# Derivative of Riemann integral [of a distribution function].

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. 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 $a < 0$. – Dilip Sarwate Aug 23 '12 at 20:02
@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 en.wikipedia.org/wiki/… ... – infoholic_anonymous Aug 24 '12 at 0:55