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The following is given in my textbook and I cannot see how this is proved. Could someone provide me with some guidance please.

f(theta| x) is a pdf btw

I understand the integral gives us the CDF and then its differentiation gives us the pdf of y. My confusion comes from how this is applied given there is another parameter theta multiplying by the density

EDIT: Here is rule 9.1

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    $\begingroup$ Could you show differentiation rule 9.1? Completely unrelated: why in the world did the textbook choose $d$ to denote a variable?? $\endgroup$
    – Isaac Ren
    Commented May 3, 2020 at 17:51
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    $\begingroup$ See my extension to the question. I believe they have chosen 'd' to denote the 'decision' for the absolute loss function $\endgroup$ Commented May 3, 2020 at 17:59
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    $\begingroup$ Just added.. Sorry did it after my reply to you $\endgroup$ Commented May 3, 2020 at 18:01

2 Answers 2

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All you have to do is apply rule 9.1. To make it clearer, let's change the notation of rule 9.1: $$\frac d{dy}\left[\int_{t=-\infty}^yg(t)dt\right]=g(y). \tag{9.1}$$ To apply to your statement, replace $y$ with $d$, $t$ with $\theta$, and $g(y)$ with $df(d|x)$.

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  • $\begingroup$ But given the integral is of theta x f(theta|x) (which are two functions of theta) do we not need to apply the product rule for differentiation? I understand the integral gives us the CDF and then its differentiation gives us the pdf of y. My confusion comes from how this is applied given there is another parameter theta multiplying by the density $\endgroup$ Commented May 3, 2020 at 18:12
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    $\begingroup$ Multiplying the density by $\theta$ is not an issue. If you look carefully, $\theta f(\theta|x)$ appears on the left hand side, and $d f(d|x)$ appears on the right hand side, so we can hide the multiplication of these two functions under another name, such as $g$, which sends $\theta$ to $\theta f(\theta|x)$ (and $d$ to $d f(d|x)$. $\endgroup$
    – Isaac Ren
    Commented May 3, 2020 at 18:18
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    $\begingroup$ Another way of looking at it: what if the density function $f$ itself was a multiplication of two functions? The product of two functions is still a function, and can be manipulated as such. $\endgroup$
    – Isaac Ren
    Commented May 3, 2020 at 18:19
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Let $g(\theta) = \theta f(\theta|x)$, then apply the fundamental theorem of calculus (rule 9.1 in your post).

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  • $\begingroup$ But given the integral is of theta x f(theta|x) (which are two functions of theta) do we not need to apply the product rule for differentiation? I understand the integral gives us the CDF and then its differentiation gives us the pdf of y. My confusion comes from how this is applied given there is another parameter theta multiplying by the density $\endgroup$ Commented May 3, 2020 at 18:13

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