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Suppose I have the pdf:

$f_{X}(x) = c, 0 \leq x \leq 2$

What would be the pdf of $X+2$ be?

What I initially thought was:

$ Y = X+2; R_{Y} = \{2,4\} $

$f_{Y}(x)$ = $P(2 \leq Y \leq 4) = P(2 \leq X + 2 \leq 4) = P(0 \leq X \leq 2) = f_{X}(x)$

I'm pretty sure I'm wrong in my reasoning. Any help would be appreciated.

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  • $\begingroup$ I have no idea what $R_Y$ is and is it wrong, but I have no doubt that your last equations are wrong, as you mix probability function with distribution function! $\endgroup$
    – vermator
    Mar 8 '19 at 6:16
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You correctly found the $Y$'s interval HOWEVER your last equations are wrong! You mix probability function with distribution function. $f_Y(x) = P(Y=x)$!

If you wanted to show where $Y$ is non-zero you could use $supp Y = [2,4]$ or just write Y's pdf similarly as you wrote $f_X(x)$ $(f_Y(x) = c, 2\leq x \leq 4)$.

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It is not correct to say $f_Y(x)=P(2\le Y\le 4)$, because the latter is just a number (namely $1$). The answer to your question is clear. The pdf of $Y$ is $1/2$ over the interval $(2,4)$ and zero elsewhere, since it is a uniform variable, just like $X$ but shifted by two units.

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