How to find CDF of $Y=|X|wedge 2$ with $Xsim Laplace(lambda)$












1












$begingroup$


Given $X$ a bilateral exponential with density $f_X(x)=frac{1}{2}e^{-|x|}, forall xin mathbb{R}$ and $lambda=1$, i have to find CDF of $Y=|X|wedge 2$.



I know that $Y$ is not a monotonic function, so i can't apply the law of transformation of random variables.
I know that the support of $Y$ is $[0,2]$, so i can write $F_Y(y)=left{begin{matrix}
0 & if & y<0\
? & if & 0leq yleq 2\
1 & if &y>2
end{matrix}right.$



Then:



$F_y(y)=P(Yleq y)=P(|X|wedge 2leq y)=...$



How can I continue?






probability random-variables linear-transformations density-function monotone-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Remember $lvert Xrvertwedge 2=min(lvert Xrvert,2)$, so this is $leq y$ when $lvert Xrvertleq y$ (since $yleq 2$).
    $endgroup$
    – user10354138
    Dec 11 '18 at 13:06
















1












$begingroup$


Given $X$ a bilateral exponential with density $f_X(x)=frac{1}{2}e^{-|x|}, forall xin mathbb{R}$ and $lambda=1$, i have to find CDF of $Y=|X|wedge 2$.



I know that $Y$ is not a monotonic function, so i can't apply the law of transformation of random variables.
I know that the support of $Y$ is $[0,2]$, so i can write $F_Y(y)=left{begin{matrix}
0 & if & y<0\
? & if & 0leq yleq 2\
1 & if &y>2
end{matrix}right.$



Then:



$F_y(y)=P(Yleq y)=P(|X|wedge 2leq y)=...$



How can I continue?






probability random-variables linear-transformations density-function monotone-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Remember $lvert Xrvertwedge 2=min(lvert Xrvert,2)$, so this is $leq y$ when $lvert Xrvertleq y$ (since $yleq 2$).
    $endgroup$
    – user10354138
    Dec 11 '18 at 13:06














1












1








1


1



$begingroup$


Given $X$ a bilateral exponential with density $f_X(x)=frac{1}{2}e^{-|x|}, forall xin mathbb{R}$ and $lambda=1$, i have to find CDF of $Y=|X|wedge 2$.



I know that $Y$ is not a monotonic function, so i can't apply the law of transformation of random variables.
I know that the support of $Y$ is $[0,2]$, so i can write $F_Y(y)=left{begin{matrix}
0 & if & y<0\
? & if & 0leq yleq 2\
1 & if &y>2
end{matrix}right.$



Then:



$F_y(y)=P(Yleq y)=P(|X|wedge 2leq y)=...$



How can I continue?






probability random-variables linear-transformations density-function monotone-functions







share|cite|improve this question











$endgroup$




Given $X$ a bilateral exponential with density $f_X(x)=frac{1}{2}e^{-|x|}, forall xin mathbb{R}$ and $lambda=1$, i have to find CDF of $Y=|X|wedge 2$.



I know that $Y$ is not a monotonic function, so i can't apply the law of transformation of random variables.
I know that the support of $Y$ is $[0,2]$, so i can write $F_Y(y)=left{begin{matrix}
0 & if & y<0\
? & if & 0leq yleq 2\
1 & if &y>2
end{matrix}right.$



Then:



$F_y(y)=P(Yleq y)=P(|X|wedge 2leq y)=...$



How can I continue?






probability random-variables linear-transformations density-function monotone-functions




probability random-variables linear-transformations density-function monotone-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 12:54







Marco Pittella

















asked Dec 11 '18 at 12:46









Marco PittellaMarco Pittella

1338




1338












  • $begingroup$
    Remember $lvert Xrvertwedge 2=min(lvert Xrvert,2)$, so this is $leq y$ when $lvert Xrvertleq y$ (since $yleq 2$).
    $endgroup$
    – user10354138
    Dec 11 '18 at 13:06


















  • $begingroup$
    Remember $lvert Xrvertwedge 2=min(lvert Xrvert,2)$, so this is $leq y$ when $lvert Xrvertleq y$ (since $yleq 2$).
    $endgroup$
    – user10354138
    Dec 11 '18 at 13:06
















$begingroup$
Remember $lvert Xrvertwedge 2=min(lvert Xrvert,2)$, so this is $leq y$ when $lvert Xrvertleq y$ (since $yleq 2$).
$endgroup$
– user10354138
Dec 11 '18 at 13:06




$begingroup$
Remember $lvert Xrvertwedge 2=min(lvert Xrvert,2)$, so this is $leq y$ when $lvert Xrvertleq y$ (since $yleq 2$).
$endgroup$
– user10354138
Dec 11 '18 at 13:06










1 Answer
1






active

oldest

votes


















2












$begingroup$

You are doing this for $0leq yleq 2$ so $|X|wedge2leq y$ is the same statement as $|X|leq y$.



That means that you can continue with:$$cdots=P(|X|leq y)=cdots$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer, again! So: $P(|X|wedge 2leq y)=P(|X|leq y)=P(-yleq |X|leq y)=int_{-y}^{y}f_X(x)dx=int_{-y}^{0}f_X(x)dx+int_{0}^{y}f_X(x)dx=(frac{1}{2}-frac{1}{2}e^{-y})+(-frac{1}{2}e^{-y}-frac{1}{2})=-e^{-y}$. But solution says that $F_Y(y)=left{begin{matrix} 0 & if & yleq 0\ 1-e^{-y} & if & 0<yleq 2\ 1 & if & y>2 end{matrix}right.$ Where I wrong?
    $endgroup$
    – Marco Pittella
    Dec 11 '18 at 15:55












  • $begingroup$
    I've bisected $|x|=left{begin{matrix}x & for & xgeq0\ -x & for & x< 0end{matrix}right.$. So I obtain $int_{-y}^{0}frac{1}{2}e^{-|x|}dx=int_{-y}^{0}frac{1}{2}e^{x}dx=frac{1}{2}[e^x]_{-y}^0=frac{1}{2}-frac{1}{2}e^{-y}$ and $int_{0}^{y}frac{1}{2}e^{-|x|}dx=int_{0}^{y}frac{1}{2}e^{-x}dx=frac{1}{2}[-e^x]_{0}^y=-frac{1}{2}e^{-y}-frac{1}{2}$.
    $endgroup$
    – Marco Pittella
    Dec 11 '18 at 16:06












  • $begingroup$
    You forgot a minus sign in e-power in last line: $int_0^yfrac12e^{-x}dx=frac12[-e^{-x}]^y_0=frac12(1-e^{-y})$
    $endgroup$
    – drhab
    Dec 12 '18 at 9:17










  • $begingroup$
    Sure, I was wrong to transcribe. I edit!
    $endgroup$
    – Marco Pittella
    Dec 12 '18 at 9:29












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1 Answer
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2












$begingroup$

You are doing this for $0leq yleq 2$ so $|X|wedge2leq y$ is the same statement as $|X|leq y$.



That means that you can continue with:$$cdots=P(|X|leq y)=cdots$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer, again! So: $P(|X|wedge 2leq y)=P(|X|leq y)=P(-yleq |X|leq y)=int_{-y}^{y}f_X(x)dx=int_{-y}^{0}f_X(x)dx+int_{0}^{y}f_X(x)dx=(frac{1}{2}-frac{1}{2}e^{-y})+(-frac{1}{2}e^{-y}-frac{1}{2})=-e^{-y}$. But solution says that $F_Y(y)=left{begin{matrix} 0 & if & yleq 0\ 1-e^{-y} & if & 0<yleq 2\ 1 & if & y>2 end{matrix}right.$ Where I wrong?
    $endgroup$
    – Marco Pittella
    Dec 11 '18 at 15:55












  • $begingroup$
    I've bisected $|x|=left{begin{matrix}x & for & xgeq0\ -x & for & x< 0end{matrix}right.$. So I obtain $int_{-y}^{0}frac{1}{2}e^{-|x|}dx=int_{-y}^{0}frac{1}{2}e^{x}dx=frac{1}{2}[e^x]_{-y}^0=frac{1}{2}-frac{1}{2}e^{-y}$ and $int_{0}^{y}frac{1}{2}e^{-|x|}dx=int_{0}^{y}frac{1}{2}e^{-x}dx=frac{1}{2}[-e^x]_{0}^y=-frac{1}{2}e^{-y}-frac{1}{2}$.
    $endgroup$
    – Marco Pittella
    Dec 11 '18 at 16:06












  • $begingroup$
    You forgot a minus sign in e-power in last line: $int_0^yfrac12e^{-x}dx=frac12[-e^{-x}]^y_0=frac12(1-e^{-y})$
    $endgroup$
    – drhab
    Dec 12 '18 at 9:17










  • $begingroup$
    Sure, I was wrong to transcribe. I edit!
    $endgroup$
    – Marco Pittella
    Dec 12 '18 at 9:29
















2












$begingroup$

You are doing this for $0leq yleq 2$ so $|X|wedge2leq y$ is the same statement as $|X|leq y$.



That means that you can continue with:$$cdots=P(|X|leq y)=cdots$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer, again! So: $P(|X|wedge 2leq y)=P(|X|leq y)=P(-yleq |X|leq y)=int_{-y}^{y}f_X(x)dx=int_{-y}^{0}f_X(x)dx+int_{0}^{y}f_X(x)dx=(frac{1}{2}-frac{1}{2}e^{-y})+(-frac{1}{2}e^{-y}-frac{1}{2})=-e^{-y}$. But solution says that $F_Y(y)=left{begin{matrix} 0 & if & yleq 0\ 1-e^{-y} & if & 0<yleq 2\ 1 & if & y>2 end{matrix}right.$ Where I wrong?
    $endgroup$
    – Marco Pittella
    Dec 11 '18 at 15:55












  • $begingroup$
    I've bisected $|x|=left{begin{matrix}x & for & xgeq0\ -x & for & x< 0end{matrix}right.$. So I obtain $int_{-y}^{0}frac{1}{2}e^{-|x|}dx=int_{-y}^{0}frac{1}{2}e^{x}dx=frac{1}{2}[e^x]_{-y}^0=frac{1}{2}-frac{1}{2}e^{-y}$ and $int_{0}^{y}frac{1}{2}e^{-|x|}dx=int_{0}^{y}frac{1}{2}e^{-x}dx=frac{1}{2}[-e^x]_{0}^y=-frac{1}{2}e^{-y}-frac{1}{2}$.
    $endgroup$
    – Marco Pittella
    Dec 11 '18 at 16:06












  • $begingroup$
    You forgot a minus sign in e-power in last line: $int_0^yfrac12e^{-x}dx=frac12[-e^{-x}]^y_0=frac12(1-e^{-y})$
    $endgroup$
    – drhab
    Dec 12 '18 at 9:17










  • $begingroup$
    Sure, I was wrong to transcribe. I edit!
    $endgroup$
    – Marco Pittella
    Dec 12 '18 at 9:29














2












2








2





$begingroup$

You are doing this for $0leq yleq 2$ so $|X|wedge2leq y$ is the same statement as $|X|leq y$.



That means that you can continue with:$$cdots=P(|X|leq y)=cdots$$






share|cite|improve this answer









$endgroup$



You are doing this for $0leq yleq 2$ so $|X|wedge2leq y$ is the same statement as $|X|leq y$.



That means that you can continue with:$$cdots=P(|X|leq y)=cdots$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 11 '18 at 13:06









drhabdrhab

104k545136




104k545136












  • $begingroup$
    Thanks for your answer, again! So: $P(|X|wedge 2leq y)=P(|X|leq y)=P(-yleq |X|leq y)=int_{-y}^{y}f_X(x)dx=int_{-y}^{0}f_X(x)dx+int_{0}^{y}f_X(x)dx=(frac{1}{2}-frac{1}{2}e^{-y})+(-frac{1}{2}e^{-y}-frac{1}{2})=-e^{-y}$. But solution says that $F_Y(y)=left{begin{matrix} 0 & if & yleq 0\ 1-e^{-y} & if & 0<yleq 2\ 1 & if & y>2 end{matrix}right.$ Where I wrong?
    $endgroup$
    – Marco Pittella
    Dec 11 '18 at 15:55












  • $begingroup$
    I've bisected $|x|=left{begin{matrix}x & for & xgeq0\ -x & for & x< 0end{matrix}right.$. So I obtain $int_{-y}^{0}frac{1}{2}e^{-|x|}dx=int_{-y}^{0}frac{1}{2}e^{x}dx=frac{1}{2}[e^x]_{-y}^0=frac{1}{2}-frac{1}{2}e^{-y}$ and $int_{0}^{y}frac{1}{2}e^{-|x|}dx=int_{0}^{y}frac{1}{2}e^{-x}dx=frac{1}{2}[-e^x]_{0}^y=-frac{1}{2}e^{-y}-frac{1}{2}$.
    $endgroup$
    – Marco Pittella
    Dec 11 '18 at 16:06












  • $begingroup$
    You forgot a minus sign in e-power in last line: $int_0^yfrac12e^{-x}dx=frac12[-e^{-x}]^y_0=frac12(1-e^{-y})$
    $endgroup$
    – drhab
    Dec 12 '18 at 9:17










  • $begingroup$
    Sure, I was wrong to transcribe. I edit!
    $endgroup$
    – Marco Pittella
    Dec 12 '18 at 9:29


















  • $begingroup$
    Thanks for your answer, again! So: $P(|X|wedge 2leq y)=P(|X|leq y)=P(-yleq |X|leq y)=int_{-y}^{y}f_X(x)dx=int_{-y}^{0}f_X(x)dx+int_{0}^{y}f_X(x)dx=(frac{1}{2}-frac{1}{2}e^{-y})+(-frac{1}{2}e^{-y}-frac{1}{2})=-e^{-y}$. But solution says that $F_Y(y)=left{begin{matrix} 0 & if & yleq 0\ 1-e^{-y} & if & 0<yleq 2\ 1 & if & y>2 end{matrix}right.$ Where I wrong?
    $endgroup$
    – Marco Pittella
    Dec 11 '18 at 15:55












  • $begingroup$
    I've bisected $|x|=left{begin{matrix}x & for & xgeq0\ -x & for & x< 0end{matrix}right.$. So I obtain $int_{-y}^{0}frac{1}{2}e^{-|x|}dx=int_{-y}^{0}frac{1}{2}e^{x}dx=frac{1}{2}[e^x]_{-y}^0=frac{1}{2}-frac{1}{2}e^{-y}$ and $int_{0}^{y}frac{1}{2}e^{-|x|}dx=int_{0}^{y}frac{1}{2}e^{-x}dx=frac{1}{2}[-e^x]_{0}^y=-frac{1}{2}e^{-y}-frac{1}{2}$.
    $endgroup$
    – Marco Pittella
    Dec 11 '18 at 16:06












  • $begingroup$
    You forgot a minus sign in e-power in last line: $int_0^yfrac12e^{-x}dx=frac12[-e^{-x}]^y_0=frac12(1-e^{-y})$
    $endgroup$
    – drhab
    Dec 12 '18 at 9:17










  • $begingroup$
    Sure, I was wrong to transcribe. I edit!
    $endgroup$
    – Marco Pittella
    Dec 12 '18 at 9:29
















$begingroup$
Thanks for your answer, again! So: $P(|X|wedge 2leq y)=P(|X|leq y)=P(-yleq |X|leq y)=int_{-y}^{y}f_X(x)dx=int_{-y}^{0}f_X(x)dx+int_{0}^{y}f_X(x)dx=(frac{1}{2}-frac{1}{2}e^{-y})+(-frac{1}{2}e^{-y}-frac{1}{2})=-e^{-y}$. But solution says that $F_Y(y)=left{begin{matrix} 0 & if & yleq 0\ 1-e^{-y} & if & 0<yleq 2\ 1 & if & y>2 end{matrix}right.$ Where I wrong?
$endgroup$
– Marco Pittella
Dec 11 '18 at 15:55






$begingroup$
Thanks for your answer, again! So: $P(|X|wedge 2leq y)=P(|X|leq y)=P(-yleq |X|leq y)=int_{-y}^{y}f_X(x)dx=int_{-y}^{0}f_X(x)dx+int_{0}^{y}f_X(x)dx=(frac{1}{2}-frac{1}{2}e^{-y})+(-frac{1}{2}e^{-y}-frac{1}{2})=-e^{-y}$. But solution says that $F_Y(y)=left{begin{matrix} 0 & if & yleq 0\ 1-e^{-y} & if & 0<yleq 2\ 1 & if & y>2 end{matrix}right.$ Where I wrong?
$endgroup$
– Marco Pittella
Dec 11 '18 at 15:55














$begingroup$
I've bisected $|x|=left{begin{matrix}x & for & xgeq0\ -x & for & x< 0end{matrix}right.$. So I obtain $int_{-y}^{0}frac{1}{2}e^{-|x|}dx=int_{-y}^{0}frac{1}{2}e^{x}dx=frac{1}{2}[e^x]_{-y}^0=frac{1}{2}-frac{1}{2}e^{-y}$ and $int_{0}^{y}frac{1}{2}e^{-|x|}dx=int_{0}^{y}frac{1}{2}e^{-x}dx=frac{1}{2}[-e^x]_{0}^y=-frac{1}{2}e^{-y}-frac{1}{2}$.
$endgroup$
– Marco Pittella
Dec 11 '18 at 16:06






$begingroup$
I've bisected $|x|=left{begin{matrix}x & for & xgeq0\ -x & for & x< 0end{matrix}right.$. So I obtain $int_{-y}^{0}frac{1}{2}e^{-|x|}dx=int_{-y}^{0}frac{1}{2}e^{x}dx=frac{1}{2}[e^x]_{-y}^0=frac{1}{2}-frac{1}{2}e^{-y}$ and $int_{0}^{y}frac{1}{2}e^{-|x|}dx=int_{0}^{y}frac{1}{2}e^{-x}dx=frac{1}{2}[-e^x]_{0}^y=-frac{1}{2}e^{-y}-frac{1}{2}$.
$endgroup$
– Marco Pittella
Dec 11 '18 at 16:06














$begingroup$
You forgot a minus sign in e-power in last line: $int_0^yfrac12e^{-x}dx=frac12[-e^{-x}]^y_0=frac12(1-e^{-y})$
$endgroup$
– drhab
Dec 12 '18 at 9:17




$begingroup$
You forgot a minus sign in e-power in last line: $int_0^yfrac12e^{-x}dx=frac12[-e^{-x}]^y_0=frac12(1-e^{-y})$
$endgroup$
– drhab
Dec 12 '18 at 9:17












$begingroup$
Sure, I was wrong to transcribe. I edit!
$endgroup$
– Marco Pittella
Dec 12 '18 at 9:29




$begingroup$
Sure, I was wrong to transcribe. I edit!
$endgroup$
– Marco Pittella
Dec 12 '18 at 9:29


















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