Delta Method with Ratio that Cancels 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.
 A: 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. 
