Say I have

Gas Consumption = Energy Used Per Hour * Heating Hours

and I want to construct a CI for Gas consumption.

Do I have to use the delta method and write:

$$\mathrm{Var}(GC) = \mathbb{E}(\textrm{Energy Used Per Hour})^2*\mathrm{Var}(\textrm{Heating Hours}) + \mathbb{E}(\textrm{Heating Hours})^2*\mathrm{Var}(\textrm{Energy Used Per Hour}) - \mathrm{Cov}(\textrm{Heating Hours, Energy Used per Hour})$$

or may I use the fact that the hours cancel in some way?

Both the heating hours and energy use per hour are random, and potentially correlated.

  • $\begingroup$ Could you explain what you mean by "hours cancel"? $\endgroup$
    – whuber
    Aug 8, 2017 at 20:34
  • $\begingroup$ @whuber -> I think the OP means that the units of measurement would cancel? However, it is not clear to me if both the heating hours and the energy user per hour are treated as random or is only energy used per hour is random? $\endgroup$ Aug 8, 2017 at 20:37
  • $\begingroup$ So both the heating hours and energy use per hour are random, and potentially correlated. $\endgroup$ Aug 8, 2017 at 20:39
  • $\begingroup$ @Lucas That's a reasonable guess, but it's rather strange to suppose that the units of measurement involved would have any effect on the result. $\endgroup$
    – whuber
    Aug 8, 2017 at 20:40
  • $\begingroup$ @user2958701 -> it is not the unit of measurement (hours) that is random rather the duration (number/quantity) of hours that is random. $\endgroup$ Aug 9, 2017 at 1:24

1 Answer 1


For the product of two estimates in a delta method you have $h(x,y)=x*y$ as your vector valued function. For this question lets assume $X$ is the Energy Used Per Hour and $Y$ is the number of hours. The gradient is

$\nabla h = \begin{bmatrix} y \\ x \\ \end{bmatrix}$ and then if your variance-covariance matrix is:

$$\begin{bmatrix} \sigma_{yy} & \sigma_{yx} \\ \sigma_{yx} & \sigma_{xx} \\ \end{bmatrix} $$

Then according to the formula for a multivariate delta method

$$\mathbb{V}ar\left(h(B)\right) \approx \nabla h(\beta)^T \cdot (\Sigma / n) \cdot \nabla h(\beta)$$

So the variance will be:

$$\begin{bmatrix} y \\ x \\ \end{bmatrix}^T \begin{bmatrix} \sigma_{yy} & \sigma_{yx} \\ \sigma_{yx} & \sigma_{xx} \\ \end{bmatrix} \begin{bmatrix} y \\ x \\ \end{bmatrix}.$$

So that the variance overall (denoted $\sigma_d$) is:

$$ \hat{\sigma}_d = \sqrt{\hat{y}^2\hat{\sigma_{xx}} + 2\hat{yx}\hat{\sigma_{xy}} + \hat{x}^2\hat{\sigma_{xx}} },$$ where $\sigma_{xy}$ denotes the covariance of the 2 estimates, and similarly for the two estimates' variances. Then the confidence interval will be

$$\hat{x}\hat{y} \pm 1.96 \times \hat{\sigma}_d.$$ Assuming you want a 95% interval.


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