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
asked Dec 11 '18 at 12:46
Marco PittellaMarco Pittella
1338
$endgroup$
add a comment |
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
asked Dec 11 '18 at 12:46
Marco PittellaMarco Pittella
1338
$endgroup$
$begingroup$
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
add a comment |
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
asked Dec 11 '18 at 12:46
Marco PittellaMarco Pittella
1338
$endgroup$
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
asked Dec 11 '18 at 12:46
Marco PittellaMarco Pittella
1338
asked Dec 11 '18 at 12:46
Marco PittellaMarco Pittella
1338
edited Dec 11 '18 at 12:54
Marco Pittella
asked Dec 11 '18 at 12:46
Marco PittellaMarco Pittella
1338
asked Dec 11 '18 at 12:46
Marco PittellaMarco Pittella
1338
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
add a comment |
$begingroup$
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
add a comment |
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:$$cdots=P(|X|leq y)=cdots$$
answered Dec 11 '18 at 13:06
drhabdrhab
104k545136
$endgroup$
$begingroup$
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
add a comment |
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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:$$cdots=P(|X|leq y)=cdots$$
answered Dec 11 '18 at 13:06
drhabdrhab
104k545136
$endgroup$
$begingroup$
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
add a comment |
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:$$cdots=P(|X|leq y)=cdots$$
answered Dec 11 '18 at 13:06
drhabdrhab
104k545136
$endgroup$
$begingroup$
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
add a comment |
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:$$cdots=P(|X|leq y)=cdots$$
answered Dec 11 '18 at 13:06
drhabdrhab
104k545136
$endgroup$
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:$$cdots=P(|X|leq y)=cdots$$
answered Dec 11 '18 at 13:06
drhabdrhab
104k545136
answered Dec 11 '18 at 13:06
drhabdrhab
104k545136
answered Dec 11 '18 at 13:06
drhabdrhab
104k545136
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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