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given that $X_1 \sim U[-2,1]. X_2=0.5e^{-|t|}, -\infty<t<\infty$. find $F_Y$ if:

$$ Y = \begin{cases} X_2, & \text{$X_1<-1$} \\ X_1, & \text{$-1 \le X_1<0$} \\ 3, & \text{$0 \le X_1$} \\ \end{cases}$$

I have problem solving for $-1\le X_1<0$

gt6989b
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user107761
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1 Answers1

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Hint

Recall that $F_Y(y) = \mathbb{P}[Y \leq y]$. So according to your definition, you want to find $\mathbb{P}[Y \leq y]$ for $y \in [-1,0]$. How do you do that?

EDIT

In response to the comment below. Note that if $y \in [-1,0]$, $$ F_Y(y) = \int_{-\infty}^y f_Y(z) dz = \int_{-\infty}^{-1} f_Y(z) dz + \int_{-1}^{y} f_Y(z) dz = F_{X_2}(-1) + \left( F_{X_1}(y) - F_{X_1}(-1) \right). $$

gt6989b
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  • well, I say that it'll be $F_{X1}(t)-F_{X1}(-1) + F_Y(-1)$. but that ain't the correct answer. – user107761 Feb 24 '14 at 14:28
  • @user107761 note the edit. $F_Y(-1)$ can be simplified, it is really $F_{X_2}(-1)$, and now you just plug in, since you know both $F_{X_1}$ and $F_{X_2}$. – gt6989b Feb 24 '14 at 14:34