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Question: Let Random variable $X$ with $PDF$ = $f_{X}(x)$ be the input to device with input output characteristics as shown below then sketch the $PDF$ of $Y$ i.e, $f_{Y}(y)$

enter image description here

enter image description here

My attempt:

for X>0

,$Y=X+1\implies f_{Y}(y)=f_{X}(y-1)$

for X<0

,$Y=X-1\implies f_{Y}(y)=f_{X}(y+1)$

but at

X=0 there is $2$ units jump from $-1$ to $1$ and derivative of jump discontinuity is impulse so i thought there should be presence of impulse in PDF of Y but i'm not sure about it .

any help in sketching O/P PDF Y will be greatly appreciated, and it is not a homework problem but from local author textbook

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the PDF of X can be written as

$$ f_X(x) = \begin{cases} \frac{1}{4} \text{ for } -2 \leq x < 0 \\ \frac{1}{2} \text{ for } 0 \leq x < 1 \\ 0 \text{ otherwise,} \end{cases}$$ but for the edges it remains unclear whether the intervals should or should not include their borders as Dilip pointed out. Now the function $Y(X)$ applies an offset of $\pm 1$ to any $X$ value corresponding to its sign: $$Y(X) = \text{sign}(X) + X$$. Using this, we can write the output PDF as $$f_Y(y) = \begin{cases} \frac{1}{4} \text{ for } -3 \leq x < 1 \\ \frac{1}{2} \text{ for } 1 \leq x < 2 \\ 0 \text{ otherwise.} \end{cases}$$

If my thoughts are correct, the resulting PDF should look like this:

comparison of PDFs in question

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There is no impulse because there is zero probability mass at $0$. As the problem itself shows, it is perfectly all right for a pdf to make sudden jumps in value; note that the graph of $f_X$ (yes, I do mean $X$, not $Y$) jumps in value at $0$ and the value of $f_X(0)$ cannot be determined from the graph. $f_X(0)$ could have value $\frac 14$ or $\frac 12$ or any value in between (or any nonnegative value for that matter)

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