# What would the pdf of X + c be?

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.

• 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! – vermator Mar 8 '19 at 6:16

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 $$f_X(x)$$ $$(f_Y(x) = c, 2\leq x \leq 4)$$.
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.