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I am having a trouble solving a derivative of a Riemann integral while trying to obtain a distribution of a variable being a function of another random variable.

Let $ X $ be a random variable with density function $ f_X $ and $ Y = g(X) = aX+b $.

The calculations up to the point of confusion:

$ P( Y \leq y ) = P( aX+b \leq y ) $ which makes:

$ P(X \leq (y-b)/a ) $ for $ a > 0 $

$ 1 - P(X \leq (y-b)/a ) $ for $ a < 0 $

We have a general rule:

$ \displaystyle\int_{a(x)}^{b(x)} f(y,x)\,\mathrm{d}x = b^\prime(x) f( b(x), x ) - a^\prime(x) f( a(x), x ) + \displaystyle\int_{a(x)}^{b(x)} f_x^\prime(y,x)\,\mathrm{d}t $

Which in our case resolves to:

$ f( a(x), x ) = 0 $ since $ a(x) = -\infty $

$ - a^\prime(x) f( b(x), x ) = |a|^{-1} f_X( (y-b)/a ) $

But what does $ \displaystyle\int_{a(x)}^{b(x)} f_x^\prime(y,x)\,\mathrm{d}t $ account for? It seems from my answer sheet that the whole expression equals to $|a|^{-1} f_X( (y-b)/a )$, but I do not understand why.

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    Are you sure you have transcribed the rule for "differentiating an integral" correctly? Look, for example, at my comment following [this answer](http://math.stackexchange.com/a/77906/15941). In any case, you don't need anything as fancy to get the answer you need: just the chain rule for derivatives suffices. Since $f_X$ is the derivative of $F_X$, then for $a > 0$, $$\frac{\mathrm d}{\mathrm dy}F_Y(y)=\frac{\mathrm d}{\mathrm dy}P\{X\leq(y-b)/a\}=\frac{\mathrm d}{\mathrm dy}F_X\left(\frac{y-b}{a}\right)=f_X\left(\frac{y-b}{a}\right)\cdot \frac{1}{a}$$ and similarly for $a < 0$.2012-08-23
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    @DilipSarwate Thank you very much, indeed your solution is the proper way. | Concerning making life harder though, as it goes to the rule - I think its accurate, I used http://en.wikipedia.org/wiki/Leibniz_integral_rule#General_form_with_variable_limits ...2012-08-24

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