# 17.5. Basics of Prediction Intervals#

Confidence intervals convey the accuracy of an estimator, but sometimes we want the accuracy of a prediction for a future observation. For example, someone might say: half the time my bus arrives at most three-quarters of a minute late, but how late might it get? As another example, the California Department of Fish and Wildlife sets the minimum catch size for Dungeness crabs at 146 mm, and a recreational fishing company might wonder how much bigger than 146 mm their customer’s catch might be when they bring them fishing. And for another example, a vet estimates the weight of a donkey to be 169 kg based on its length and girth and uses this estimate to administer medication. For the donkey’s safety, the vet is keen to know how different the donkey’s real weight might be from this estimate.

What these examples have in common is an interest in the prediction of a future observation and the desire to quantify how far that future observation might be from this prediction. Just like with confidence intervals, we compute the statistic (the estimator) and use it in making the prediction, but now we’re interested in typical deviations of future observations from the prediction. In the following sections, we work through examples of prediction intervals based on quantiles, standard deviations, and those conditional on covariates. Along the way, we provide additional information about the typical variation of observations about a prediction.

## 17.5.1. Example: Predicting Bus Lateness#

Chapter 4 models the lateness of a Seattle bus in arriving at a particular stop. We observed that the distribution was highly skewed and chose to estimate the typical lateness by the median, which was 0.74 minutes. We reproduce the sample histogram from that chapter here.

times = pd.read_csv("data/seattle_bus_times_NC.csv")
fig = px.histogram(times, x="minutes_late", width=350, height=250)
fig.update_xaxes(range=[-12, 60], title_text="Minutes late")
fig


The prediction problem addresses how late a bus might be. While the median is informative, it doesn’t provide information about the skewness of the distribution. That is, we don’t know how late the bus might be. The 75th percentile, or even the 95th percentile, would add useful information to consider. We compute those percentiles here:

print(f"median:          {times['minutes_late'].median():.2f} mins late")
print(f"75th percentile: {np.percentile(times['minutes_late'], 75.0, method='lower'):.2f} mins late")
print(f"95th percentile: {np.percentile(times['minutes_late'], 95.0, method='lower'):.2f} mins late")

median:          0.74 mins late
75th percentile: 3.78 mins late
95th percentile: 13.02 mins late


From these statistics, we learn that while more than half the time the bus is not even a minute late, one-quarter of the time it’s almost four minutes late, and with some regularity it can happen that the bus is nearly 15 minutes late. These three values together help us make plans.

## 17.5.2. Example: Predicting Crab Size#

Fishing for Dungeness crabs is highly regulated, including limiting the shell size to 146 mm in width for crabs caught for recreation. To better understand the distribution of shell size of Dungeness crabs, the California Department of Fish and Wildlife worked with commercial crab fishers from Northern California and Southern Oregon to capture, measure, and release crabs. Here is a histogram of crab shell sizes for the approximately 450 crabs caught:

def subset_and_rename(df):
df = df[["presz", "inc"]]
df.columns = ["shell", "inc"]
return df

crabs = (
.pipe(subset_and_rename)
.query("shell > 100 and inc > 8")
)


The distribution is somewhat skewed left, but the average and standard deviations are reasonable summary statistics of the distribution:

crabs['shell'].describe()[:3]

count    452.00
mean     131.53
std       11.07
Name: shell, dtype: float64


The average, 132 mm, is a good prediction for the typical size of a crab. However, it lacks information about how far an individual crab may vary from the average. The standard deviation can fill in this gap.

In addition to the variability of individual observations about the center of the distribution, we also take into account the variability in our estimate of the mean shell size. We can use the bootstrap to estimate this variability, or we can use probability theory (we do this in the next section) to show that the standard deviation of the estimator is $$SD(pop)/\sqrt{n}$$. We also show, in the next section, that these two sources of variation combine as follows:

$\sqrt{SD(pop)^2 + \frac {SD(pop)^2}{n}} ~=~ SD(pop) \sqrt{1 + \frac {1}{n}}$

We substitute $$SD(sample)$$ for $$SD(pop)$$ and apply this formula to our crabs:

np.std(crabs['shell']) * np.sqrt(1 + 1/len(crabs))

11.073329460297957


We see that including the SE of the sample average essentially doesn’t change the prediction error because the sample is so large. We conclude that crabs routinely differ from the typical size of 132 mm by 11 to 22 mm. This information is helpful in developing policies around crab fishing to maintain the health of the crab population and to set expectations for the recreational fisher.

## 17.5.3. Example: Predicting the Incremental Growth of a Crab#

After Dungeness crabs mature, they continue to grow by casting off their shell and building a new, larger one to grow into each year; this process is called molting. The California Department of Fish and Wildlife, wanted a better understanding of crab growth so that they could set better limits on fishing that would protect the crab population. The crabs caught in the study mentioned in the previous example were about to molt, and in addition to their size, the change in shell size from before to after molting was also recorded:

crabs.corr()

shell inc
shell 1.0 -0.6
inc -0.6 1.0

These two measurements are negatively correlated, meaning that the larger the crab, the less they grow when they molt. We plot the growth increment against the shell size to determine whether the relationship between these variables is roughly linear:

px.scatter(crabs, y='inc', x= 'shell', width=350, height=250,
labels=dict(shell='Dungeness crab shell width (mm)',
inc='Growth (mm)'),)


The relationship appears linear, and we can fit a simple linear model to explain the growth increment by the pre-molt size of the shell. For this example, we use the statsmodels library, which provides prediction intervals with get_prediction. We first set up the design matrix and response variable, and then we use least squares to fit the model:

import statsmodels.api as sm

y = crabs['inc']

inc_model = sm.OLS(y, X).fit()

print(f"Increment estimate = {inc_model.params[0]:0.2f} + ",
f"{inc_model.params[1]:0.2f} x Shell Width")

Increment estimate = 29.80 +  -0.12 x Shell Width


When modeling, we create prediction intervals for given values of the explanatory variable. For example, if a newly caught crab is 120 mm across, then we use our fitted model to predict its shell’s growth.

As in the previous example, the variability of our prediction for an individual observation includes the variability in our estimate of the crab’s growth and the crab-to-crab variation in shell size. Again, we can use the bootstrap to estimate this variation, or we can use probability theory to show that these two sources of variation combine as follows:

$SD(\mathbf{e}) \sqrt{1 + \mathbf{x}_0 (\textbf{X}^\top \textbf{X})^{-1}\mathbf{x}_0^\top}$

Here $$\textbf{X}$$ is the design matrix that consists of the original data, $$\mathbf{e}$$ is the $$n \times 1$$ column vector of residuals from the regression, and $$\mathbf{x}_0$$ is the $$1 \times (p + 1)$$ row vector of features for the new observation (in this example, these are $$\left[1, 120\right]$$):

new_data = dict(const=1, shell=120)
new_X = pd.DataFrame(new_data, index=[0])
new_X

const shell
0 1 120

We use the get_prediction method in statsmodels to find a 95% prediction interval for a crab with a 120 mm shell:

pred = inc_model.get_prediction(new_X)
pred.summary_frame(alpha=0.05)

mean mean_se mean_ci_lower mean_ci_upper obs_ci_lower obs_ci_upper
0 15.86 0.12 15.63 16.08 12.48 19.24

Here we have both a confidence interval for the average growth increment for a crab with a 120 mm shell, [15.6, 16.1], and a prediction interval for the growth increment, [12.5, 19.2]. The prediction interval is quite a bit wider because it takes into account the variation in individual crabs. This variation is seen in the spread of the points about the regression line, which we approximate by the SD of the residuals. The correlation between shell size and growth increment means that the variation in a growth increment prediction for a particular shell size is smaller than the overall SD of the growth increment:

print(f"Residual SD:    {np.std(inc_model.resid):0.2f}")
print(f"Crab growth SD: {np.std(crabs['inc']):0.2f}")

Residual SD:    1.71
Crab growth SD: 2.14


The intervals provided by get_prediction rely on the normal approximation to the distribution of growth increment. That’s why the 95% prediction interval endpoints are roughly twice the residual SD away from the prediction. In the next section, we dive deeper into these calculations of standard deviations, estimators, and predictions. We also discuss some of the assumptions that we make in calculating them.