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Let $X$ and $Y$ be independent, continuous random variables.

The PDF of a sum $W=X+Y$ is given by the convolution formula $$ f_{X+Y} (w) = \int_{-\infty}^{\infty} f_X(x) f_Y(w-x)\,\,dx$$


In this lecture, it is shown a more general way to use the convolution formula to get the PDF of $Z=2X-Y$.

It goes as follows. If $$Z = (2X)+(-Y)$$ Then $$f_{2X-Y}(z) = \int_{-\infty}^{\infty} f_{2X}(x) f_{-Y}(z-x)\,\,dx$$


What I don't really understand is why the argument of the PDF of $-Y$ should be kept in the same form. Shouldn't it be

$$f_{2X-Y}(z) = \int_{-\infty}^{\infty} f_{2X}(x) f_{-Y}(2x-z)\,\,dx \,\,\,\,\text{?}$$


... since $Z=2X-Y$ and then $Y=2X-Z$ ?



Thanks in advance,
SB

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$X\neq x$ --> The error in your reasoning is to think that $x$ and $X$ stand for the same object: $x$ is just the variable of integration, we can use any letter. In the first case $x$ is a realization/value of $X$ but in the second case $x$ is a realization/value of $2X$ when you write $f_{2X}(x)$.

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    $\begingroup$ ... guess I'll stop writing my answer (+1)! $\endgroup$
    – jbowman
    Commented Jul 10, 2014 at 1:48
  • $\begingroup$ Thanks. I got it for $f_{2X}$, but I still didnt get it for $f_{-Y} (z-x)$. What do you guys recommend? Reviewing convolution derivation? $\endgroup$
    – Samuel Barbosa
    Commented Jul 10, 2014 at 1:54
  • $\begingroup$ @SamuelBarbosa: it is just notation. Write $a=2x$. It is now easy to accept that $f_{2X-Y}(z)=\int_{-\infty}^{+\infty} f_{2X}(a)\cdot f_{-Y}(z-a)da$? $\endgroup$
    – Sergio Parreiras
    Commented Jul 10, 2014 at 2:19
  • $\begingroup$ Got it from discrete case notation. $$P(Z=z) = \sum P(2X=x).P(-Y=z-x)$$ Now I understand what you were trying to say. Thanks! $\endgroup$
    – Samuel Barbosa
    Commented Jul 10, 2014 at 3:05

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