Common Errors In Mean-Variance Calculations

Why Mean-Variance Errors Are Easy to Make

A mean-variance calculation can look plausible and still be wrong. Common causes include applying a formula for a proportion to a count, dropping the dispersion factor, using a mean outside the response family’s domain, or reporting a variance function as though it were the response variance. These are not cosmetic distinctions: they change the numerical variance and can make a model impossible under the stated assumptions.

A reliable check begins by identifying the random variable whose variance is requested. Is it a single binary response, a count from several trials, or a proportion? Then identify its mean parameterization and the family’s variance law. In the usual dispersion notation, the response variance is written as $\operatorname{Var}(Y)=\phi V(\mu)$, where $V$ is the variance function and $\phi$ is the dispersion parameter. The formula applies only with the response and parameterization for which $V$ was defined.

Definition: For a response $Y$ with finite mean $\mu=E[Y]$ and finite variance, a variance function $V(\mu)$ describes how variance depends on the mean in a specified family and response parameterization. Under the dispersion convention used here, the full response variance is $\operatorname{Var}(Y)=\phi V(\mu)$. The variance function alone is not generally the response variance.

Keep Counts and Proportions on Their Own Scales

A binomial count and the corresponding proportion contain the same information, but they are different random variables. Let $Y$ be the number of successes in $m$ independent Bernoulli trials, each with success probability $p$. Then $Y\sim\operatorname{Binomial}(m,p)$, its mean is $mp$, and its variance is $mp(1-p)$. The proportion $\bar{Y}=Y/m$ has mean $p$ and variance $p(1-p)/m$. The factor $m$ in one variance becomes a factor $1/m$ in the other because dividing a random variable by $m$ divides its variance by $m^2$.

Result: If $Y\sim\operatorname{Binomial}(m,p)$ and $\bar{Y}=Y/m$, then $$ E[Y]=mp,\qquad \operatorname{Var}(Y)=mp(1-p), $$ and $$ E[\bar{Y}]=p,\qquad \operatorname{Var}(\bar{Y})=\frac{p(1-p)}{m}. $$

To verify these formulas, write $Y=\sum_{j=1}^{m}X_j$, where $X_j$ is the indicator of success on trial $j$. Each $X_j$ has $E[X_j]=p$ and $\operatorname{Var}(X_j)=p(1-p)$. Independence gives $$ E[Y]=\sum_{j=1}^{m}E[X_j]=mp,\qquad \operatorname{Var}(Y)=\sum_{j=1}^{m}\operatorname{Var}(X_j)=mp(1-p). $$ Since $\bar{Y}=Y/m$, linearity of expectation and the rule $\operatorname{Var}(aY)=a^2\operatorname{Var}(Y)$ give the stated formulas. $\square$

Worked Example: A Binomial Count Is Not a Proportion

Suppose $Y$ is the number of positive tests among $m=20$ independent tests, each with probability $p=0.30$ of a positive result. Find the mean and variance of the count and of the proportion of positive tests.

For the count, $E[Y]=mp=20(0.30)=6$, and \[ \operatorname{Var}(Y)=mp(1-p)=20(0.30)(0.70)=4.2. \] For the proportion $\bar{Y}=Y/20$, $E[\bar{Y}]=p=0.30$, and \[ \operatorname{Var}(\bar{Y})=\frac{p(1-p)}{m} =\frac{(0.30)(0.70)}{20}=0.0105. \] As a check, $4.2/20^2=4.2/400=0.0105$. Using $p(1-p)=0.21$ as the count variance would be an error: that quantity is the variance of one Bernoulli response, not the variance of the count of 20 responses.

Include Dispersion, and Know What the Variance Function Describes

The variance function is only one part of the variance formula when dispersion is not fixed at one. In the convention $\operatorname{Var}(Y)=\phi V(\mu)$, omitting $\phi$ gives the wrong variance unless $\phi=1$. The response’s scale also matters: a variance function for a count is not automatically the variance function for its proportion.

Here are representative mean-variance laws, with the response parameterization made explicit. For a binomial count, $Y\sim\operatorname{Binomial}(m,p)$, the mean is $\mu=mp$ and its variance is $V_m(\mu)=\mu(1-\mu/m)$, with $\phi=1$. For a Poisson response, $V(\mu)=\mu$, also with $\phi=1$. For a Gamma response under the stated dispersion convention, $V(\mu)=\mu^2$, so the full variance is $\phi\mu^2$. The same symbol $V$ should not be detached from the family and response scale used to define it.

Worked Example: Restore the Dispersion Factor

Suppose a Gamma response has mean $\mu=5$ and dispersion $\phi=0.2$, with variance function $V(\mu)=\mu^2$. Find its response variance.

First evaluate the variance function: $V(5)=5^2=25$. Then multiply by dispersion: \[ \operatorname{Var}(Y)=\phi V(\mu)=0.2(25)=5. \] The value 25 is the variance function at this mean; it is not the response variance under this dispersion parameterization. Omitting $\phi$ would report 25 instead of 5.

Changing units provides another useful check. If $Z=cY$ for a nonzero constant $c$, then $E[Z]=c\mu$ and $\operatorname{Var}(Z)=c^2\operatorname{Var}(Y)$. Thus, when $\operatorname{Var}(Y)=\phi V_Y(\mu)$, the variance function for $Z$, expressed in terms of its own mean $\nu=E[Z]$, is $V_Z(\nu)=c^2 V_Y(\nu/c)$, with the same $\phi$. This follows by substituting $\mu=\nu/c$ into the variance of $Z$: $$ \operatorname{Var}(Z)=c^2\phi V_Y(\mu) =\phi c^2V_Y(\nu/c). $$ This identity assumes $c\ne 0$ and that the original variance law applies; it shows explicitly why a change of response scale changes the variance function. $\square$

Worked Example: Rescaling a Response Changes Its Variance

Let $Y$ be a Bernoulli response with success probability $p=0.4$, so $E[Y]=0.4$ and $\operatorname{Var}(Y)=p(1-p)=0.24$. Define $Z=100Y$, which records the response in percentage-point units. Find the mean and variance of $Z$.

The mean is $E[Z]=100E[Y]=40$. The variance is \[ \operatorname{Var}(Z)=100^2\operatorname{Var}(Y) =10{,}000(0.24)=2{,}400. \] It would be wrong to calculate $40(1-40)=-1{,}560$ by inserting the rescaled mean into the Bernoulli formula. $Z$ is not a Bernoulli response: it takes values 0 and 100. The Bernoulli variance function $p(1-p)$ applies to the original 0/1 response, not to the rescaled variable.

Check the Admissible Mean Domain

A mean must be compatible with the distribution family. For a Bernoulli response, $\mu\in[0,1]$; for a binomial count with $m$ trials, $\mu\in[0,m]$; for a Poisson response, $\mu\in[0,\infty)$. Normal means may be any real number, while Gamma and inverse-Gaussian means must be strictly positive. These are the domains of admissible means for the distributions themselves.

Boundary cases need care. A Poisson distribution with mean zero is valid: it places probability one on zero. In contrast, a finite linear predictor under the log link gives $\mu=\exp(\eta)>0$, so that particular parameterization does not attain zero at any finite $\eta$. The family’s admissible mean domain and the range generated by a particular link are related, but they are not interchangeable statements.

Worked Example: A Valid Poisson Boundary Mean

For $Y\sim\operatorname{Poisson}(\mu)$, consider $\mu=0$. At the boundary, the distribution is $P(Y=0)=1$, so $E[Y]=0$ and $\operatorname{Var}(Y)=0$. This agrees with the Poisson variance function $V(\mu)=\mu$, evaluated at zero.

Now compare this family statement with a log-link parameterization $\mu=\exp(\eta)$. For every finite real $\eta$, $\exp(\eta)>0$, so the model’s fitted mean cannot equal zero at a finite predictor. It can approach zero as $\eta$ decreases without bound, but that is not the same as attaining zero for a finite $\eta$. Thus, saying “Poisson means must be strictly positive” is too strong when describing the distribution family, though it is correct for a finite log-link predictor.

A Practical Error-Checking Sequence

Before accepting a mean-variance calculation, work through the following checks in order. They are deliberately basic: each catches a different common source of error.

Name the random variable.
State whether it is a binary response, a count, a sample proportion, or a rescaled response. Do not transfer a formula between these variables without deriving the scale change.

Identify the mean and its domain.
Check that the proposed mean is admissible for the response family. Distinguish a family boundary value from values attainable under a particular link with a finite predictor.

Use the matching variance function.
Confirm that $V(\mu)$ belongs to the same family and response parameterization as the random variable in the question.

Apply dispersion and check the scale.
Compute $\phi V(\mu)$, then check any count-to-proportion or unit conversion using $\operatorname{Var}(cY)=c^2\operatorname{Var}(Y)$.

Common Mistakes and Corrections

Using a single-trial variance for a count. The reasoning “a Bernoulli trial has variance $p(1-p)$, so the count has that variance too” overlooks that the count sums $m$ independent trials. The correction is $\operatorname{Var}(Y)=mp(1-p)$.
Using the count formula for a proportion. A proportion is $Y/m$, so its variance is the count variance divided by $m^2$, giving $p(1-p)/m$.
Dropping dispersion. Evaluating $V(\mu)$ and stopping there is wrong when the model specifies $\phi\ne1$. Multiply by $\phi$ to obtain the response variance.
Calling a variance function the full variance. The function describes mean dependence under a specified family and response scale. Use the stated dispersion convention and parameterization before interpreting it as a variance.
Excluding zero from the Poisson family’s mean domain. A Poisson variable with mean zero is degenerate at zero and is valid. A finite log-link predictor is a separate matter: it produces a strictly positive mean.
Checking only whether a formula gives a nonnegative number. A numerically nonnegative result can still come from applying the wrong family’s variance function or using an inadmissible mean. Check both the arithmetic and the model assumptions.

Key takeaway: A correct mean-variance calculation must match the random variable, response scale, family, mean domain, and dispersion. Write down each of these before substituting numbers; the final variance is not always just the variance function evaluated at the mean.

Check Your Understanding

For each question, identify the response scale and show which variance formula applies.

If $Y\sim\operatorname{Binomial}(12,0.25)$, calculate the mean and variance of $Y$, then calculate the variance of $Y/12$.
A Gamma response has mean 3, dispersion 0.5, and variance function $V(\mu)=\mu^2$. What is its response variance, and what incorrect answer results from omitting dispersion?
Is a Poisson response with mean zero valid? Can a finite real-valued predictor under a log link produce that mean? Explain the distinction.
A Bernoulli response $Y$ is replaced by $Z=10Y$. If $E[Y]=0.2$, find $E[Z]$ and $\operatorname{Var}(Z)$.
Explain why knowing $V(\mu)$ is not always enough to report the response variance.