Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Probably it is trivial but I need to be sure about the correctness of the following expression:

$$\frac{d}{dk}\int_{-k}^k \sqrt{f_0(y)f_1(y)} \, \mathrm{d}y=\sqrt{f_0(k)f_1(k)}-\sqrt{f_0(-k)f_1(-k)}$$

I dont know if necessary but $f_i$ are some density functions.

Thanks!

share|improve this question
add comment

2 Answers 2

up vote 1 down vote accepted

You had a minus where you need a plus.

Here's the plodding traditional way: $$ \frac{d}{dk} \int_0^k \sqrt{f_0(y)f_1(y)} \,dy = \sqrt{f_0(k)f_1(k)}, $$ and $$ \frac{d}{dk} \int_{-k}^0 \sqrt{f_0(y)f_1(y)} \,dy = \frac{d}{dk} \int_k^0 \sqrt{f_0(-u)f_1(-u)} \, (-du) = \frac{d}{dk} \int_0^k \sqrt{f_0(-u)f_1(-u)} \, du $$ $$ = +\sqrt{f_0(-k)f_1(-k)}. $$ Now add the two together.

Here's another way. The area under the function between $-k$ and $k$ has two boundaries that move as $k$ changes. The size of the boundary times the rate at which the boundary moves equals the rate at which the area changes. The sizes of the boundaries are the two terms being added in our bottom-line answer. The two moving boundaries are moving at the same rate: the rate of change of $k$. So the sum of the sizes times the rate of change of $k$ equals the rate of change of area. Hence the sum of the sizes is the rate of change of area with respect to $k$.

share|improve this answer
    
PS: The second method I give above is how I saw instantly that it should have been a plus. It took me a minute to see that by the first method. So the second point of view can be quite useful. –  Michael Hardy Mar 2 '13 at 23:26
add comment

Let $g(y) = \sqrt{f_0(y)f_1(y)}$. Let $G(y)$ be a primitive of $g$. Then, $$ \frac{d}{dk}\int_{-k}^k g(y) \mathrm{d} y = \frac{d}{dk}(G(k) - G(-k)) $$ Taking the derivative and heeding the chain rule, we have: $$ G'(k) - (-1)G'(-k) = \sqrt{f_0(k)f_1(k)} + \sqrt{f_0(-k)f_1(-k)} $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.