Expected value of product of 2 random variables from a uniform closure

Question

I am interested in finding an approximate expression as a function of 'n' (for a large 'n') that represents the expected value of product of 2 random variables from a uniform closure.

i.e., Find an approximate function $$ f(n) = E( V_1.V_2)$$ where

$$ V_j = \frac{U_j (0,1)}{\sum_{i=1}^n U_i (0,1)} $$

Answer 1

Since

$$ V_1V_2 = \frac {U_1U_2} {\displaystyle \left(\sum_{i=1}^n U_i\right)^2} = \frac {U_1U_2} {\displaystyle n^2 \left(\frac {1} {n} \sum_{i=1}^n U_i\right)^2}$$

By Law of Large Number, the sample mean converge to the true mean, so for large $n$, we expect

$$ E[V_1V_2] = E\left[\frac {U_1U_2} {\displaystyle n^2 \left(\frac {1} {n} \sum_{i=1}^n U_i\right)^2} \right] \approx E\left[\frac {U_1U_2} {\displaystyle n^2 E[U_1]^2 }\right] = \frac {E[U_1]E[U_2]} {n^2 E[U_1]^2} = \frac {1} {n^2}$$

Edit: I have run the following VBA code several time, the coefficient is close to $1$, although it is slightly downward biased:

Sub RandomTest()
Const m = 40
Const n = 100000
Dim i As Long
Dim j As Long
Dim U1 As Double
Dim U2 As Double
Dim sum As Double
Dim total As Double

total = 0
For i = 1 To n
    U1 = Rnd()
    U2 = Rnd()
    sum = U1 + U2

    For j = 3 To m
        sum = sum + Rnd()
    Next j
    total = total + U1 * U2 / sum ^ 2
Next i
MsgBox total / n * m ^ 2`
End Sub

And actually the approximation only holds for a sufficiently large $n$, so in general we do not expect it equal to it exactly for $n = 4$.

Let $\displaystyle X = U_1U_2, Y = \sum_{i=1}^n U_i$. Then $\displaystyle V_1V_2 = \frac {X} {Y^2}$ and we can do a further analysis using the Taylor Expansion around the mean $(E[X], E[Y])$, up to second order:

$$ \begin{align} E[V_1V_2] &= E\left[\frac {X} {Y^2} \right] \\ &\approx E\left[\frac {E[X]} {E[Y]^2} - (X - E[X])(Y - E[Y]) \frac {2} {E[Y]^3} + \frac {1} {2} (Y - E[Y])^2 \frac {6E[X]} {E[Y]^4}\right] \\ &= \frac {E[X]} {E[Y]^2} - \frac {2Cov[X, Y]} {E[Y]^3} + \frac {3E[X]Var[Y]} {E[Y]^4} \end{align}$$

So now we just compute the moments and plug in:

$$ E[X] = \frac {1} {4}, E[Y] = \frac {n} {2}, Var[Y] = \frac {n} {12}$$

A little bit more computation for the covariance term:

$$ \begin{align} Cov[X, Y] &= Cov[U_1U_2, U_1] + Cov[U_1U_2, U_2] \\ &= 2Cov[U_1U_2, U_1] \\ &= 2Cov[E[U_1U_2 \mid U_1], E[U_1 \mid U_1]] + 2E[Cov[U_1U_2, U_1 \mid U_1]] \\ &= 2Cov[U_1E[U_2], U_1] + 0 \\ &= Var[U_1] \\ &= \frac {1} {12}\end {align}$$

As a result,

$$ \begin{align} E[V_1V_2] &\approx \frac {1/4} {(n/2)^2} - \frac {2(1/12)} {(n/2)^3} + \frac{3(1/4)(n/12)} {(n/2)^4} \\ & = \frac {1} {n^2} - \frac {4} {3n^3} + \frac {1} {n^3} \\ & = \frac {1} {n^2} - \frac {1} {3n^3} \\ \end{align}$$

And this explain the slightly downward bias observed