I have a set of independent random variables $\{A_1, A_2, B_1, B_2\}$. All of them have the same distribution function $F(x)$. I want to find distribution function of a variable $C$, where $C=max(A_1 + B_1, max(A_1, A_2) + B_2)$.

To find CDF of $max(A_1, A_2)$ I can multiply CDF's of $A_1$ and $A_2$.

To find $max(A_1, A_2) + B_2$ I can find convolution of CDF's of both terms.

But now I'm struggling to find a CDF of the terms of the outer max, because the terms of the outer max are not independent.

Could you help me, please, solving this problem? Or give a good reference for that?

If it is required, we can make an assumption of a particular distribution function for the variables $A_1, A_2, B_1, B_2$.


I made some transformations:

Consider all possible outcomes:

\begin{equation} C=\begin{cases} A_1 + B_1, &\text{if } &A_1+B_1>A_1+B_2 \wedge A_1+B_1>A_2 + B_2 & (1)\\ A_1 + B_2, &\text{if } &A_1+B_2>A_1+B_1 \wedge A_1+B_2>A_2 + B_2 & (2)\\ A_2 + B_2, &\text{if } &A_2+B_2>A_1+B_1 \wedge A_2+B_2>A_1 + B_2 & (3) \end{cases} \end{equation}

Now the expected value of C is

$$ \begin{align} E[C] = &P(A_1 + B_1 = max)\cdot E[A_1 + B_1 \mid A_1 + B_1 = max] + \\ &P(A_1 + B_2 = max)\cdot E[A_1 + B_2 \mid A_1 + B_2 = max] + \\ &P(A_2 + B_2 = max)\cdot E[A_2 + B_2 \mid A_2 + B_2 = max] \end{align} $$

Now consider probabilities for each term $(1)$, $(2)$, and $(3)$ separately.

$$ \require{cancel} \begin{align} P(A_1 + B_2 = max) &= P(\cancel{A_1}+B_2>\cancel{A_1}+B_1 \wedge A_1 + \cancel{B_2} > A_2 + \cancel{B_2}) \\ &= P(B_2 > B_1)\cdot P(A_1 > A_2) \\ &= 0.5 \cdot 0.5 = 0.25 \end{align} $$

$P(B_2>B_1) = 0.5$, because $B_2$ and $B_1$ have the same distribution.

$$ \begin{align} P(A_1 + B_1 = max) &= P(\cancel{A_1}+B_1>\cancel{A_1}+B_2\wedge A_1+B_1>A_2+B_2) \\ &= P(B_1 > B_2)\cdot P(A_1 + B_1>A_2+B_2) \\ &= P(A_1+B_1>A_2+B_2\mid B_1>B_2\wedge A_1>A_2)P(A_1>A_2)P(B_1>B_2) + \\ &\phantom{=} + P(A_1+B_1>A_2+B_2\mid B_1>B_2\wedge A_1<A_2)P(A_1<A_2)P(B_1>B_2) \\ &= 1\cdot 0.5 \cdot 0.5 + P((B_1-B_2)>(A_1-A_2))\cdot 0.25 \\ &= 0.25 + 0.5*0.25 = 0.375 \end{align} $$

$P((B_1-B_2)>(A_1-A_2)) = 0.5$, because convolution of the same distributions leads to the same convoluted distribution function.

Putting everything in $E[C]$:

$$ \begin{align} E[C] = 0.375\cdot &E[A_1 + B_1 \mid A_1 + B_1 = max] + \\ 0.25\cdot &E[A_1 + B_2 \mid A_1 + B_2 = max] + \\ 0.375\cdot &E[A_2 + B_2 \mid A_2 + B_2 = max] \end{align} $$

But at the moment I don't know how to find expected values of the terms comprising $E[C]$, and I don't know if this is helpful at all.

  • $\begingroup$ Just as a matter of notation, it is common convention to use CAPITALS for random variables, so $C$ not $c$. I would also ponder as to why you are using $a_1$, $a_2$, $b_1$ etc, if they all have the same distribution function and are independent. Why the subscripts? What are you trying to convey by this notation? Why not simply ${X_1, X_2, X_3, X_4}$, or ${X,Y,Z,W}$ for the 4 variables? $\endgroup$ – wolfies Sep 2 '15 at 8:09
  • $\begingroup$ Thanks for pointing to capitals. The reason why I distinguish a's and b's is mostly because in the terms of the outer max a's are on the left and b's are on the right. And this problem is a simplification of a bigger problem, which I'm trying to solve. There can be more A's, more B's, and more cascades (C's D's and so on). $\endgroup$ – mcsim Sep 2 '15 at 8:12
  • 1
    $\begingroup$ Nice question. A more symmetrical formulation is $P(C<x)=P(A+B<x,A+A'<x,A'+B'<x)$ hence $P(C<x)=E(F(x-A)F(x-A')\mathbf 1_{A+A'<x})$... which does not help much, I am afraid. $\endgroup$ – Did Sep 2 '15 at 9:07
  • $\begingroup$ @Did What $\mathbf{1}_{A+A'}$ notation means? $\endgroup$ – mcsim Sep 3 '15 at 8:12
  • $\begingroup$ Nothing, but $\mathbf 1_{A+A'<x}=1$ if $A+A'<x$ and $\mathbf 1_{A+A'<x}=0$ otherwise. $\endgroup$ – Did Sep 3 '15 at 8:15

If required, we can make an assumption of a particular distribution function for the variables

This tricky question has not attracted any answers. I don't know if you have tried using a computer algebra system to assist, but doing so can be helpful to find closed-form solutions, at least for simpler cases. To illustrate ...

The standard Uniform case: Solution

If the parent iid random variables are standard Uniform(0,1), then the joint pdf of $(A_1, A_2, B_1, B_2)$, say $f(a_1, a_2, b_1, b_2)$, is:

enter image description here Define:

enter image description here

Then the cdf of $CC$, namely $P(CC<c)$, is:

enter image description here

All done.

Here is a plot of the corresponding pdf:

$$ pdf(c) = \begin{cases}\frac{5 c^3}{6} & 0<c\leq 1 \\ \frac{1}{6} (c-2) ((c-16) c+10) & 1<c<2 \end{cases}$$

enter image description here

The above should also provide a framework to check your own workings.


  1. The Prob function used above is from the mathStatica package for Mathematica. As disclosure, I should add that I am one of the authors.

  2. Checked using Monte Carlo methods.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.