Optimal group size in microlending

Abstract

Microlending, where a bank lends to a small group of people without credit histories, began with the Grameen Bank in Bangladesh, and is widely seen as the creation of Muhammad Yunus, who received the Nobel Peace Prize in recognition of his largely successful efforts. Since that time the modeling of microlending has received a fair amount of academic attention. One of the issues not yet addressed in full detail, however, is the issue of the size of the group. Some attention has nevertheless been paid using an experimental and game theory approach. We, instead, take a mathematical approach to the issue of an optimal group size, where the goal is to minimize the probability of default of the group. To do this, one has to create a model with interacting forces, and to make precise the hypotheses of the model. We show that the original choice of Muhammad Yunus, of a group size of five people, is, under the right, and, we believe, reasonable hypotheses, either close to optimal, or even at times exactly optimal, i.e., the optimal group size is indeed five people.

Appendices

Appendix: Proof of the theorem

Let $\mathcal {S}(x):=\sum _{n=0}^\infty \left( \frac{1}{n+1}\right) \left( \frac{1}{n+2}\right) \left( \frac{1}{f(x)} \right) ^n$

Note:

• $\mathcal {S}(x)$ is a decreasing function in x.

• $\mathcal {S}(x)\in \left( \frac{1}{2}, 1\right) , \ \text {for all } x\in \mathbb {R}$ because:

  $$ \frac{1}{2}< \frac{1}{2} + \sum _{n=1}^\infty \left( \frac{1}{n+1}\right) \left( \frac{1}{n+2}\right) \left( \frac{1}{f(x)} \right) ^n = \mathcal {S}(x) < \sum _{n=0}^\infty \left( \frac{1}{n+1}\right) \left( \frac{1}{n+2}\right) = 1 $$

Set $h(x):=f(x)-xf'(x)$ and note that in (a, b), h(x) is a monotone function because of condition (4b) or (4d). More explicitly:

$$ h'(x)=f'(x)-\left[ f'(x)+xf''(x) \right] = -xf''(x) >0 \ (\text {or } <0) \qquad \text {for all } x\in (a, b) $$

Moreover, the monotonicity of h(x) and condition (4a) or (4c) imply $\frac{1}{2}<h(x)<1$$\text {for all } x \in (a,b)$

In this way $\text {for all } x\in (a,b)$:

• Both $\mathcal {S}(x)$ and h(x) are continuous and monotone

• $\mathcal {S}(x)$ is bounded between $\left( \frac{1}{2}, 1 \right) .$ This actually holds $\text {for all } x\in \mathbb {R}^+$

• h(x) increases from $\frac{1}{2}$ to 1 (or decreases from 1 to $\frac{1}{2}$)

Then there exists a unique $x^*\in (a,b)$ such that

$$\begin{aligned} h(x^*)=\mathcal {S}(x^*) \end{aligned}$$

(3)

We shall see that this $x^*$ is actually the unique maximizer. Thanks to equation (3), we have:

$$\begin{aligned} f(x^*) - x^* f'(x^*)&=\sum _{n=0}^\infty \left( \frac{1}{n+1}\right) \left( \frac{1}{n+2}\right) \left( \frac{1}{f(x^*)} \right) ^n \nonumber \\ \iff 0&= \sum _{n=0}^\infty \left( \frac{1}{n+1}\right) \left( \frac{1}{n+2}\right) \left( \frac{1}{f(x^*)} \right) ^n +x^*f'(x^*)-f(x^*) \nonumber \\&= \sum _{n=0}^\infty \left( \frac{1}{n+1}\right) \left( \frac{1}{f(x^*)} \right) ^n - \sum _{n=0}^\infty \left( \frac{1}{n+2}\right) \left( \frac{1}{f(x^*)} \right) ^n +x^*f'(x^*)-f(x^*) \nonumber \\&= \sum _{n=1}^\infty \left( \frac{1}{n}\right) \left( \frac{1}{f(x^*)} \right) ^{n-1} - \sum _{n=2}^\infty \left( \frac{1}{n}\right) \left( \frac{1}{f(x^*)} \right) ^{n-2} - f(x^*) +x^*f'(x^*) \nonumber \\&= \sum _{n=1}^\infty \left( \frac{1}{n}\right) \left( \frac{1}{f(x^*)} \right) ^{n-1} - \sum _{n=1}^\infty \left( \frac{1}{n}\right) \left( \frac{1}{f(x^*)} \right) ^{n-2} +x^*f'(x^*) \nonumber \\&= \sum _{n=1}^\infty \left( \frac{1}{n}\right) \left( \frac{1}{f(x^*)} \right) ^{n+1} - \sum _{n=1}^\infty \left( \frac{1}{n}\right) \left( \frac{1}{f(x^*)} \right) ^{n} + \frac{x^*f'(x^*)}{(f(x^*))^2} \end{aligned}$$

(4)

Now, recall we want to find a maxima for $\left( 1-\varphi (x)\right) ^x$. This is equivalent to maximizing $\mathcal {U}(x):= x\ln \left( 1-\varphi (x)\right)$.

Note $\mathcal {U}'(x)= \ln \left( 1- \varphi (x) \right) - \frac{x}{1-\varphi (x)} \varphi '(x)$

It suffices to find $x^*$ (the maximizer) such that $\mathcal {U}'(x^*)=0$, which is equivalent to $g(x^*)=0$ where $g(x):= \left[ 1-\varphi (x) \right] \ln \left( 1-\varphi (x)\right) - x\varphi '(x)$

Recall: $\ln (1-y) = - \sum _{n=1}^\infty \frac{y^n}{n}$, if $|y|<1$. Then:

$$\begin{aligned} g(x)&= \left[ 1-\varphi (x)\right] \left[ -\sum _{n=1}^\infty \frac{\varphi ^n(x)}{n} \right] - x\varphi '(x) \\&=\sum _{n=1}^\infty \frac{\varphi ^{n+1}(x)}{n} -\sum _{n=1}^\infty \frac{\varphi ^n(x)}{n} -x\varphi '(x) \\&=\sum _{n=1}^\infty \left( \frac{1}{n}\right) \left( \frac{1}{f(x)} \right) ^{n+1} - \sum _{n=1}^\infty \left( \frac{1}{n}\right) \left( \frac{1}{f(x)} \right) ^{n} + x \frac{f'(x)}{\left( f(x) \right) ^2} \end{aligned}$$

Finally, it is easy to see that this last line and (4) imply $g(x^*) =0$ and we can conclude $x^*$ is the unique maximizer in (a, b). If a, b are unique, it is clear that the maximizer is unique.

Appendix: Analysis of the example

Now, we show that the example given in (2) satisfies the conditions of the theorem.

Conditions 1 and 2 are immediate.

Condition (3): $f'(x)>0$ $\text {for all } x\ge 2$

Proof of (3): $f'(x)=px^{p-1} + \frac{1}{p} \left( \frac{1}{x}\right) \left( \ln x \right) ^{\frac{1}{p}-1}$       As $x\ge 2$, it is clear $f'(x)>0$ $\square$

Condition (4a): There exist a and b such that $f(a)-af'(a)=\frac{1}{2}$ and $f(b)-bf'(b)=1$

Proof of (4a):

Set $h_p(x):=f(x)-xf'(x)= x^p + (\ln x)^{\frac{1}{p}} - px^p -\frac{1}{p} \left( \ln x\right) ^{\frac{1}{p}-1} = (1-p)x^p + \left( \ln x\right) ^{\frac{1}{p}-1} \left( \ln x -\frac{1}{p} \right)$

Claim

$h_p(x)$ is increasing in x

Proof of claim

$\frac{\partial }{\partial x} h_p(x) =(1-p)px^{p-1} +\left( \frac{1}{p} -1 \right) \left( \ln x \right) ^{\frac{1}{p}-2} \left[ \ln x -\frac{1}{p} \right] \frac{1}{x} + \frac{1}{x} \left( \ln x \right) ^{\frac{1}{p}-1}$

It is clear $\left( 1- p \right) p x^{p-1} > 0$. So, it suffices to show

$$\begin{aligned} \left( \frac{1}{p} -1 \right) \left( \ln x \right) ^{\frac{1}{p}-2} \left[ \ln x -\frac{1}{p} \right] \frac{1}{x} + \frac{1}{x} \left( \ln x \right) ^{\frac{1}{p}-1} \ge 0 \end{aligned}$$

(5)

For reasons that will become clear later, we only consider $x\ge e$. As $\ln x +1 \ge 2$ and $1 \le \frac{1}{p} \le 2$, it follows that

$$\begin{aligned} \ln x +1&\ge \frac{1}{p} \\ \frac{1}{p} \left( \ln x +1 - \frac{1}{p} \right)&\ge 0 \\ \frac{1}{p} \left( \ln x - \frac{1}{p} \right) - \left( \ln x - \frac{1}{p} \right) + \ln x&\ge 0 \\ \left( \frac{1}{p} -1 \right) \left( \ln x -\frac{1}{p} \right) + \ln x&\ge 0 \end{aligned}$$

This shows (5) and thus that $h_p(x)$ is increasing for $x \ge e$ $\square$

Claim

$h_p(e)=(1-p) e^p + 1 - \frac{1}{p}$ is concave in p and hence there exists a local maxima, namely $p^*$. (Recall $\frac{1}{2} \le p \le 1$)

Proof of claim

$\frac{\partial }{\partial p} h_p(e)= -e^p + (1-p) e^p + \frac{1}{p^2} = -pe^p + \frac{1}{p^2}$

$\frac{\partial ^2}{\partial p^2} h_p(e) = -e^p -pe^p - \frac{2}{p^3} < 0 \Longrightarrow \ \ h_p(e)$ is concave

Now, to find the maxima, we set the derivative equal to 0, i.e. $\frac{\partial }{\partial p} h_p(e) = 0$

$$\begin{aligned} 0&= -pe^p + \frac{1}{p^2} \\ 1&= p^3 e^p \\ 1&= p e^{\frac{1}{3}p} \end{aligned}$$

Set $u = \frac{1}{3} p$, we need to solve $u e^u = \frac{1}{3}$, which we do by using the product logarithm.

Hence $u=W\left( \frac{1}{3} \right)$ and thus $p^*=3W\left( \frac{1}{3} \right) \approx 0.772883 \Longrightarrow h_p(e)|_{p=0.773} \approx 0.1981 < \frac{1}{2}$    $\square$

Claim

$h_p(e) < \frac{1}{2} \ \ \text {for all } p \in \left[ \frac{1}{2}, 1 \right]$

Proof of claim

As we have shown that $h_p(e)$ is concave in p and that $h_{p^*}(e) \approx 0.1981 < \frac{1}{2}$, it follows that $h_p(e) < \frac{1}{2} \ \ \text {for all } p \in \left( \frac{1}{2}, 1 \right)$

We only need to check the endpoints, $h_p(e)|_{p=\frac{1}{2}} \approx -0.1756$ and $h_p(e)|_{p=1}=0$    $\square$

Hence $h_p(e) < \frac{1}{2} \ \ \text {for all } p\in \left[ \frac{1}{2}, 1 \right]$. This, along with $h_p(x)$ being continuous, increasing in $x\ge e$ and $\lim _{x\rightarrow \infty } h_p(x)= \infty$, imply that there exists a such that $h_p(a) = \frac{1}{2}$ and that $a \ge e \approx 2.7 \Rightarrow a \ge 3$ $\square$

Claim

$h_p(e^2) = (1-p) e^{2p} + 2^{\frac{1}{p}-1} \left( 1-\frac{1}{p} \right)$ is concave in p

Proof of claim

$$\begin{aligned}&\frac{\partial }{\partial p} h_p(e^2) = -e^{2p} + 2(1-p)e^{2p} +\left( \frac{1}{p^3} -\frac{2}{p^2} \right) (\ln 2) 2^{\frac{1}{p}-1} + \left( \frac{1}{p^2}\right) 2^{\frac{1}{p}-1}\\&\frac{\partial ^2}{\partial p^2} h_p(e^2) = -4e^{2p} + 4(1-p)e^{2p} - 2^{\frac{1}{p}} \frac{1}{p^3} + 2^{\frac{1}{p}} \left( 2 -\frac{1}{p} \right) \frac{\ln 2}{p^3} - 2^{\frac{1}{p}} \frac{\ln 2}{p^4} \\&\qquad + 2^{\frac{1}{p}-1} \left( 2-\frac{1}{p} \right) \frac{(\ln 2)^2}{p^4} \end{aligned}$$

Now we show that $\frac{\partial ^2}{\partial p^2} h_p(e^2)<0$

• 1. It is clear that $-4e^{2p} + 4(1-p)e^{2p} <0$

• 2. As $\ln 2 <1$ and $2-\frac{1}{p} \le 1$

  $$\begin{aligned} \left( 2-\frac{1}{p}\right) \ln 2<&\ 1\\ -1 +\left( 2-\frac{1}{p}\right) \ln 2<&\ 0 \\ - 2^{\frac{1}{p}} \frac{1}{p^3} + 2^{\frac{1}{p}} \left( 2 -\frac{1}{p} \right) \frac{\ln 2}{p^3} <&\ 0 \end{aligned}$$

• 3. Similarly

  $$\begin{aligned} \left( 2-\frac{1}{p}\right) \frac{\ln 2}{2}<&\ 1 \\ -1 +\left( 2-\frac{1}{p}\right) \frac{\ln 2}{2}<&\ 0 \\ 2^{\frac{1}{p}} \frac{\ln 2}{p^4} + 2^{\frac{1}{p}-1} \left( 2-\frac{1}{p} \right) \frac{(\ln 2)^2}{p^4}<&\ 0 \\ \end{aligned}$$

  $\square$

Claim

$h_p(e^2) \ge 1 \ \ \text {for all } p\in \left[ \frac{1}{2}, 1\right]$

Proof of claim

As we have shown that $h_p(e^2)$ is concave in p, it suffices to show $h_p(e^2)|_{p=\frac{1}{2}}\ge 1$ and $h_p(e^2)|_{p=1}\ge 1$

• 1. $h_p(e^2)|_{p=\frac{1}{2}} =\frac{1}{2} e > 1$ as $e>2$

• 2. $h_p(e^2)|_{p=1} =1$    $\square$

Hence $h_p(e^2) \ge 1 \ \ \text {for all } p\in \left[ \frac{1}{2}, 1 \right]$. This, along with $h_p(x)$ being continuous and increasing in x implies that there exists b such that $h_p(b) = 1$ and that $b \le e^2 \approx 7.4 \Rightarrow b \le 7$    $\square$

Condition (4b): $f''(x)<0$ $\text {for all } x\in (a,b)$ for some $a, b>0$ such that $e\le a<b$

Proof of (4b):

$$\begin{aligned}&f''(x)=p(p-1)x^{p-2} +\frac{1}{p}\left[ \left( \frac{1-p}{p} \right) \left( \frac{1}{x} \right) ^2 \left( \ln x \right) ^{\frac{1}{p}-2} - \left( \frac{1}{x}\right) ^2 \left( \ln x \right) ^{\frac{1}{p}-1} \right] \\&\quad = p(p-1) x^{p-2} + \frac{1}{p} \left( \frac{1}{x} \right) ^2 \left( \ln x \right) ^{\frac{1}{p}-2} \left[ \frac{1-p}{p} -\ln x \right] \end{aligned}$$

• 1. $p(p-1)x^{p-2} \le 0$ because, by assumption $p\le 1$.

• 2. By assumption $1\le \frac{1}{p} \le 2$ and as $\ln x +1 \ge 2$ for $x\ge e$, we get $\frac{1}{p} \le \ln x +1$ . Hence, $\frac{1-p}{p} - \ln x \le 0$, which implies the 2nd term is non-positive for all $x\ge e$             $\square$

Hence for $f(x)=x^p +[\ln x]^{\frac{1}{p}}$, using our theorem, we can claim that the maximizer $x^*\in [e, e^2] \ \text {for all } p\in \left[ \frac{1}{2}, 1\right]$. It is worth noticing that for this particular f(x), we can obtain a narrower interval in the following way:

To find the maximizer $x^*$ of $\left( 1-\varphi (x) \right) ^x$, we need to set the derivative equal to 0, which, as noted in the proof of the theorem, is equivalent to solving $\left[ 1-\varphi (x) \right] \ln \left( 1-\varphi (x)\right) - x\varphi '(x) = 0$. Using $\varphi (x) = \frac{1}{f(x)}= \frac{1}{x^p +[\ln x]^{1/p}}$, let us define:

$$\begin{aligned} \mathcal {H}(x,p):&= \left[ 1-\varphi (x) \right] \ln \left( 1-\varphi (x)\right) - x\varphi '(x) \\&= \left( 1 - \frac{1}{x^p +[\ln x]^{\frac{1}{p}}} \right) \ln \left( 1 - \frac{1}{x^p +[\ln x]^{\frac{1}{p}}} \right) + \frac{p^2 x^p \ln x + \left( \ln x \right) ^{\frac{1}{p}}}{p \ln x \left[ x^p + \left( \ln x \right) ^{\frac{1}{p}} \right] ^2} \end{aligned}$$

Then, after fixing p, we need to find x such that $\mathcal {H}(x, p) = 0$. As our theorem allows us to conclude that the maximizer $x^*\in [e, e^2]$, we only need to analyze the behaviour of $\mathcal {H}(x,p)$ when x is in such interval.

Note $\mathcal {H}(x, p)$ is decreasing in x on the interval $x\in [e, e^2]$ for all fixed $p \in \left[ \frac{1}{2}, 1 \right]$. Now, we want to find a value of $x\in [e, e^2]$ for which $\mathcal {H}(x, p) \ge 0$ for all $p \in \left[ \frac{1}{2}, 1 \right]$. Since this is not the case for $x=3.5$, we choose $x=3.4$ as such bound gives us uniform positivity for all $p \in \left[ \frac{1}{2}, 1 \right]$

More precisely, for fixed $p \in \left[ \frac{1}{2}, 1\right]$ and $\text {for all } x \le 3.4$, we have $\mathcal {H}(x,p) \ge \mathcal {H}(3.4,p)$, which implies

$$\begin{aligned} \mathcal {H}(x,p) \ge \min _{p} \mathcal {H}(x,p) \ge \min _{p}\mathcal {H}(3.4,p) > 0 \end{aligned}$$

Hence $\mathcal {H}(x,p) = 0$ does not have a solution when $x \le 3.4$ and $p \in \left[ \frac{1}{2}, 1\right]$. So, ${x^*}$ must be in the interval $(3.4, \infty )$.

Similarly, we want to find a value of $x\in [3.4, e^2]$ for which $\mathcal {H}(x, p) \le 0$ for all $p \in \left[ \frac{1}{2}, 1 \right]$. We can check that $x=5.2$ gives us uniform negativity for all $p \in \left[ \frac{1}{2}, 1 \right]$. In this way we get $\text {for all } x\ge 5.2$ and $\text {for all } p \in \left[ \frac{1}{2}, 1\right]$:

$$ \mathcal {H}(x,p) \le \max _p\mathcal {H}(x, p) \le \max _p \mathcal {H}(5.2, p) < 0 $$

This implies that $\mathcal {H}(x,p) = 0$ does not have a solution when $x \ge 5.2$ and $p \in \left[ \frac{1}{2}, 1\right]$. Hence, we finally get a narrower interval $x^* \in (3.4, 5.2)$ $\text {for all } p\in \left[ \frac{1}{2}, 1\right]$

Note that the choice of $x=3.4$ and $x=5.2$ as a comparison points was arbitrary as all we require is $\mathcal {H}(x,p)$ to be positive or negative (respectively) for all $p\in \left[ \frac{1}{2}, 1\right]$. Another options that work are any $e\le x\le 3.486$ and $5.135 \le x \le e^2$ respectively, but as we will round up or down to the nearest integer, we believe that using one decimal place is enough.