n = 1000$ will more than do the trick.

• If $Z_n = X_1 X_2 \dots X_n$, then by Monte Carlo simulation, as $n \rightarrow \infty$, $E[Z_n] \approx 0.367$.

• The following diagram compares the simulated Monte Carlo pdf of $Z_n$ to a Power Function distribution [ i.e. a Beta(a,1) pdf) ]

... here with parameter $a=0.57$:

_{(source: tri.org.au)}

where:

• the blue curve denotes the Monte Carlo 'empirical' pdf of $Z_n$
• the red dashed curve is a PowerFunction pdf.

The fit appears pretty good.

Code

Here are 1 million pseudorandom drawings of the product $Z_n$ (say with $n = 1000$), here using Mathematica:

data = Table[Times @@ NestList[RandomReal[{#, 1}] &, RandomReal[], 1000], {10^6}];

The sample mean is:

 Mean[data]

0.367657

Purely intuitively, and based on Rusty's other answer, I think the answer should be something like this:

n = 1:1000
x = (1 + (n^2 - 1)/(n^2)) / 2
prod(x)

Which gives us 0.3583668. For each $X$, you are splitting the $(a,1)$ range in half, where $a$ starts out at $0$. So it's a product of $1/2, (1 + 3/4)/2, (1 + 8/9)/2$, etc.

This is just intuition.

The problem with Rusty's answer is that U[1] is identical in every single simulation. The simulations are not independent. A fix for this is easy. Move the line with U[1] = runif(1,0,1) to inside the first loop. The result is:

set.seed(3)    #Just for reproducibility of my solution

n = 1000    #Number of random variables
S = 1000    #Number of Monte Carlo samples

Z = rep(NA,S)
U = rep(NA,n)

for(j in 1:S){
    U[1] = runif(1,0,1)
    for(i in 2:n){
        U[i] = runif(1,U[i-1],1)
    }
    Z[j] = prod(U)
}

mean(Z)

This gives 0.3545284.