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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.

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Didn't you forget to differentiate $(m-d-\mu)/\sigma$ with respect to $\sigma$? –  yohBS Sep 27 '12 at 7:38
    
@yohBS Sorry, I did that. There is $(1/σ^2)$ multiplying my answer. Apart from that, do you think it is correct? –  Daniel Lårs Sep 27 '12 at 7:46
    
Perhaps the sign? –  Willie Wong Sep 27 '12 at 7:46
    
As for the sign, there is a minus sign from the rule and the derivative is negative, I think -*- is positive. –  Daniel Lårs Sep 27 '12 at 7:48
    
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 $\sigma^{-2}$. –  Willie Wong Sep 27 '12 at 7:53
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1 Answer

up vote 2 down vote accepted

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. :-)

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