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 Bernoulli random variables:

$V_{\mathrm{XY}}$ 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).
$V_{\mathrm{YX}}$ denotes the space for vulnerability that Person X receives from Person Y.

Define the mutual space for vulnerability between X and Y by

$V := \min\{ V_{\mathrm{XY}} , V_{\mathrm{YX}}\}.$

Problem 1. Suppose that $V_{\mathrm{XY}} \sim \mathrm{Ber}(p_{\mathrm X \mathrm Y})$ and $V_{\mathrm{YX}} \sim \mathrm{Ber}(p_{\mathrm Y \mathrm X})$ are independent. Determine the distribution of $V$ .

(Click for Solution)

Solution. We observe that

$\{V = 1\} = \{V_{\mathrm{XY}} = 1\} \cap \{V_{\mathrm{YX}} = 1\}.$

and $V = 0$ otherwise. Under independence,

$\begin{aligned} \mathbb P(V=1) &= \mathbb P(V_{\mathrm{XY}} = 1) \cdot \mathbb P(V_{\mathrm{XY}} = 1) = p_{\mathrm X \mathrm Y} \cdot p_{\mathrm Y \mathrm X}. \end{aligned}$

Therefore, $V \sim \mathrm{Ber}(p_{\mathrm X \mathrm Y} \cdot p_{\mathrm Y \mathrm X})$ .

Definition 2. Define the following continuous random variables bounded in $[0, 1]$ :

$A_{\mathrm{XY}}$ denotes the access that Person X gives to Person Y (i.e. the amount of personal information that Person X discloses to Person Y).
$A_{\mathrm{YX}}$ denotes the access that Person X receives from Person Y.

Define the mutual access between X and Y by

$A := \min\{ A_{\mathrm{XY}} , A_{\mathrm{YX}}\}.$

Problem 2. Suppose $A_{\mathrm{XY}},A_{\mathrm{YX}}$ are independent with c.d.f.s $F_{\mathrm X \mathrm Y}, F_{\mathrm X \mathrm Y}$ . Determine the c.d.f. of $A$ .

(Click for Solution)

Solution. Denote the c.d.f. of $A$ by $F$ . We observe that for any $t$ ,

$\{A > t\} = \{A_{\mathrm{XY}} > t\} \cap \{A_{\mathrm{YX}} > t\}.$

Under independence, since $\mathbb P(A > t) = 1 - F(t)$ ,

$\begin{aligned} \mathbb P(A > t) &= \mathbb P(A_{\mathrm{XY}} > t) \cdot \mathbb P(A_{\mathrm{YX}} > t) \\ 1 - F(t) &= (1 - F_{\mathrm{XY}}(t)) \cdot (1 - F_{\mathrm{YX}}(t)) \\ &= 1 - (F_{\mathrm{XY}}(t) + F_{\mathrm{YX}}(t) - (F_{\mathrm{XY}} \cdot F_{\mathrm{YX}})(t)). \end{aligned}$

Therefore, $F = F_{\mathrm{XY}}+F_{\mathrm{YX}}- F_{\mathrm{XY}} \cdot F_{\mathrm{YX}}$ .

Definition 3. Define the mutual closeness $C$ that X feels he shares with Y by the equation

$C = V \cdot A.$

Remark 1. 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).

Suppose furthermore that $V, A$ are independent. Denote $\mathbb E[A_{\mathrm{XY}}] = \mu_{\mathrm{XY}}$ .

Problem 3. Prove that

$\displaystyle \mathbb E[C] = \frac 12 \cdot p_{\mathrm{XY}} \cdot p_{\mathrm{YX}} \cdot \left( \mu_{\mathrm{XY}} + \mu_{\mathrm{YX}} - \mathbb E[|A_{\mathrm{XY}} - A_{\mathrm{YX}}|]\right).$

(Click for Solution)

Solution. Observe that

$\displaystyle A = \min\{A_{\mathrm{XY}}, A_{\mathrm{YX}}\} = \frac{ A_{\mathrm{XY}} + A_{\mathrm{YX}} - |A_{\mathrm{XY}} - A_{\mathrm{YX}}|}{2}.$

Taking expectations on all sides,

$\displaystyle \mathbb E[A] = \frac{ \mu_{\mathrm{XY}} + \mu_{\mathrm{YX}} - \mathbb E[ |A_{\mathrm{XY}} - A_{\mathrm{YX}}| ] }{2}.$

Using Problem 1, $\mathbb E[V] = p_{\mathrm X \mathrm Y} \cdot p_{\mathrm Y \mathrm X}$ . Since $V,A$ are independent,

$\displaystyle \mathbb E[C] = p_{\mathrm X \mathrm Y} \cdot p_{\mathrm Y \mathrm X} \cdot \frac{ \mu_{\mathrm{XY}} + \mu_{\mathrm{YX}} - \mathbb E[ |A_{\mathrm{XY}} - A_{\mathrm{YX}}| ] }{2}.$

Remark 2. Problem 3 tells us that if we are willing to create more space for vulnerability for others (thereby increasing $\mathbb E[V_{\mathrm{XY}}] = p_{\mathrm{XY}}$ ), on average, $\mathbb E[C]$ increases. That is, the closeness level across our friendships would be, on average, higher.

Remark 3. By the triangle inequality and the monotonicity of expectation,

$\mathbb E[|A_{\mathrm{XY}} - A_{\mathrm{YX}}|]\leq \mathbb E[A_{\mathrm{XY}}] + \mathbb E[A_{\mathrm{YX}} ] = \mu_{\mathrm{XY}} + \mu_{\mathrm{YX}}.$

Therefore, the greater the access given (i.e. $\mu_{\mathrm{XY}}$ ), the higher the tolerance for $\mathbb E[|A_{\mathrm{XY}} - A_{\mathrm{YX}}|]$ , i.e. the absolute difference in access given between both directions, matching the reality of heightened risk in giving others access. Nevertheless, if both quantities are close to each other, corresponding to the imprecise sense of “feeling the vibe”, then $\mathbb E[|A_{\mathrm{XY}} - A_{\mathrm{YX}}|] \approx 0$ , yielding

$\displaystyle \mathbb E[C] \approx \frac 12 \cdot p_{\mathrm{XY}} \cdot p_{\mathrm{YX}} \cdot \left( \mu_{\mathrm{XY}} + \mu_{\mathrm{YX}}\right).$

—Joel Kindiak, 24 Jun 26, 1937H

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