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.

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

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

Remark 1. In my original formulation, $Q, R \in \{ 0,1\}$ . This means that mutual space for vulnerability is binary—either absent if at least one person withholds space for vulnerability (yielding 0) or present if both persons provide space for vulnerability for the other (yielding 1).

Remark 2. Note that Person X determines these quantities, and may differ from Person Y’s evaluation (explaining why X can feel close to Y but not likewise in the opposite direction; i.e. the feelings are not necessarily mutual).

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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