$P[|X-1| < c]$ of a marginal PDF

Question

How do I compute $P[|X-1| < c]$ of $f_x(x) = e^{-x}$ for $0<x<\infty$? My initial idea was to solve

$\int_0^{c+1}e^{-x}$

Yet I don't think the bounds of my integral are correct, as I don't know how to proceed with the absolute value. Given normal calculus of the absolute value, the upper bound should be both, $c+1$ and $c-1$. But that's not really possible (I think).

Answer 1

We have $\{ |X-1| < c \} = \{ X - 1 < c \cap X-1 > -c \} = \{ X \in (1-c,1+c) \}$

So we have to integrate $f_X(x) = e^{-x}\chi_{(0,+\infty)}(x)$ over set $(1-c,1+c)$.

That is: $$\mathbb P(|X-1| < c) = \int_{1-c}^{1+c} f_X(x) dx = \int_{1-c}^{1+c} e^{-x}\chi_{(0,+\infty)(x)}dx$$

Well, I'm assuming $c\ge 0$ due to absolute value being non-negative. (For $c<0$ our probability is $0$)

If $ c \in [0,1] $ then lower bound is $1-c$, because both $x > 0$ and $x>1-c$ must hold.

So for case $c\in[0,1]$, we have :$$ \mathbb P(|X-1| < c) = \int_{1-c}^{1+c} e^{-x}dx = e^{c-1} - e^{-c-1} $$

And for the case $c\in(1,+\infty)$, lower bound is $0$, because then $1-c < 0$, so we get: $$\mathbb P(|X-1|<c) = \int_0^{1+c} e^{-x}dx = 1 - e^{-c-1} $$

Answer 2

$$\begin{aligned}P(|X-1|<c) &= 1_{c>0}P(X\in (-c+1,c+1))\\ &=1_{c>0}\int_{\max(0,-c+1)}^{c+1} e^{-x} dx\\ &=1_{c> 0} (e^{-\max(0,-c+1)} - e^{-(c+1)}) \end{aligned}$$