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I am not entirely sure how the derivative follows from the preceding line in this example.

$f(x)$ is a PDF. You are supposed to set the derivative to 0 as the expectation needs to be minimised.

Could someone please explain a bit?

$$ C_t(X) = \begin{cases} c(t-X), & \text{if }\ X \le t \\ k(X-t), & \text{if }\ X \ge t \end{cases} $$

$$ E[C_t(X)] = ct\int_0^tf(x)dx \ -c\int_0^txf(x)dx \ +k\int_t^\infty xf(x)dx \ -kt\int_t^\infty f(x)dx\ $$

$$ \cfrac {d}{dt}E[C_t(X)] = ctf(t) \ + cF(t) \ -ctf(t) \ -ktf(t) \ +ktf(t) \ - k[1-F(t)] $$

I have only managed to get the first half of the derivative so far.

$$ ct\int_0^tf(x)dx = ctF(t) \ - \ ctF(0) $$

Since the function is a CDF, it evaluates to 0 at 0 for this distribution. So by differentiating and using the product rule we get:

$$ \cfrac{d}{dt}ctF(t)=ctf(t) +cF(t) $$

For the next part: $$ -c \int_0^t xf(x)dx = -c[xF(x)|_0^t \ - \ \int_0^tF(x)] $$ The derivative will be: $$ -c[tf(t)+F(t)-F(t)] \ = \ -ctf(t) $$

Im not sure how to proceed from here, since the upper limit is infinite for the next two integrals. I get stuck when I have to evaluate the function after using integration by parts because of the upper limit.

Shreesha
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    Please add the [tag:self-study] tag & read its [wiki](https://stats.stackexchange.com/tags/self-study/info). Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. – kjetil b halvorsen Nov 21 '21 at 03:09
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    An equivalent question is answered at https://stats.stackexchange.com/questions/251600, so that thread might help you. In the present case, all you need is the sum rule of differentiation, integration by parts, and the Fundamental Theorem of Calculus. – whuber Nov 21 '21 at 03:19
  • You have to differentiate $\int_0^t g(x)\,\text dx$ as a function of $t$, see eg https://math.stackexchange.com/a/781433/14893. – Xi'an Nov 21 '21 at 06:08
  • @kjetilbhalvorsen Apologies for that. I did indeed try to solve it myself, and was unable to complete the answer. It is not actually classwork. – Shreesha Nov 21 '21 at 09:50
  • Thank you for the other answers. I did check, but I'm afraid I still don't get how to finish my problem. – Shreesha Nov 21 '21 at 09:52
  • If the infinite upper limit bothers you, you may either (a) negate the variable of integration, thereby making it the lower limit; or (b) express the integral as a sum of a part with a finite upper limit and the rest--because the rest is a constant, it will disappear when you differentiate. For more help on differentiating, please post your questions on [math.se]. – whuber Nov 21 '21 at 15:24

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