Histogram Bias

To understand the quality of the approximation that the histogram provides, we look at its bias:

Bias[ˆf(x)]=E[ˆf(x)]−f(x)$$\text{Bias}\left[\hat{f}\left(x\right)\right]=E\left[\hat{f}\left(x\right)\right]-f\left(x\right)$$

Where the expectation on the right-hand side is taken with respect to all samples D$D$ of size n$n$. To understand how to compute the expectation let us assume, without loss of generality, that x$x$ belongs to some interval Ak$A_k$. For a given sample D$D$ drawn from f$f$, let Nk$N_k$ be the random variable which represents the number of points that ended up in bin Ak$A_k$. We have that:

E[ˆf(x)]=E[Nknh]=1nhE[Nk]$$E\left[\hat{f}\left(x\right)\right]=E\left[\frac{N_k}{nh}\right]=\frac{1}{nh}E\left[N_k\right]$$

Suppose we knew that the probability that a single sample ends up in bin Ak$A_k$ is pk$p_k$ (which is different from its estimator, ˆpk${\hat{p}}_k$). The probability that we get Nk=m$N_k=m$ samples out of n$n$ (the size of D$D$) in Ak$A_k$ is then just the probability that we get m$m$ heads out of n$n$ coin flips for a coin where the probability of getting heads is pk$p_k$. This means that Nk$N_k$ follows a binomial distribution; i.e., Nk∼Binomial(n,pk)$N_k\sim \text{Binomial}(n,p_k)$. E[Nk]$E\left[N_k\right]$ therefore is equal to npk$n{p}_k$, so that:

1nhE[Nk]=npknh=pkh$$\frac{1}{nh}E\left[N_k\right]=\frac{n{p}_k}{nh}=\frac{p_k}{h}$$

Now note that pk$p_k$ is the probability that one sample falls in bin Ak$A_k$; i.e., that it falls between tk$t_k$ and tk+1$t_{k+1}$.

Figure 3: Histogram bins and the underlying density.

Fig. 3 tells us, however, that pk$p_k$ can be expressed in terms of f$f$ as:

pk=P[tk≤X≤tk+1]=∫tk+1tkf(x)dx=F(tk+1)−F(tk)$$\begin{array}{}{p}_k=P[t_k\le X\le t_{k+1}]={\int }_{t_k}^{t_{k+1}}f\left(x\right)dx=F\left(t_{k+1}\right)-F\left(t_k\right)& \text{(1)}\end{array}$$

Where F$F$ is the corresponding cumulative density function for f$f$. The final equation for the bias is, then:

Bias[ˆf(x)]=pkh−f(x)$$\begin{array}{}\text{Bias}\left[\hat{f}\left(x\right)\right]=\frac{p_k}{h}-f\left(x\right)& \text{(2)}\end{array}$$

What Eq. (2) tells us is that the bias for the histogram depends on h$h$ but not on n$n$, which is reasonable. In particular, the bias will be small at x$x$ as pk$p_k$ approaches hf(x)$hf\left(x\right)$. Further, Eq. (1) tells us that the bias will be small for any x$x$ as h$h$ tends to zero, given that:

limh→0pkh=limh→0F(tk+1)−F(tk+1−h)h=f(tk+1)∗=f(x)$$\begin{array}{}\underset{h\to 0}{lim}\frac{p_k}{h}=\underset{h\to 0}{lim}\frac{F\left(t_{k+1}\right)-F(t_{k+1}-h)}{h}=f\left(t_{k+1}\right)\stackrel{\ast }{=}f\left(x\right)& \text{(3)}\end{array}$$

where the last equality results from the fact that x$x$ is by construction contained in Ak$A_k$ and, therefore, as h$h$ approaches zero we get that tk=tk+1=x$t_k=t_{k+1}=x$.

Great! So we just have to make our h$h$ very small and we will get a great histogram, right? Well, not so fast…

Histogram Variance

As is often the case, a lower bias does not come for free. Indeed, if we look at the variance for Nk$N_k$, we get:

Var[ˆf(x)]=Var[Nknh]=pk(1−pk)nh2$$\begin{array}{}\text{Var}\left[\hat{f}\left(x\right)\right]=\text{Var}\left[\frac{N_k}{nh}\right]=\frac{p_k(1-p_k)}{n{h}^2}& \text{(4)}\end{array}$$

which does not look like good news: to prevent variance from blowing up, it would appear that we have to increase n$n$ faster than we decrease h$h$, and that the required increase in n$n$ becomes more dramatic as h$h$ approaches zero.

I say that it would appear to be that way because what happens to variance as we keep n$n$ fixed and let h$h$ go to zero is not immediately obvious: as we narrow down our bin width h$h$, pk$p_k$, the probability that a point falls in our bin, also decreases towards zero. If pk$p_k$ manages to compensate the decrease in h$h$ somehow, perhaps variance does not blow up after all.

We can solve this mistery by rewriting Eq. (4) and looking at its limit as we let h→0$h\to 0$:

limh→01nh(pkh−p2kh)$$\underset{h\to 0}{lim}\frac{1}{nh}\left(\frac{p_k}{h}-\frac{p_k^2}{h}\right)$$

We know from Eq. (3) that limh→0pkh=f(x)${lim}_{h\to 0}\frac{p_k}{h}=f\left(x\right)$, which is a finite non-negative real number. Since pk<1$p_k<1$, we also know that 0≤p2k≤pk$0\le p_k^2\le p_k$, meaning that limh→0(pk/h−p2k/h)${lim}_{h\to 0}\left(p_k/h-p_k^2/h\right)$ must converge to a non-negative real number as well (limits are linear). This leaves us with limh→01nh${lim}_{h\to 0}\frac{1}{nh}$, which, barring an equivalent increase in n$n$, will blow up to +∞$+\infty$. As h$h$ tends to zero, therefore, Var[ˆf(x)]$\text{Var}\left[\hat{f}\left(x\right)\right]$ tends to infinity.

We are, therefore, stuck with this fact of life: when using histograms we can either choose to have wide but imprecise (biased) bins, or narrow but unreliable ones.

More on Bias

Another interesting point to consider is the relationship between bin width, bias and the shape of f$f$. Clearly, an f$f$ which contains narrow peaks will be much harder to approximate with a wide rectangle than an f$f$ which is completely flat (think uniform distribution). This intuition can be made formal if we look at the bias equation (Eq. (2)) again, but transform it slightly. Specifically, we will rewrite the term pk/h$p_k/h$ in another way.

Recall that Eq. (1) tells us that:

pk=∫tk+1tkf(x)dx$$p_k={\int }_{t_k}^{t_{k+1}}f\left(x\right)dx$$

The mean value theorem for definite integrals, then, tells us that if we were to draw a rectangle with base in Ak=[tk,tk+1)$A_k=[t_k,t_{k+1})$ and make it tall enough so that its area is equal to pk$p_k$, then its height must be equal to f(ξ)$f\left(\xi \right)$ for some ξ∈Ak$\xi \in A_k$.

The intuition for this is depicted in Fig. 4: if we draw a rectangle with base in Ak=[tk,tk+1)$A_k=[t_k,t_{k+1})$ and area ak$a_k$, we have three options. Either the top side of the rectangle:

1. is completely under f(x)$f\left(x\right)$, in which case ak<pk$a_k<p_k$;
2. is completely above f(x)$f\left(x\right)$, in which case ak>pk$a_k>p_k$;
3. is neither above nor below f(x)$f\left(x\right)$, in which case we can pick tk≤ξ<tk+1$t_k\le \xi <t_{k+1}$, such that ak=pk$a_k=p_k$.

Figure 4: Mean value theorem for definite integrals.

Recalling that tk+1−tk=h$t_{k+1}-t_k=h$, we can therefore say that:

pk=∫tk+1tkf(x)dx=hf(ξ)$$p_k={\int }_{t_k}^{t_{k+1}}f\left(x\right)dx=hf\left(\xi \right)$$

for some tk≤ξ<tk+1$t_k\le \xi <t_{k+1}$. We can therefore rewrite the bias from Eq. (2) as:

Bias[ˆf(x)]=pkh−f(x)=f(ξ)−f(x)$$\text{Bias}\left[\hat{f}\left(x\right)\right]=\frac{p_k}{h}-f\left(x\right)=f\left(\xi \right)-f\left(x\right)$$

If we now make the rather mild assumption that f$f$ is Lipschitz continuous on Ak$A_k$; i.e., that there exists some γk∈R${\gamma }_k\in R$ such that, for every x,y∈R$x,y\in R$ we have:

|f(y)−f(x)|<γk|y−x|$$\left|f\left(y\right)-f\left(x\right)\right|<{\gamma }_k\left|y-x\right|$$

We get that the absolute value for the bias is also bounded:

|Bias[ˆf(x)]|=|f(ξ)−f(x)|≤γk|ξ−x|≤γkh$$\begin{array}{}\left|\text{Bias}\left[\hat{f}\left(x\right)\right]\right|=\left|f\left(\xi \right)-f\left(x\right)\right|\le {\gamma }_k|\xi -x|\le {\gamma }_{k}h& \text{(5)}\end{array}$$

where the last inequality holds because both x$x$ and ξ$\xi$ are in Ak$A_k$ and, therefore, cannot be farther apart by more than h$h$, the width of the bin Ak$A_k$.

One way of looking at what Eq. (5) is saying is that the closer the slope of f$f$ over Ak$A_k$ is to zero, the wider the rectangles we can use to approximate f$f$ acceptably. Although this is only an upper bound, the result is intuitive: a flatter function is easier to represent with a rectangle than a function that varies steeply, up or down¹.

Histogram Overplots

One simple way of improving the situation is by plotting superimposed, transparent versions of our shifted histograms on top of each other. We call these shifted histogram overplots, or SHOs. The idea with SHOs is that we can amortize the effects of variations due to shifting by preemptively shifting them ourselves and “visually averaging” the effects of this shifting.

A SHO is specified by adding an extra parameter, m$m$, which controls the number of shifted copies we plot. For a given bin width h$h$, we plot m$m$ histograms, shifting bin alignment for the jth$j^{th}$ histogram (0≤j<m$0\le j<m$) by j×hm$j\times \frac{h}{m}$. For simplicity we assume that, in the absence of shifting, the first bin of the histogram aligns with 0$0$.

In R:

sho <- function(X, width, m, alpha, ...) {
    for (shift in 0:(m - 1)) {
      offset <- shift * (width / m)
      min_k <- floor((min(X) - offset) / width)
      max_k <- ceiling((max(X) - offset) / width)
      hist(
        X,
        breaks = offset + width * seq(
          from = min_k, to = max_k
        ),
        freq = FALSE,
        add = (shift != 0),
        col = scales::alpha('gray', alpha = alpha),
        lty = 'blank',
        ...
      )
  } 
}

We then plot our data again and simulate what would happen had we gotten our data with a different shift each time: shifting the data changes the data/bin alignment just as shifting bin boundaries did, and so generates similar disturbances. The results are shown in Fig. 6.

par(mfrow = c(1, 3))
for (offset in c(0, 0.2, 0.34)) {
  sho(
    geyser$duration + offset,  
    width = 0.9,
    m = 5, 
    alpha = 0.3, 
    main = sprintf('m = 5, shift %f', offset),
    xlab = 'duration', 
    ylim = c(0, 0.7),
    xlim = c(0, 7)
  )
}

Figure 6: Shifted Histogram Overplots with h=0.9$h=0.9$, m=5$m=5$.

As we can see, although we still get some variability, the main features of the density remain largely unchanged. SHOs are therefore more resilient to changing data/bin alignment than plain histograms.

The issue with SHOs is that they are essentially a visual tool. While we could take the maximum of all shifted histograms and let this be our estimate for f$f$, this would fail to take into account that some regions of the SHO are less transparent (and thus, more likely to be part of the actual density) than others.

The final step in our histogram ladder, then, is a technique which takes these factors into account while constructing a numeric density estimate for f$f$.

Average Shifted Histograms

An Average Shifted Histogram [1] (ASH) constructs a density estimate by averaging together estimates produced by a set of shifted histograms just like the ones we had built for our SHOs. Formally, if we name the estimates produced by the m$m$ shifted histograms at x$x$ as f0(x),⋯,fm−1(x)$f_0\left(x\right),\cdots ,f_{m-1}\left(x\right)$, the ASH estimate ^fash(x)$\widehat{f_{ash}}\left(x\right)$ for x$x$ can be written as:

^fash(x)=1mm−1∑j=0^fj(x)$$\widehat{f_{ash}}\left(x\right)=\frac{1}{m}\sum _{j=0}^{m-1}\widehat{f_j}\left(x\right)$$

Or, in R:

# Average Shifted Histogram
ash <- function(X, nbins, m) {
  width <- diff(range(X)) / nbins
  f_ash <- rep(0, length(X))
    
  for (shift in 1:m) {
    f_ash <- f_ash + 
      f_j(
        X = X, 
        breaks = shift * (width / m) + seq(
          from = min(X) - width, to = max(X) + 1, by = width
        )
      )
  }
  
  i <- order(X)
  ash <- tibble(f = f_ash[i]/m, x = X[i])
  class(ash) <- c('ash', class(ash))
  ash
}

# Histogram Density Estimate
f_j <- function(X, breaks) {
  n <- length(X)
  f_dens <- rep(NA, n)
  for (i in 1:(length(breaks) - 1)) {
    low <- breaks[i]
    high <- breaks[i + 1]
    elements <- (X >= low) & (X < high)
    f_dens[elements] <- sum(elements) / ((high - low) * n)
  }
  stopifnot(all(!is.na(f_dens)))
  f_dens
}

# Plot ASH
plot.ash <- function(ash) {
  ggplot(ash) + 
    geom_line(aes(x = x, y = f)) +
    ylab('density')
}

Let’s see how the plot looks for our Old Faithful’s data:

plot(ash(geyser$duration, m = 5, nbins = 10)) + 
  xlab('duration') +
  ggtitle('Duration of Old Faithful\'s Eruptions (ASH, m = 5, 10 bins)')

As we can see, the result is a much smoother function, with features that resemble the ones we could see in the SHOs of Fig. 6.

Connection to Kernel Methods

ASHes have some nice properties which they share with kernel methods. Scott [1] describes them as a middle ground between a kernel estimator in terms of statistical efficiency (which I presume to mean that their variance decreases faster with an equivalent number of samples than that of a histogram of equivalent bias) and yet are much cheaper to compute due to their relative simplicity.

While discussing matters of statistical efficiency is beyond the scope of this post (though I am certain to revisit it in later posts), it is nevertheless interesting to discuss the shape of the weight function and its relationship to the triangle kernel.

We start by dividing each of our intervals Ak$A_k$ (our bins of length h$h$) into m$m$ subparts, as depicted in Fig. 7. Each subpart is actually a smaller bin, which we refer to as Bi$B_i$. For ease of notation, we assume that x$x$ is contained in small bin B0$B_0$, and give negative indices to the small bins to the left of x$x$. Finally, let vi$v_i$ be the number of points in our sample that fall in small bin Bi$B_i$ (∑vi=∑ni=n$\sum v_i=\sum n_i=n$).

Figure 7: Shifted histograms and their partial estimates.

We know that the density estimate at x$x$ is the average density coming from m$m$ overlapping histograms. Looking at Fig. 7, we see that we can express those in terms of the vi$v_i$ as follows:

^f0(x)=1nh(v0+⋯+vm−1)^f1(x)=1nh(v−1+v0+⋯vm−2)^f2(x)=1nh(v−2+v−1+v0+⋯vm−3)⋯^fi(x)=1nh(i∑j=1v−j+v0+m−i+1∑k=1vk)⋯ˆfm−1(x)=1nh(m−1∑j=1v−j+0)$$\begin{array}{cc}\widehat{f_0}\left(x\right)=& \frac{1}{nh}(v_0+\cdots +v_{m-1})\\ \widehat{f_1}\left(x\right)=& \frac{1}{nh}\left(v_{-1}+v_0+\cdots v_{m-2}\right)\\ \widehat{f_2}\left(x\right)=& \frac{1}{nh}\left(v_{-2}+v_{-1}+v_0+\cdots v_{m-3}\right)\\ \cdots & \\ \widehat{f_i}\left(x\right)=& \frac{1}{nh}\left(\sum _{j=1}^i{v}_{-j}+v_0+\sum _{k=1}^{m-i+1}{v}_k\right)\\ \cdots & \\ {\hat{f}}_{m-1}\left(x\right)=& \frac{1}{nh}\left(\sum _{j=1}^{m-1}{v}_{-j}+0\right)\end{array}$$

We have, therefore that:

1mm−1∑j=0^fj(x)=1m1nh[mv0+(m−1)(v−1+v1)+⋯1(vm−1+v−(m−1))]==1mm−1∑j=0(m−j)nh(vj+v−j)

We can see Eq. (6) as a weighted sum, where the underlying weight function assigns weights m−jnh to both vj and v−j. If we plot these weights as a function of vi, we can observe the shape of the weight function. We plot those in Fig. 8 for m∈{5,15,100}. As we increase m, the weight function approaches a continuous triangle and the behavior of ASHes approaches that of kernel density estimation with a triangular kernel.

Figure 8: The ASH weight function tends to the triangle kernel as m→+∞.