The Friendship Formula

Friendship is a massively important topic of contemplation in my life, so much to the point I have proposed an equation to quantify the closeness between two friends. Suppose Persons X and Y are friends, and Person X wants to evaluate his closeness with Person Y.

Definition 1. Define the following random variables bounded in $[0, 1]$ :

$Q$ denotes the space for vulnerability that Person X gives to Person Y (i.e. the extent to which Person Y does not need to feel defensive when interacting with Person X).
$R$ denotes the space for vulnerability that Person X receives from Person Y.
$S$ denotes the access that Person X gives to Person Y (i.e. the amount of personal information that Person X discloses to Person Y).
$T$ denotes the access that Person X receives from Person Y.

Assume the random variables $Q,R,S,T$ are jointly independent continuous random variables.

Note that Person X determines these quantities, and may differ from Person Y’s evaluation (explaining why X can feel close to Y but the feelings are not mutual).

Axiom 1. The mutual closeness $C$ that X feels he shares with Y is defined by the equation

$C := \min\{ Q, R \} \cdot \min\{ S, T \}.$

Problem 1. Prove that $\mathbb E[C]$ is given by the quantity

$\displaystyle \begin{aligned} \left(\mathbb E[Q] + \mathbb E[R] - 1 + \int_{[0, 1]} F_Q \cdot F_R\, \mathrm d\lambda \right) \cdot \left(\mathbb E[S] + \mathbb E[T] - 1 + \int_{[0, 1]} F_S \cdot F_T\, \mathrm d\lambda\right). \end{aligned}$

This quantity evaluates the expected closeness that X feels with Y.

(Click for Solution)

Solution. We notice that all random variables have expectation bounded in $[0, 1]$ . Define

$V := \min\{ Q, R \},\quad A := \min\{ S, T \}.$

We first evaluate $\mathbb E[V]$ . By definition,

$\begin{aligned} \mathbb P(V > v) &= \mathbb P(\min\{Q, R\} > v) \\ &= \mathbb P( Q > v, R > v ) \\ &= \mathbb P( Q > v ) \cdot \mathbb P( R > v ). \end{aligned}$

In particular,

$\begin{aligned}(F_Q \cdot F_R)(v) &= F_Q(v) \cdot F_R(v) \\ &= (1 - \mathbb P(Q > v)) \cdot (1 - \mathbb P(R > v)) \\ &= 1 - \mathbb P(Q > v) - \mathbb P(R > v) + \mathbb P(Q > v) \cdot \mathbb P(R > v) \\ &= 1 - \mathbb P(Q > v) - \mathbb P(R > v) + \mathbb P(V > v). \end{aligned}$

Taking integrals on all sides, using the tail-probability characterisation of the expectation, and performing algebruh,

$\displaystyle \mathbb E[V] = \mathbb E[Q] + \mathbb E[R] - 1 + \int_{[0, 1]} (F_Q \cdot F_R)(v)\, \mathrm dv.$

Similarly,

$\displaystyle \mathbb E[A] = \mathbb E[S] + \mathbb E[T] - 1 + \int_0^1 (F_S \cdot F_T)(a)\, \mathrm da.$

Since $Q,R,S,T$ are jointly independent, $V = \min\{Q, R\}$ and $A = \min\{S, T\}$ are also independent, and

$\mathbb E[C] = \mathbb E[V \cdot A] = \mathbb E[V] \cdot \mathbb E[A],$

giving the desired result, after replacing relevant dummy variables.

—Joel Kindiak, 7 Aug 25, 2324H

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