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I have an indicator function $I(D\leq Q)$which is equal to $1$ if $D\leq Q$ and $0$ otherwise. What would be derivative of this function with respect to different variables such as $D$ or $Q$ or $P$ ($D$ is a function of $P$).

Clarification to what I am trying to do:

  • $D$ represents demand which is a function of price, assume $D=a-bp$
  • $Q$ represents quantity or supply, which is assumed to be fixed

$$\text{profit} = p\min(D,Q)= PDI(D\lt Q)+PQI(Q\leq D)$$

I want to take derivative of profit with respect to price.

Thanks in advance

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On what domain is your indicator function defined? What do the variables D, Q, and P represent? Have you verified wether you can derive such function? –  Vitalij Zadneprovskij May 9 '12 at 22:29
    
    
Thanks for the answers. Please see the clarification made to original question. Also I would be grateful if you briefly explain what you mean by derivative in the sense of distribution. Thanks –  Eln May 14 '12 at 19:34
    
@Elnaz: Please consider registering, so that you don't log in as two different people and have to get "permission" to edit your own post. Thank you. –  Arturo Magidin May 14 '12 at 19:42
    
Thank you. I just registered. Also I am trying to delete my answer but can't find how to do it. Sorry –  Eln May 14 '12 at 19:51

1 Answer 1

The derivative in the usual sense does not exist at a discontinuity, and is $0$ everywhere else. If you're talking about a derivative in the sense of distributions, $\dfrac{\partial}{\partial D} I(D \le Q) = -\delta(D-Q)$.

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Thanks alot. Can you please explain more about your answer. why -(D-Q) and not just (Q-D). Is there a general rule for such derivatives? thanks again –  Eln May 14 '12 at 20:01
    
$\delta$ is symmetric, so $\delta(D-Q) = \delta(Q-D)$. $\delta$ can be thought of as the derivative of the Heaviside function $H(x) = 1$ for $x > 0$, $0$ for $x < 0$, so if $f(x)$ is smooth except for a jump discontinuity at $x = a$ with $\lim_{x \to a+} f(x) - \lim_{x \to a-} f(x) = c$, then $\dfrac{\partial f}{\partial x} = c \delta(x-a) + $ (some smooth function). –  Robert Israel May 14 '12 at 21:44
    
Thanks, very helpful. Just I am not sure if I understand what you mean by the last part "+ (some smooth function). –" –  Eln May 14 '12 at 22:34
    
The second part is the derivative of $f(x)$ away from $x=a$. –  Robert Israel May 14 '12 at 23:24

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