Say I have a cummulative distribution function of a random variable that depends on two variables $a_i \ \& \ a_j$:

$$ H({a}_i - {a}_j)$$

I write it's partial derivative as $h_{a_i}({a}_i - {a}_j) = \frac{\partial H({a}_i - {a}_j)}{\partial a_i}$

Now, if there's a formula for $a_i$ that depends on another variable $b_i$ and a number $\delta \in (-1;1)$ so that: $$ a_i = \frac{b_i}{1- \delta^2} $$

then I can write

$$ H( \frac{b_i}{1-\delta^2} - a_j)$$

I can now write a partial derivate $ h_{b_i}(\frac{b_i}{1- \delta^2} - {a}_j) = \frac{\partial H( \frac{b_i}{1-\delta^2} - a_j)}{\partial b_i}$

What would be the relationship between $h_{a_i}(0)$ and $h_{b_i}(0)$? (The inside of the parenthesis is zero because of something outside of the scope of the question, I just need to be able to compare the partial derivatives under the same circumstances). Does the chain rule apply here so that $h_{b_i} = h_{a_i} \cdot \frac{\partial a_i}{\partial b_i}$?


1 Answer 1


Yes, assuming that $a_i=\dfrac{b_i}{1-\delta^2}$ for all $i$, the chain rules gives

$$\begin{align}\dfrac{\partial H((b_i-b_j)/(1-\delta^2))}{\partial b_i}&=\dfrac{\partial[(b_i-b_j)/(1-\delta^2)]}{\partial b_i}\cdot\left.\dfrac{\partial H(a_i-a_j)}{\partial a_i}\right\rvert_{a_i:=b_i/(1-\delta^2)\\a_j:=b_j/(1-\delta^2)}\\[2ex]&=\dfrac{1}{1-\delta^2}\cdot\left.\dfrac{\partial H(a_i-a_j)}{\partial a_i}\right\vert_{a_i:=b_i/(1-\delta^2)\\a_j:=b_j/(1-\delta^2)}\end{align}$$

PS: I would avoid using the designations $h_{a_i}$ or $h_{b_i}$ as they are a source of confusion, since $H$ is actually a monovariate function which is being composed with a bivariate argument $(a_i-a_j)$.

At the very least, designate $G(a_i,a_j):=H(a_i-a_j)$ and $F(b_i,b_j,\delta):=H((b_i-b_j)/(1-\delta^2))$ so you can have expression for the partial derivatives with respect to the arguments of these multivariate functions. $$F^{(1,0,0)}(b_i,b_j,\delta)= \dfrac{G^{(1,0)}(b_i/(1-\delta^2),b_j/(1-\delta_2))}{1-\delta^2}$$

PPS: The chain rules for the partial derivative with respect to $\delta$ is:

$$\small F^{(0,0,1)}(b_i,b_j,\delta) = \dfrac{2\delta}{(1-\delta^2)^2}\left[b_i~G^{(1,0)}\big(b_i/(1-\delta^2), b_j/(1-\delta^2)\big)-b_j~G^{(0,1)}\big(b_i/(1-\delta^2), b_j/(1-\delta^2)\big)\right]$$


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