0
$\begingroup$

Differentiating

$$\int_{(m-d-\mu)/\sigma}^{\infty}xf(x)dx$$

with respect to $\sigma$, where $\sigma$ is the standard deviation of the standardarized random variable $x$ and $\mu$ its mean. I guess that it is

$$\frac{m-d-\mu}{\sigma}f\left(\frac{m-d-\mu}{\sigma}\right).$$

Correct me if I am wrong.

$\endgroup$
  • 1
    $\begingroup$ Didn't you forget to differentiate $(m-d-\mu)/\sigma$ with respect to $\sigma$? $\endgroup$ – yohBS Sep 27 '12 at 7:38
  • $\begingroup$ @yohBS Sorry, I did that. There is $(1/σ^2)$ multiplying my answer. Apart from that, do you think it is correct? $\endgroup$ – Daniel Lårs Sep 27 '12 at 7:46
  • $\begingroup$ Perhaps the sign? $\endgroup$ – Willie Wong Sep 27 '12 at 7:46
  • $\begingroup$ As for the sign, there is a minus sign from the rule and the derivative is negative, I think -*- is positive. $\endgroup$ – Daniel Lårs Sep 27 '12 at 7:48
  • $\begingroup$ Ah, sorry, I wrote my comment before I saw your response to yohBS. But in the end you are missing also a factor of $m-d-\mu$ in addition to the $\sigma^{-2}$. $\endgroup$ – Willie Wong Sep 27 '12 at 7:53
2
$\begingroup$

Let $F(y) = \int_y^\infty x f(x)~ \mathrm{d}x$, we can write it also as

$$ F(y) = - \int_{\infty}^y x f(x)~ \mathrm{d}x$$

so by the fundamental theorem of calculus

$$ \frac{\mathrm{d}}{\mathrm{d}y} F(y) = - y f(y) $$

Let $G(\sigma) = F( (m-d-\mu)/\sigma )$. Then we have that by the Chain rule (not the Leibniz rule!) that

$$ \frac{\mathrm{d}}{\mathrm{d}\sigma} G(\sigma) = \frac{\mathrm{d}}{\mathrm{d}\sigma} F\left( \frac{m - d - \mu}{\sigma}\right) = F'\left( \frac{m - d - \mu}{\sigma}\right) \cdot \frac{\mathrm{d}}{\mathrm{d}\sigma} \frac{m-d-\mu}{\sigma} $$

and you can take it from here. :-)

$\endgroup$
  • $\begingroup$ ♦ thank you a lot. You suggested me the simplest way. $\endgroup$ – Daniel Lårs Sep 27 '12 at 7:54
  • $\begingroup$ I want to be certain about the result, would the final result look like this?: $(((m-d-μ))/(σ²))(((m-d-μ))/σ)f(((m-d-μ)/σ))$ ? $\endgroup$ – Daniel Lårs Sep 27 '12 at 8:34
  • $\begingroup$ That looks okay to me, provided all the parentheses are correctly closed (I didn't check that). $\endgroup$ – Willie Wong Sep 27 '12 at 10:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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