Log Transformation for an Exponential Mean

Why Transform the Exponential Sample Mean?

For an exponential distribution, the sample mean estimates a positive population mean, but its uncertainty depends on the value of that mean. Taking the logarithm changes the scale: the delta method shows that the transformed estimator has an asymptotic variance that does not depend on the unknown mean. This makes it possible to build a confidence interval on the log scale and then express it in the original units.

The interval developed here is an asymptotic confidence interval. Its coverage is justified by a limiting normal distribution as the sample size grows; it is not generally an exact finite-sample interval.

Model and notation: Let $X_1,\ldots,X_n$ be independent and identically distributed observations with exponential density $f(x;\mu)=\mu^{-1}\exp(-x/\mu)$ for $x>0$, where $\mu>0$ is the population mean. Write $\bar X=n^{-1}\sum_{i=1}^n X_i$. For this parameterization, $E[X_i]=\mu$ and $\operatorname{Var}(X_i)=\mu^2$.

Limiting Distribution on the Log Scale

The central limit theorem gives the large-sample distribution of $\bar X$ around $\mu$. Applying the logarithm changes the scale of its fluctuations. Since the derivative of $g(x)=\log x$ at $x=\mu$ is $1/\mu$, the first-order change in $\log \bar X$ is approximately $(\bar X-\mu)/\mu$. The following derivation makes that approximation precise.

Theorem: Under the stated independent, identically distributed exponential model with fixed $\mu>0$, $$ \sqrt n\bigl(\log\bar X-\log\mu\bigr)\xrightarrow{d}N(0,1). $$ Thus the asymptotic variance of $\log\bar X$, to first order, is $1/n$, and its asymptotic standard error is $1/\sqrt n$.

Proof. The observations have finite mean $\mu$ and variance $\mu^2$, so the central limit theorem gives

\sqrt n(\bar X-\mu)\xrightarrow{d}N(0,\mu^2).

Also, Chebyshev’s inequality gives, for every $\varepsilon>0$,

P(|\bar X-\mu|>\varepsilon) \leq \frac{\operatorname{Var}(\bar X)}{\varepsilon^2} =\frac{\mu^2}{n\varepsilon^2}\longrightarrow 0.

Therefore $\bar X\xrightarrow{p}\mu$. Each $X_i$ is positive with probability one, so $\bar X>0$ with probability one and its logarithm is defined. By the mean value theorem, for each sample there is a value $\xi_n$ between $\bar X$ and $\mu$ such that

\log\bar X-\log\mu =\frac{1}{\xi_n}(\bar X-\mu).

Because $\bar X\xrightarrow{p}\mu$, we have $\xi_n\xrightarrow{p}\mu$, and hence $1/\xi_n\xrightarrow{p}1/\mu$. Slutsky’s theorem now gives

\sqrt n(\log\bar X-\log\mu) =\frac{1}{\xi_n}\sqrt n(\bar X-\mu) \xrightarrow{d}N(0,1).

The limiting variance is $1$ for the scaled quantity. Equivalently, the first-order variance approximation for the unscaled log estimator is $1/n$. This is an asymptotic variance statement, not a claim that $\operatorname{Var}(\log\bar X)$ equals $1/n$ exactly at every sample size. $\square$

Worked Example: Reading the Asymptotic Standard Error

Suppose the exponential mean is $\mu=4$ hours and the sample size is $n=25$. The sample mean has variance $\mu^2/n=16/25=0.64$, so its standard error is $0.8$ hours. For the logarithm, the first-order variance approximation is

\operatorname{Var}(\log\bar X)\approx\frac{1}{n} =\frac{1}{25}=0.04, \qquad \operatorname{SE}(\log\bar X)\approx\sqrt{0.04}=0.2.

The approximate standard error $0.2$ does not depend on whether the mean is 4 hours or another positive value. The logarithm changes the measurement scale, so this standard error is in log units, not hours.

Constructing a Confidence Interval

Let $0<\alpha<1$, and let $z_{1-\alpha/2}$ denote the $1-\alpha/2$ quantile of the standard normal distribution. The limiting distribution implies that the standardized log difference is approximately between $-z_{1-\alpha/2}$ and $z_{1-\alpha/2}$. Solving those inequalities for $\mu$ gives a positive interval.

Asymptotic confidence interval: An approximate two-sided $100(1-\alpha)\%$ confidence interval for $\mu$ is $$ \left[ \bar X\exp\left(-\frac{z_{1-\alpha/2}}{\sqrt n}\right), \; \bar X\exp\left(\frac{z_{1-\alpha/2}}{\sqrt n}\right) \right]. $$ The lower endpoint is the smaller one, since $z_{1-\alpha/2}>0$ for the usual two-sided confidence levels.

Coverage justification. The theorem gives $T_n=\sqrt n(\log\bar X-\log\mu)\xrightarrow{d}N(0,1)$. The event that the interval contains $\mu$ is exactly

\bar X\exp\left(-\frac{z_{1-\alpha/2}}{\sqrt n}\right) \leq \mu \leq \bar X\exp\left(\frac{z_{1-\alpha/2}}{\sqrt n}\right) \quad\Longleftrightarrow\quad |T_n|\leq z_{1-\alpha/2}.

The standard normal distribution is continuous, so convergence in distribution implies

P_\mu(\mu\text{ is in the interval}) =P(|T_n|\leq z_{1-\alpha/2}) \longrightarrow P(|Z|\leq z_{1-\alpha/2}) =1-\alpha, \qquad Z\sim N(0,1).

This establishes the interval’s asymptotic coverage. It does not establish exact coverage for a particular finite $n$. $\square$

Worked Example: A 95% Interval With 36 Observations

An independent exponential sample has $n=36$ observations and sample mean $\bar x=2.4$. Construct the large-sample 95% confidence interval for the population mean. For 95% confidence, $z_{0.975}\approx1.96$, so

\frac{z_{0.975}}{\sqrt n} =\frac{1.96}{6} \approx0.3267.

The multiplicative factors are

\exp(-0.3267)\approx0.7213, \qquad \exp(0.3267)\approx1.3864.

Multiplying the lower factor by $2.4$ gives $2.4(0.7213)\approx1.731$. Multiplying the upper factor by $2.4$ gives $2.4(1.3864)\approx3.327$. Thus the interval is approximately $[1.731,\,3.327]$, in the same units as the observations. Its endpoints are positive, and the upper endpoint exceeds the lower endpoint.

Multiplicative Interpretation

The interval’s two endpoints are the sample mean multiplied by reciprocal factors. If $c=\exp(z_{1-\alpha/2}/\sqrt n)>1$, it can be written $[\bar X/c,\,\bar Xc]$. It is therefore symmetric around $\log\bar X$ on the log scale, but generally not symmetric around $\bar X$ on the original scale. The ratio of the upper endpoint to the lower endpoint is $c^2$, and both endpoints remain positive.

Worked Example: Interpreting the Interval as a Multiplicative Range

For a 90% interval with $n=64$ and observed mean $\bar x=3$, use $z_{0.95}\approx1.645$. The log-scale margin is

\frac{1.645}{\sqrt{64}}=\frac{1.645}{8}=0.205625.

The factors are $\exp(-0.205625)\approx0.8141$ and $\exp(0.205625)\approx1.2283$. Thus the interval is

[3(0.8141),\,3(1.2283)] \approx[2.442,\,3.685].

The observed mean is multiplied by about $0.8141$ for the lower endpoint and $1.2283$ for the upper endpoint. These factors are reciprocals apart from rounding, which illustrates the interval’s symmetry after taking logarithms.

Common Pitfalls and Practical Meaning

A common mistake is to treat the two endpoints as $[\bar Xc,\bar X/c]$ when $c>1$. That reverses their order: $\bar Xc$ is larger than $\bar X/c$. The correctly ordered interval is $[\bar X/c,\bar Xc]$, or equivalently the exponential-endpoint form given above.

Another mistake is to report $1/\sqrt n$ as the exact standard error of $\log\bar X$. It is the asymptotic standard error under this model. The finite-sample distribution of the logarithm need not be normal, and the derivation does not make its variance exactly $1/n$. Similarly, the interval’s stated confidence level describes limiting coverage, not a guarantee of exact coverage at every sample size.

The practical benefit is a simple uncertainty calculation for a positive mean that respects its parameter space. The interval is easy to compute from $\bar X$ and $n$, and its relative spread is governed by $z_{1-\alpha/2}/\sqrt n$, rather than by the unknown value of $\mu$. Whether the normal approximation is adequate for a particular sample size remains a separate practical consideration.

Check Your Understanding

Try these questions using the model and interval derived above.

For an exponential distribution parameterized by its mean $\mu$, what are $E[X_i]$ and $\operatorname{Var}(X_i)$?
Why does the first-order asymptotic variance of $\log\bar X$ not depend on $\mu$?
Write the asymptotic 90% confidence interval for $\mu$ in terms of $\bar X$, $n$, and $z_{0.95}$.
If $c=\exp(z_{1-\alpha/2}/\sqrt n)>1$, which is the correctly ordered interval: $[\bar X/c,\bar Xc]$ or $[\bar Xc,\bar X/c]$? Explain.
Does the limiting normal argument guarantee exact $1-\alpha$ coverage for every finite sample size? Why or why not?