I'm trying to find the variance of $L$, $Var(L)$, using the delta method (I want to find a closed form). $L$ is defined as: $$L = \frac{A}{B} + \frac{C}{D}$$ All $A$, $B$, $C$, and $D$ are dependent.
But I'm not sure how to proceed.
$\begingroup$
$\endgroup$
-
2$\begingroup$ The answer to stats.stackexchange.com/questions/21875/… will get you started; your problem has four variables and will therefore be more complicated, but the core approach is the same. $\endgroup$ – jbowman Apr 22 '12 at 17:03
-
1$\begingroup$ Note, too, that the delta method is an approximation, so please don't overinterpret any "closed form" result! $\endgroup$ – whuber♦ Apr 22 '12 at 21:06
$\begingroup$
$\endgroup$
For notational simplicity define $X_1=A$, $X_2=B$, $X_3=C$, $X_4=D$ and $L=f(X_1,X_2,X_3,X_4)$ with $f(x_1,x_2,x_3,x_4) = \frac{x_1}{x_2}+\frac{x_3}{x_4}$. A first order expansion around $\mu = (\mu_1, \mu_2,\mu_3, \mu_4)$ where $\mu_i=\mathbb{E}[X_i]$ shows that the following approximation holds $$L \approx f(\mu) + \sum_{i=1}^4 \partial_i f(\mu) \, (X_i-\mu_i)$$ as soon as the quantities $X_i-\mu_i$ are small enough. Since $\mathbb{E}\big[\sum_{i=1}^4 \partial_i f(\mu) \, (X_i-\mu_i) \big]=0$ it thus follows that the variance of $L$ can also be approximated by $$\textrm{var}(L) \approx \mathbb{E} \Big[ \Big\{\sum_{i=1}^4 \partial_i f(\mu) \, (X_i-\mu_i) \Big\}^2 \Big] \\ = \sum_{1 \leq i,j \leq 4} \textrm{Cov}(X_i,X_j) \partial_i f(\mu) \partial_j f(\mu).$$
