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Note: I'm not even sure how to best title this question, so if anyone has any ideas, please edit!

Consider six independent draws from a distribution defined by a cdf $F(x)$ over 0-1. Let’s call them $X1, X2 . . . X6$. I want to define the distribution for X1+X3, if I know the following information:

  • $X1 +X4 < X2 + X6$
  • $X3 < X4$
  • $X5 < X6$

It may help to think of the draws as being arranges in a tree structure. The image below is my best attempt to illustrate what we know (first row) and then what distribution I'm looking for (2nd row).

Note that I don't care specifically if $X3$ is the 3rd variable drawn. More accurately I might say that, given 2 draws from $F(x)$ for the first level of the tree, and then given 4 draws for the second level of the tree, label the draws $X1$ to $X6$ such that the conditions specified are true.

tree visualization

This question is similar to this question , except that sums take away independence. I am specifically interested in F(x) = x (uniform distribution) but an answer to the general case would be very interesting.

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By symmetry, $H(t)=P(X_{final}<t)=$
$4P(X_1+X_3<t,X_3<X_4,X_2+\max(X_5,X_6)>X_1+X_4)$
Let $Y=X_2+\max(X_5,X_6)$ and note that $Y$ is independent of $X_3,X_4,X_1$.
Suppose the support of the distribution is $(a,b)$, where $a,b$ can be infinite. The above probability becomes $$P(a<X_3<t,X_3<X_4<b,a<X_1<t-X_3,X_1+X_4<Y<b)\hbox{ }(1)$$. The distribution of $y$ is $$g(y)=\int_a^b 2f(x)f(z-x)F(z-x) dx \hbox{ } $$ Let $m(x_1,x_3,x_4,y)=f(x_1)f(x_3)f(x_4)g(y)$ be the joint probability. So the probability (1) becomes $$H(t)=\int_a^t dx_3 \int_{x_3}^b dx_4 \int_a^{t-x_3} dx_1 \int_{x_1+x_4}^b m dy\hbox{ }(2)$$ Now if $b$ is infinite, the above probability can be calculated directly. When both $a$ and $b$ are finite however, $g(y)$ is a piecewise function.

In the case of a location-scale family of distributions, such as the uniform distribution note the following: if $U_1$ has support $[0,1]$ and $U_2$ has support $[b,a+b]$, then $P(U_{2,final}<t)=P(U_{1,final}<(t-2b)/a)$ hence the general case results from $[0,1]$ case.

Let me show for the uniform distribution on [0,1].We get $g(y)=\begin{cases} g_1(y)=y^2& 0\leq y\leq 1 \\ g_2(y)=-y^2+2y& 1\leq y\leq 2 \end{cases}$

Now since we have a piecewise function, I had to consider cases $t\leq 1,t>1,X_1+X_4\leq 1,X_1+X_4>1$ so I got a bunch of integrals.
It turns out when $t\leq 1$
$H(t)=\int_0^t dx_3\int_0^{t-x_3} dx_1\int_{1-x_1}^{x_3} dx_4\int_{x_1+x_4}^1 g_1 dy+$ $\int_0^t dx_3\int_0^{t-x_3}dx_1\int_{1-x_1}^{x_3}dx_4\int_1^2 g_2 dy+$
$\int_0^t dx_3\int_0^{t-x_3}dx_1\int_{1-x_1}^1 dx_4\int_{x_1+x_4}^2 g_2 dy$

and when $t>1$, $H(t)=\int_0^1dx_3\int_0^{1-x_3}dx_1\int_{x_3}^{1-x_1}dx_4\int_{x_1+x_4}^1 g_1dy+$ $\int_0^1dx_3\int_0^{1-x_3}dx_1\int_{x_3}^{1-x_1}dx_4\int_1^2 g_2dy$+
$\int_0^{t-1}dx_3\int_{x_3}^1dx_4\int_{1-x_4}^1 dx_1\int_{x_1+x_4}^2 g_2dy$+ $\int_{t-1}^1 dx_3\int_{x_3}^1 dx_4\int_{1-x_4}^{t-x_3}dx_1\int_{x_1+x_4}^2 g_2dy$.

So using a CAS we obtain $H(t)=(1/15)*t^6-(1/6)*t^4-(8/9)*t^3+(11/6)*t^2,0\leq t<1$ $(2/45)*t^6-(1/3)*t^5+(2/3)*t^4+(8/9)*t^3-(16/3)*t^2+(112/15)*t-23/9$ $,1\leq t<2$
The mean of it is 139/210=.6619 .
So when $X$ is the uniform on $[b,a+b]$, the resulting cdf is $H((t-2b)/a)$, where $2b<t<2a+2b$. This has mean $2b+139/210*a$.
Example: the mean of the result when $X$ is uniform [-7,11] is -14+18*139/210=-2.09 .

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  • $\begingroup$ Wow. Thank you so much! I really appreciate your walking through a specific example as well as providing the general solution. Thank you! $\endgroup$
    – OctaviaQ
    Aug 2 '11 at 19:27

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