The Fibonacci sequence and the distribution of Mammen (1993)

Getting started with Fibonacci numbers

The Fibonacci sequence is ubiquitous in mathematics and nature. Sunflower seeds, pinecones, and even hurricanes resemble the Fibonacci pattern.

Sunflower, F1b

The first ten numbers of the Fibonacci sequence are \(0,1,1,2,3,5,8,13,21,34\). We can generate as many Fibonacci numbers as we want, since they nicely follow the following linear recurrence relation,

\[F_0 = 0, F_1 = 1, F_{n+1} = F_n + F_{n-1}\ \forall\ i \geq 1\]

Interestingly, the closed-form expression for a Fibonnaci number is,

\[F_n = \frac{\phi^n-\bar{\phi}^n}{\sqrt{5}}\]

wherein $\phi=\frac{1+\sqrt{5}}{2}$ and $\bar{\phi}=\frac{1-\sqrt{5}}{2}$. Note that $\phi + \bar{\phi} = 1$ and $\phi\cdot \bar{\phi} = -1$. In mathematical circles, $\phi$ is named golden number or golden ratio.

One approach to verify this formula is through first-order difference equation techniques. An application of strong induction is enough on our end.

Proof. The base case follows straightforwardly. As our induction hypothesis, suppose that the relation above holds for $n$ and $n-1$. Let us show that it also holds for $n+1$. By recurrence, we already know that $F_{n+1} = F_n + F_{n-1}$. Consequently,

\[\begin{align*} F_{n+1} &= \frac{\phi^n-\bar{\phi}^n}{\sqrt{5}} + \frac{\phi^{n-1}-\bar{\phi}^{n-1}}{\sqrt{5}}\\ &= \frac{\phi^n+\bar{\phi}^{n-1}}{\sqrt{5}} - \frac{\bar{\phi}^{n}+\bar{\phi}^{n-1}}{\sqrt{5}}\\ &= \frac{\phi^{n-1}}{\sqrt{5}}(1+\phi) - \frac{\bar{\phi}^{n-1}}{\sqrt{5}}(1+\bar{\phi})\\ &= \frac{\phi^n}{\sqrt{5}}(1 - \bar{\phi}) - \frac{\bar{\phi}^n}{\sqrt{5}}(1 - \phi)\\ &= \frac{\phi^{n+1} - \bar{\phi}^{n+1}}{\sqrt{5}} \end{align*}\]

which completes the inductive step and hence the proof. \(\square\)

Wild bootstrap and the two-point distribution of Mammen (1993)

Bootstrapping is a powerful method in econometrics. It is a resampling technique with replacement (i.e., i.i.d. sampling) applied to a single dataset, which allows us to estimate sampling distributions, standard errors, confidence intervals, and hypothesis tests for various statistics, often used when traditional methods rely on strong assumptions about the population distribution. On the other hand, the empirical bootstrap estimates sampling distributions and statistics by resampling with replacement from the empirical distribution of the data.

Consider the regression model

\[Y_i=Z_i^{\top} \theta_0+U_i\]

for \(i=1,2, \ldots, n\), where \(\left\{Y_i, Z_i\right\}\) are i.i.d. and $\mathbb{E}\left[Z_i U_i\right]=0$. Let \(\hat{U}_i=Y_i-Z_i^{\top} \hat{\theta}_n\). The residual bootstrap consists of two steps. First, draw a sample from the empirical distribution of \(\left\{\hat{U}_i\right\}_{i=1}^n\). Second, obtain the bootstrap sample \(\left\{Y_i^*, Z_i\right\}_{i=1}^n\) given by

\[Y_i^*=Z_i^{\top} \hat{\theta}_n+\hat{U}_i^* .\]

The wild bootstrap improves on the residual bootstrap in that it is robust to conditional heteroskedasticity. One may set \(\hat{U}_i^*=\hat{U}_i W_i\) for a sequence of i.i.d. multipliers \(\left\{W_i\right\}_{i=1}^n\), which also have zero mean, unit variance, and are independent of \(\left\{Y_i, Z_i\right\}_{i=1}^n\). One immediate distribution for \(W_i\) is the normal distribution. Another less known alternative is the two-point Mammen distribution

\[W_i \in \left\{\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}\right\},\quad \mbox{Pr}\left(W_i = \frac{1+\sqrt{5}}{2} \right) = -\frac{1-\sqrt{5}}{2\sqrt{5}}\]

It is routine to check that $\mathbb{E}[W_i] = 0$ and $\mbox{Var}(W_i) = 1$ by using elementary expansions. Note that the values that $W_i$ can take on are nothing else but the golden ratio and its additive inverse. Thus, it is natural to conjecture the existence of a relationship between the moments of $W_i$ and the golden number. This is indeed the case.

The result

Lemma. Let \(X\) be a random variable that follows the two-point distribution of Mammen (1993), and $n$ any positive integer. Therefore,

\[\mathbb{E}{X^n} = F_{n-1}\]

where $F_i$ is the $i-$th Fibonacci number.

Proof. Recall the definitions of $\phi$ and $\bar{\phi}$, and that $\phi\cdot \bar{\phi} = -1$. Operate accordingly on the two-point distribution of Mammen (1993) to infer that,

\[\begin{aligned} \mathbb{E}{X^n} &= \bar{\phi}^n\left(\frac{1+\sqrt{5}}{2 \sqrt{5}}\right)+\phi^n\left(\frac{-1+\sqrt{5}}{2 \sqrt{5}}\right) \notag\\ &=\frac{1}{\sqrt{5}}\left(\frac{(-1)^n}{\phi^{n-1}}+\phi^{n-1}\right)\notag\\ &= \frac{1}{\sqrt{5}}\left((-1)^{2n-1} \bar{\phi}^{n-1}+\phi^{n-1}\right)\notag \\ &= \frac{1}{\sqrt{5}}\left(\phi^{n-1} - \bar{\phi}^{n-1}\right) \end{aligned}\]

which is precisely the closed-form expression of the $(n-1)-$th Fibonacci number. \(\square\)

Thus, the above lemma implies that $\mathbb{E}{X} = 0$, $\mathbb{E}{X^2} = 1$, and $\mathbb{E}{X^{n+1}} = \mathbb{E}{X^{n}} + \mathbb{E}{X^{n-1}}\ \forall\ n \geq 2$. It is immediate that $\mathbb{E}{W_i} = 0$ and $\mbox{Var}(W_i) = 1$, as per the requirements of the i.i.d. multipliers of the wild bootstrap.

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