Midterm 2 Review Session

Published

Last modified: 2026-06-09: 7:57:49 (UTC)

Configuring R

Functions from these packages will be used throughout this document:

Show R code
library(conflicted) # check for conflicting function definitions
# library(printr) # inserts help-file output into markdown output
library(rmarkdown) # Convert R Markdown documents into a variety of formats.
library(pander) # format tables for markdown
library(ggplot2) # graphics
library(ggfortify) # help with graphics
library(dplyr) # manipulate data
library(tibble) # `tibble`s extend `data.frame`s
library(magrittr) # `%>%` and other additional piping tools
library(haven) # import Stata files
library(knitr) # format R output for markdown
library(tidyr) # Tools to help to create tidy data
library(plotly) # interactive graphics
library(dobson) # datasets from Dobson and Barnett 2018
library(parameters) # format model output tables for markdown
library(haven) # import Stata files
library(latex2exp) # use LaTeX in R code (for figures and tables)
library(fs) # filesystem path manipulations
library(survival) # survival analysis
library(survminer) # survival analysis graphics
library(KMsurv) # datasets from Klein and Moeschberger
library(parameters) # format model output tables for
library(webshot2) # convert interactive content to static for pdf
library(forcats) # functions for categorical variables ("factors")
library(stringr) # functions for dealing with strings
library(lubridate) # functions for dealing with dates and times
library(broom) # Summarizes key information about statistical objects in tidy tibbles
library(broom.helpers) # Provides suite of functions to work with regression model 'broom::tidy()' tibbles

Here are some R settings I use in this document:

Show R code
rm(list = ls()) # delete any data that's already loaded into R

conflicts_prefer(dplyr::filter)
ggplot2::theme_set(
  ggplot2::theme_bw() + 
        # ggplot2::labs(col = "") +
    ggplot2::theme(
      legend.position = "bottom",
      text = ggplot2::element_text(size = 12, family = "serif")))

knitr::opts_chunk$set(message = FALSE)
options('digits' = 6)

panderOptions("big.mark", ",")
pander::panderOptions("table.emphasize.rownames", FALSE)
pander::panderOptions("table.split.table", Inf)
conflicts_prefer(dplyr::filter) # use the `filter()` function from dplyr() by default
legend_text_size = 9
run_graphs = TRUE

1 Introduction

This chapter walks through the most common mistakes that students made on Midterm 2, which covered survival analysis and Cox proportional hazards models. Each section states a mistake, explains why it is wrong, and works through the correct approach.

The exam had two parts:

  • Part 1 (Kaplan-Meier and Nelson-Aalen estimators) used a small data set of survival times: 10, 14, 14+, 15+, 18, 21, 25 months (the + marks censored observations).

  • Part 2 (Cox proportional hazards regression) used a model fit to \(n = 3{,}142\) men from the Western Collaborative Group Study (WCGS), relating time to incident coronary heart disease (CHD) to baseline cigarette smoking and other covariates.

Tip

The mistakes below are organized so that the most consequential and most frequent errors come first within each part. If you only have time to review a few things, start at the top of each part.

2 Part 1: Kaplan-Meier and Nelson-Aalen

Show R code
library(survival)
library(dplyr)

surv_data <- tibble(
  time  = c(10, 14, 14, 15, 18, 21, 25),
  death = c(1,  1,  0,  0,  1,  1,  1),
  surv  = Surv(time, death)
)

Here is the full table that everyone was trying to reproduce; we will refer back to it throughout this part.

Show R code
KM_est <- survfit(surv ~ 1, type = "kaplan-meier", data = surv_data)

surv_table <-
  KM_est |>
  summary(censored = TRUE, data.frame = TRUE) |>
  as_tibble() |>
  mutate(
    hazard       = n.event / n.risk,
    nonhazard    = 1 - hazard,
    cusum_hazard = cumsum(hazard),
    surv_KM      = cumprod(nonhazard),
    surv_NA      = exp(-cusum_hazard)
  ) |>
  select(time, n.risk, n.event, hazard, surv_KM, cusum_hazard, surv_NA)

surv_table |> pander::pander()
Table 1: Kaplan-Meier and Nelson-Aalen calculations for the Part 1 data
time n.risk n.event hazard surv_KM cusum_hazard surv_NA
10 7 1 0.1429 0.8571 0.1429 0.8669
14 6 1 0.1667 0.7143 0.3095 0.7338
15 4 0 0 0.7143 0.3095 0.7338
18 3 1 0.3333 0.4762 0.6429 0.5258
21 2 1 0.5 0.2381 1.143 0.3189
25 1 1 1 0 2.143 0.1173

2.1 The risk set shrinks by the number of subjects who leave

Exam reference: Exercise 1.1 (6 points) — compute the KM and Nelson-Aalen estimates of \(S(t)\).

WarningCommon mistake

Reducing the number at risk by one at every row, regardless of how many subjects actually left the study. With this mistake, the number at risk after the two censored subjects (at \(t = 14^+\) and \(t = 15^+\)) was computed as \(5\) instead of \(4\).

Solution. The number at risk \(n_j\) is the number of subjects still under observation just before time \(t_j\) — that is, everyone who has neither had the event nor been censored yet.

Track the cohort of 7:

time what happens number at risk before this time
10 1 death 7
14 1 death and 1 censored 6
15 1 censored 4
18 1 death 3
21 1 death 2
25 1 death 1

At \(t = 14\), one subject dies and a different subject is censored (the 14+), so two subjects leave the risk set; the count drops from 6 to 4, not from 6 to 5. Every subject who has an event or is censored is removed from all later risk sets.

2.2 Nelson-Aalen estimates the cumulative hazard first, then exponentiates

Exam reference: Exercise 1.1 (6 points) — compute the KM and Nelson-Aalen estimates of \(S(t)\).

WarningCommon mistake

Two versions of this error were common:

  1. Computing a running sum of arbitrary quantities (e.g. summing survival probabilities, or summing \(1 - \hat\lambda_j\)) instead of summing the hazards \(\hat\lambda_j = d_j / n_j\).

  2. Computing the cumulative hazard correctly but then forgetting the final step, \(\hat S_{NA}(t) = \operatorname{exp}\mathopen{}\left\{-\hat\Lambda(t)\right\}\mathclose{}\), and reporting the cumulative hazard itself (or the KM product) as the Nelson-Aalen survival estimate.

Solution. The Nelson-Aalen estimator builds the cumulative hazard by adding up the instantaneous hazard contributions at each event time:

\[\hat\Lambda(t) = \sum_{j: t_j \le t} \frac{d_j}{n_j}\]

and then converts it to a survival estimate using \(\operatorname{S}(t) = \operatorname{exp}\mathopen{}\left\{-{\Lambda}(t)\right\}\mathclose{}\):

\[\hat S_{NA}(t) = \operatorname{exp}\mathopen{}\left\{-\hat\Lambda(t)\right\}\mathclose{}\]

For example, at \(t = 10\):

\[\hat\Lambda(10) = \frac{1}{7} = 0.143, \qquad \hat S_{NA}(10) = \operatorname{exp}\mathopen{}\left\{-0.143\right\}\mathclose{} = 0.867\]

Contrast this with Kaplan-Meier, which multiplies conditional survival probabilities:

\[\hat S_{KM}(t) = \prod_{j: t_j \le t}\left(1 - \frac{d_j}{n_j}\right), \qquad \hat S_{KM}(10) = 1 - \tfrac{1}{7} = 0.857\]

The two estimators are close but not identical, and \(\hat S_{NA}(t) \ge \hat S_{KM}(t)\) always. Reporting the same numbers for both, or reporting \(\hat\Lambda\) where \(\hat S_{NA}\) was asked for, loses credit.

2.3 Survival estimates are step functions: use the most recent event time

Exam reference: Exercise 1.3 (1 point) — the KM and NA survival estimates for \(t = 17\) months.

WarningCommon mistake

For \(\hat S(17)\), looking at the row for \(t = 18\) (the next event after 17), or interpolating between rows. Some answers gave only a verbal description (“it’s flat there”) without the numeric value.

Solution. \(\hat S(t)\) is a right-continuous step function: it only changes at observed event times and stays flat in between. To evaluate it at \(t = 17\), find the most recent event time at or before 17, which is \(t = 14\) (the subsequent \(t = 15^+\) is a censoring, not an event, so the curve does not step there). So:

\[\hat S_{KM}(17) = \hat S_{KM}(14) = 0.714, \qquad \hat S_{NA}(17) = \hat S_{NA}(14) = 0.734\]

Give the actual numbers, not just “it’s flat.”

2.4 Median survival time is read off the curve, not modeled

Exam reference: Exercise 1.4 (1 point) — the KM and NA estimates of median survival time.

WarningCommon mistake

Estimating the median by fitting an exponential model (\(\hat\lambda = \text{events} / \text{total follow-up}\), then \(\hat{\text{E}}[T] = 1/\hat\lambda\)). That computes an exponential-model mean, not the nonparametric median, and answers a different question. A second error was reading the median off the wrong curve or the wrong row.

Solution. The median survival time is the smallest \(t\) at which the estimated survival curve drops to (or below) \(0.5\):

\[\hat t_{\text{median}} = \min\{t : \hat S(t) \le 0.5\}\]

Reading from Table 1:

  • \(\hat S_{KM}(t)\) first reaches \(\le 0.5\) at \(t = 18\) (where it drops to \(0.476\)), so the KM median is 18 months.
  • \(\hat S_{NA}(t)\) first reaches \(\le 0.5\) at \(t = 21\) (where it drops to \(0.319\)), so the NA median is 21 months.

We can confirm this in R:

Show R code
quantile(KM_est, p = 0.5)$quantile
#> 50 
#> 18

No distributional assumption (exponential or otherwise) is used: this is a nonparametric read-off of the estimated curve.

2.5 How censored times are used: denominator yes, numerator no

Exam reference: Exercise 1.5 (2 points) — describe how the censored time(s) are utilized in the Kaplan-Meier estimate.

WarningCommon mistake

Describing only the symptom (“the curve doesn’t drop at a censoring time”) without the mechanism, or implying that censored subjects are simply dropped from the analysis entirely.

Solution. A censored subject contributes to the risk-set denominators (\(n_j\)) for every event time up to and including their censoring time, because we know they survived event-free until then. They never contribute to an event numerator (\(d_j\)), because no event was observed for them.

In this example, subjects #3 and #4 (censored at \(14^+\) and \(15^+\)) are counted in the number at risk at \(t = 10\) and \(t = 14\), which is why those denominators are 7 and 6. After their censoring times they drop out of all later risk sets. This is exactly what lets Kaplan-Meier use partial information from censored subjects instead of discarding them.

3 Part 2: Cox proportional hazards models

For reference, here is the fitted model from the exam. The primary exposure is baseline smoking category (nonsmoker = reference, light smoker \(L=1\), heavy smoker \(H=1\)), adjusting for age in decades (\(A\)), behavior pattern (\(P\)), overweight (\(B\)), and high cholesterol (\(C\)).

Table 2: Cox PH regression for the WCGS CHD data
Characteristic \(\log(\widehat{HR})\) \(\widehat{HR}\) 95% CI for HR \(p\)
nonsmoker (ref.) – – – –
light smoker (\(L\)) 0.36 1.43 (1.05, 1.95) 0.023
heavy smoker (\(H\)) 0.78 2.18 (1.63, 2.91) < 0.001
age per decade (\(A\)) 0.66 1.94 (1.56, 2.40) < 0.001
behavior type A (\(P\)) 0.73 2.06 (1.58, 2.70) < 0.001
overweight (\(B\)) 0.30 1.34 (1.05, 1.72) 0.019
high cholesterol (\(C\)) 0.75 2.12 (1.65, 2.71) < 0.001

The exam also provided the estimated covariance matrix of the coefficient estimates (on the \(\log(HR)\) scale). The entries we need below are:

\[{\operatorname{Var}}(\hat\beta_L) = 0.024840, \quad {\operatorname{Var}}(\hat\beta_H) = 0.021905, \quad \operatorname{Cov}(\hat\beta_L, \hat\beta_H) = 0.010504\]

3.1 Writing the model: name every assumption and show where it is used

Exam reference: Exercise 2.1 (10 points) — write the mathematical form of the proportional hazards model corresponding to Table 1.

WarningCommon mistake

Several patterns lost the most points here:

  • Treating the question as “interpret the coefficients” and writing only the linear predictor, omitting the likelihood, the distribution functions, and the assumptions.
  • Listing assumption names without showing where they are used, or writing the math without naming the assumption.
  • Adding a spurious “baseline hazard is exponential” assumption. The Cox model leaves the baseline hazard \(\lambda_0(t)\) unspecified; assuming a parametric form is wrong.
  • Omitting one or more of the standalone distribution functions (survival, density, hazard, log-hazard, cumulative hazard).

Solution. A full-credit answer connects the likelihood to the linear predictor through a chain of named components and assumptions. Using \(\tilde X = (L, H, A, P, B, C)\):

Joint likelihood of the data set: \[\mathscr{L}\stackrel{\text{def}}{=}\operatorname{p}(\tilde Y = \tilde y, \tilde D = \tilde d \mid \mathbf X = \mathbf x)\]

Marginal likelihood contribution of observation \(i\): \[\mathscr{L}_i \stackrel{\text{def}}{=}\operatorname{p}(Y_i = y_i, D_i = d_i \mid \tilde X_i = \tilde x_i)\]

Independent-observations assumption (used to factor the likelihood): \[\mathscr{L}= \prod_{i=1}^n \mathscr{L}_i\]

Non-informative censoring assumption (used so each contribution reduces to survival and hazard terms): \(T_i \perp\!\!\!\perp C_i \mid \tilde X_i\), giving \[\mathscr{L}_i \propto \left[\operatorname{f}(y_i \mid \tilde x_i)\right]^{d_i} \left[\operatorname{S}(y_i \mid \tilde x_i)\right]^{1 - d_i} = \operatorname{S}(y_i \mid \tilde x_i)\cdot\left[{\lambda}(y_i \mid \tilde x_i)\right]^{d_i}\]

Distribution functions (define each one):

\[\operatorname{S}(t \mid \tilde x) \stackrel{\text{def}}{=}\operatorname{P}(T > t \mid \tilde X = \tilde x) = \operatorname{exp}\mathopen{}\left\{-{\Lambda}(t \mid \tilde x)\right\}\mathclose{}\]

\[\operatorname{f}(t \mid \tilde x) \stackrel{\text{def}}{=}{\lambda}(t \mid \tilde x)\,\operatorname{S}(t \mid \tilde x)\]

\[{\lambda}(t \mid \tilde x) \stackrel{\text{def}}{=}\operatorname{p}(T = t \mid T \ge t, \tilde X = \tilde x) = \frac{\operatorname{f}(t \mid \tilde x)}{\operatorname{S}(t \mid \tilde x)}\]

\[{\Lambda}(t \mid \tilde x) \stackrel{\text{def}}{=}\int_0^t {\lambda}(u \mid \tilde x)\,du = -\log\operatorname{S}(t \mid \tilde x)\]

\[\eta(t \mid \tilde x) \stackrel{\text{def}}{=}\log{\lambda}(t \mid \tilde x)\]

Proportional-hazards assumption (used to split the hazard into a baseline that depends only on time and a factor that depends only on covariates): \[{\lambda}(t \mid \tilde x) = \lambda_0(t)\cdot\theta(\tilde x)\] where \(\lambda_0(t)\) is the unspecified baseline hazard.

Logarithmic-link assumption (used to make the covariate factor a function of a linear predictor): \[\eta(t \mid \tilde x) = \eta_0(t) + \Delta\eta(\tilde x), \qquad \theta(\tilde x) = \operatorname{exp}\mathopen{}\left\{\Delta\eta(\tilde x)\right\}\mathclose{}\]

Linear functional-form assumption (used to write the covariate term as a linear combination): \[\Delta\eta(\tilde x) = \tilde x \cdot \tilde\beta = \beta_L\, l + \beta_H\, h + \beta_A\, a + \beta_P\, p + \beta_B\, b + \beta_C\, c\]

Notice that the baseline hazard \(\lambda_0(t)\) is carried along symbolically the whole time — we never assume a shape for it.

3.2 Interpreting a hazard ratio: magnitude, reference, adjustment, and significance

Exam reference: Exercise 2.2 (4 points) — summarize how baseline smoking category is associated with hazard of incident CHD.

WarningCommon mistake
  • Reporting the direction of the effect but omitting the magnitude (“X% higher/lower hazard”) or the reference group.
  • Forgetting to say the estimate is adjusted for / holding constant the other covariates.
  • Omitting statistical significance (whether the 95% CI excludes 1, or \(p < 0.05\)).
  • Calling the hazard ratio a risk or an odds ratio. It is a ratio of hazards, not of risks or odds.

Solution. A complete interpretation has four parts: magnitude, reference group, adjustment, and significance. For the smoking effect in Table 2:

Adjusting for age, behavior pattern, overweight, and cholesterol, light smokers had an estimated hazard of incident CHD about 43% higher than nonsmokers (\(\widehat{HR} = 1.43\)); this is statistically significant at the 0.05 level, since the 95% CI \((1.05, 1.95)\) excludes 1 (\(p = 0.023\)).

Heavy smokers had an estimated hazard about 118% higher (roughly double; \(\widehat{HR} = 2.18\)) than nonsmokers, also adjusting for the other covariates; this is highly significant (\(p < 0.001\), CI \((1.63, 2.91)\)).

Note “43% higher” comes from \(1.43 - 1 = 0.43\), and “118% higher” from \(2.18 - 1 = 1.18\). Interpret each smoking level separately against the reference, rather than lumping them together.

3.3 A hazard ratio for a non-unit change: exponentiate

Exam reference: Exercise 2.3 (2 points) — the hazard ratio associated with a 7.5-year increase in age.

WarningCommon mistake

For the hazard ratio associated with a 7.5-year increase in age (with the table reporting the HR per decade), the errors were:

  • Multiplying: \(1.94 \times 0.75 = 1.46\). Wrong — the HR for a multi-unit change is not linear in the HR.
  • Arithmetic slips that produced answers like 2.08.

Solution. On the log scale, the effect is linear: a change of \(c\) units multiplies the log-hazard by \(c \cdot \beta\). So the hazard ratio for a \(c\)-unit change is

\[HR(c) = \operatorname{exp}\mathopen{}\left\{c\,\beta\right\}\mathclose{} = \left(\operatorname{exp}\mathopen{}\left\{\beta\right\}\mathclose{}\right)^c = HR^{\,c}\]

A 7.5-year increase is \(c = 0.75\) decades, and the per-decade \(\widehat{HR} = 1.94\) (i.e. \(\hat\beta_A = \log 1.94 = 0.66\)):

\[\widehat{HR}(0.75) = 1.94^{\,0.75} = \operatorname{exp}\mathopen{}\left\{0.75 \times 0.66\right\}\mathclose{} = \operatorname{exp}\mathopen{}\left\{0.495\right\}\mathclose{} = 1.64\]

So a 7.5-year-older man has about 64% higher estimated hazard of CHD, all else equal. The operation is exponentiation (\(HR^{0.75}\)), not multiplication.

3.4 Comparing two coefficients: variance of a difference

Exam reference: Exercise 2.4 (2 points) — test whether the hazard of incident CHD differs between heavy and light smokers; compute the z-statistic and two-sided p-value.

WarningCommon mistake

This was the single most consequential Part 2 error. To test \(H_0: \beta_H = \beta_L\), the standard error of \(\hat\beta_H - \hat\beta_L\) was computed incorrectly in several ways:

  • Using a single covariance entry \(\operatorname{Cov}(\hat\beta_L, \hat\beta_H)\) as the variance.
  • Adding \(2\,\operatorname{Cov}\) instead of subtracting it.
  • Plugging in the hazard ratios (\(2.18\), \(1.43\)) instead of the coefficients (\(0.78\), \(0.36\)) in the numerator.
  • Guessing a value for \(z\) because the variance-of-a-difference formula was not on the formula sheet.

Solution. Work on the \(\log(HR)\) (coefficient) scale, where the estimates are approximately normal. The point estimate of the difference is

\[\widehat{\Delta} = \hat\beta_H - \hat\beta_L = 0.78 - 0.36 = 0.42\]

The variance of a difference of two estimates uses all three relevant entries of the covariance matrix:

\[{\operatorname{Var}}(\hat\beta_H - \hat\beta_L) = {\operatorname{Var}}(\hat\beta_H) + {\operatorname{Var}}(\hat\beta_L) - 2\,\operatorname{Cov}(\hat\beta_H, \hat\beta_L)\]

The \(-2\,\operatorname{Cov}\) term is essential — its sign is minus for a difference. Plugging in:

Show R code
var_H <- 0.021905
var_L <- 0.024840
cov_HL <- 0.010504

diff <- 0.78 - 0.36
var_diff <- var_H + var_L - 2 * cov_HL
se_diff <- sqrt(var_diff)

z <- diff / se_diff
p_value <- 2 * pnorm(-abs(z))

c(diff = diff, se = se_diff, z = z, p_value = p_value) |> round(4)
#>    diff      se       z p_value 
#>  0.4200  0.1604  2.6180  0.0088

So \(z = 0.42 / 0.160 = 2.62\), two-sided \(p = 0.0088\). We reject \(H_0\): the hazard of CHD differs significantly between heavy and light smokers, holding the other covariates constant.

Note

Compare with the naive standard error that ignores the covariance, \(\sqrt{{\operatorname{Var}}(\hat\beta_H) + {\operatorname{Var}}(\hat\beta_L)} = 0.216\). Because \(\hat\beta_H\) and \(\hat\beta_L\) are positively correlated, subtracting \(2\,\operatorname{Cov}\) makes the correct SE (0.160) smaller, which sharpens the test. Using a single covariance cell as the variance gave a wildly wrong SE and an absurd \(z\).

3.5 A confidence interval for a hazard ratio: build it on the log scale, then exponentiate

Exam reference: Exercise 2.5 (2 points) — a 95% confidence interval for the hazard ratio comparing heavy to light smokers.

WarningCommon mistake
  • Forming the interval directly on the HR scale, \(\widehat{HR} \pm 1.96 \cdot \text{SE}\) (e.g. \(1.52 \pm 0.31\)), instead of on the log scale.
  • Building the log-scale interval correctly but never exponentiating, then reporting a log-scale interval as if it were the HR (sometimes even declaring “significant” while the reported interval contained 1).
  • Reusing the wrong SE from the Wald-test mistake above.

Solution. A hazard ratio is positive and its sampling distribution is skewed, so the symmetric normal approximation is applied to \(\log(HR)\), and the endpoints are then exponentiated. For the heavy-vs-light comparison (\(\widehat\Delta = 0.42\), \(\text{SE} = 0.160\)):

\[\text{log-scale CI:}\quad 0.42 \pm 1.96\times 0.160 = (0.106,\ 0.734)\]

\[\text{HR CI:}\quad \left(e^{0.106},\ e^{0.734}\right) = (1.11,\ 2.08)\]

Show R code
est <- 0.42
se  <- sqrt(0.021905 + 0.024840 - 2 * 0.010504)
exp(est + c(-1, 1) * 1.96 * se) |> round(2)
#> [1] 1.11 2.08

The point estimate is \(\widehat{HR} = e^{0.42} = 1.52\), and the 95% CI \((1.11, 2.08)\) excludes 1, consistent with the significant Wald test in Section 3.4. Always exponentiate the endpoints, and check that your CI and your hypothesis test agree.

4 Summary of the most common errors

# Exam Q Topic The fix
1 1.1 Risk set after censoring (Section 2.1) Remove everyone who has an event or is censored from later risk sets.
2 1.1 Nelson-Aalen (Section 2.2) Sum hazards \(d_j/n_j\), then \(\hat S_{NA} = \operatorname{exp}\mathopen{}\left\{-\hat\Lambda\right\}\mathclose{}\).
3 1.3 Reading \(\hat S(t)\) (Section 2.3) Step function: use the most recent event time \(\le t\); give the number.
4 1.4 Median survival (Section 2.4) First \(t\) with \(\hat S(t) \le 0.5\) — not a mean, not an exponential fit.
5 1.5 Role of censoring (Section 2.5) Denominator until censoring; never the event numerator.
6 2.1 Writing the model (Section 3.1) Name and apply every assumption; baseline hazard stays unspecified.
7 2.2 Interpreting an HR (Section 3.2) Magnitude + reference + adjustment + significance; HR \(\ne\) risk/odds.
8 2.3 HR for a non-unit change (Section 3.3) \(HR^{\,c} = \operatorname{exp}\mathopen{}\left\{c\beta\right\}\mathclose{}\) — exponentiate, don’t multiply.
9 2.4 Comparing two coefficients (Section 3.4) \({\operatorname{Var}}(\hat\beta_H - \hat\beta_L) = {\operatorname{Var}}_H + {\operatorname{Var}}_L - 2\,\operatorname{Cov}\).
10 2.5 CI for an HR (Section 3.5) Build on the log scale, then exponentiate the endpoints.
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