Mixing
CDF of $max(X,X^2)$ for $X$ uniformly distributed on $[-1,1]$
$begingroup$
$X$ is uniformly distributed on $[-1,1]$. And $Y=max(X,X^2)$.
What is $F_{Y}(t)$ , the CDF of $Y$?
My attempt:
I tried to graph it, but I think I found wrong. I found the joint pdf $5/6$. Is this correct? And what can I do next?
probability-theory probability-distributions
edited Jan 11 at 21:46
Did
248k23223460
asked May 4 '14 at 19:27
SomeoneSomeone
226
$endgroup$
add a comment |
$begingroup$
$X$ is uniformly distributed on $[-1,1]$. And $Y=max(X,X^2)$.
What is $F_{Y}(t)$ , the CDF of $Y$?
My attempt:
I tried to graph it, but I think I found wrong. I found the joint pdf $5/6$. Is this correct? And what can I do next?
probability-theory probability-distributions
edited Jan 11 at 21:46
Did
248k23223460
asked May 4 '14 at 19:27
SomeoneSomeone
226
$endgroup$
add a comment |
$begingroup$
$X$ is uniformly distributed on $[-1,1]$. And $Y=max(X,X^2)$.
What is $F_{Y}(t)$ , the CDF of $Y$?
My attempt:
I tried to graph it, but I think I found wrong. I found the joint pdf $5/6$. Is this correct? And what can I do next?
probability-theory probability-distributions
edited Jan 11 at 21:46
Did
248k23223460
asked May 4 '14 at 19:27
SomeoneSomeone
226
$endgroup$
$X$ is uniformly distributed on $[-1,1]$. And $Y=max(X,X^2)$.
What is $F_{Y}(t)$ , the CDF of $Y$?
My attempt:
I tried to graph it, but I think I found wrong. I found the joint pdf $5/6$. Is this correct? And what can I do next?
probability-theory probability-distributions
probability-theory probability-distributions
edited Jan 11 at 21:46
Did
248k23223460
asked May 4 '14 at 19:27
SomeoneSomeone
226
edited Jan 11 at 21:46
Did
248k23223460
asked May 4 '14 at 19:27
SomeoneSomeone
226
edited Jan 11 at 21:46
Did
248k23223460
edited Jan 11 at 21:46
Did
248k23223460
edited Jan 11 at 21:46
Did
248k23223460
248k23223460
asked May 4 '14 at 19:27
SomeoneSomeone
226
asked May 4 '14 at 19:27
SomeoneSomeone
226
asked May 4 '14 at 19:27
SomeoneSomeone
226
226
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that $0leqslant Yleqslant1$ almost surely and that $[Yleqslant y]=[-sqrt{y}leqslant Xleqslant y]$ for every $y$ in $(0,1)$. The length of the interval $[-sqrt{y},y]$ is $y+sqrt{y}$ hence $$F_Y(y)=P(Yleqslant y)=tfrac12(y+sqrt{y}).$$
answered May 4 '14 at 19:43
DidDid
248k23223460
$endgroup$
$begingroup$
How can you understand that $[−y<X<sqrt(y)]$
$endgroup$
– Someone
May 4 '14 at 19:56
1
$begingroup$
I think your interval should be -sqrt(y) < X < y, since y^2 < y on (-1,1), but the final answer is correct.
$endgroup$
– Alex Zorn
May 4 '14 at 19:58
1
$begingroup$
Indeed: $Ylt y$ means that $Xlt y$ and $X^2lt y$, that is, $Xlt y$ and $-sqrt{y}lt Xltsqrt{y}$, and $sqrt{y}gt y$ hence this is $-sqrt{y}lt Xlt y$. Thanks for notifying me.
$endgroup$
– Did
May 4 '14 at 20:00
add a comment |
$begingroup$
Using the joint cdf may not be the easiest way.
If $Xle 0$, then $Y=X^2$. If $Xgt 0$, then $Y=X$.
Let us find the cdf of $Y$. The only interesting part is for $0le yle 1$.
With probability $frac{1}{2}$, we have $Xle 0$. Conditional on $Xle 0$, the distribution of $X$ is uniform on $[-1,0]$. Thus
$$Pr(Yle y|Xle 0)=Pr(X^2le y|Xle 0)=Pr(-sqrt{y}le Xle 0|Xle 0)=sqrt{y}.$$
With probability $frac{1}{2}$ we have $Xgt 0$. Conditional on $Xgt 0$, the distribution of $X$ is uniform on $[0,1]$. Thus
$$Pr(Yle y|Xgt 0)=Pr(Xle y|Xgt 0)=y.$$
We conclude that
$$F_Y(y)=frac{1}{2}(sqrt{y})+frac{1}{2}y$$
for $0le yle 1$.
answered May 4 '14 at 19:54
André NicolasAndré Nicolas
453k36425810
$endgroup$
1
$begingroup$
I don't think that your solution is right. Because when we put $y=1$ the result is $1/2$. But it must be $1$ I think. @Andre Nicolas
$endgroup$
– Someone
May 4 '14 at 20:00
$begingroup$
@Someone: Thanks, got tangled up with $1-(1-dots)$. Fixed.
$endgroup$
– André Nicolas
May 4 '14 at 20:07
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $0leqslant Yleqslant1$ almost surely and that $[Yleqslant y]=[-sqrt{y}leqslant Xleqslant y]$ for every $y$ in $(0,1)$. The length of the interval $[-sqrt{y},y]$ is $y+sqrt{y}$ hence $$F_Y(y)=P(Yleqslant y)=tfrac12(y+sqrt{y}).$$
answered May 4 '14 at 19:43
DidDid
248k23223460
$endgroup$
$begingroup$
How can you understand that $[−y<X<sqrt(y)]$
$endgroup$
– Someone
May 4 '14 at 19:56
1
$begingroup$
I think your interval should be -sqrt(y) < X < y, since y^2 < y on (-1,1), but the final answer is correct.
$endgroup$
– Alex Zorn
May 4 '14 at 19:58
1
$begingroup$
Indeed: $Ylt y$ means that $Xlt y$ and $X^2lt y$, that is, $Xlt y$ and $-sqrt{y}lt Xltsqrt{y}$, and $sqrt{y}gt y$ hence this is $-sqrt{y}lt Xlt y$. Thanks for notifying me.
$endgroup$
– Did
May 4 '14 at 20:00
add a comment |
$begingroup$
Note that $0leqslant Yleqslant1$ almost surely and that $[Yleqslant y]=[-sqrt{y}leqslant Xleqslant y]$ for every $y$ in $(0,1)$. The length of the interval $[-sqrt{y},y]$ is $y+sqrt{y}$ hence $$F_Y(y)=P(Yleqslant y)=tfrac12(y+sqrt{y}).$$
answered May 4 '14 at 19:43
DidDid
248k23223460
$endgroup$
$begingroup$
How can you understand that $[−y<X<sqrt(y)]$
$endgroup$
– Someone
May 4 '14 at 19:56
1
$begingroup$
I think your interval should be -sqrt(y) < X < y, since y^2 < y on (-1,1), but the final answer is correct.
$endgroup$
– Alex Zorn
May 4 '14 at 19:58
1
$begingroup$
Indeed: $Ylt y$ means that $Xlt y$ and $X^2lt y$, that is, $Xlt y$ and $-sqrt{y}lt Xltsqrt{y}$, and $sqrt{y}gt y$ hence this is $-sqrt{y}lt Xlt y$. Thanks for notifying me.
$endgroup$
– Did
May 4 '14 at 20:00
add a comment |
$begingroup$
Note that $0leqslant Yleqslant1$ almost surely and that $[Yleqslant y]=[-sqrt{y}leqslant Xleqslant y]$ for every $y$ in $(0,1)$. The length of the interval $[-sqrt{y},y]$ is $y+sqrt{y}$ hence $$F_Y(y)=P(Yleqslant y)=tfrac12(y+sqrt{y}).$$
answered May 4 '14 at 19:43
DidDid
248k23223460
$endgroup$
Note that $0leqslant Yleqslant1$ almost surely and that $[Yleqslant y]=[-sqrt{y}leqslant Xleqslant y]$ for every $y$ in $(0,1)$. The length of the interval $[-sqrt{y},y]$ is $y+sqrt{y}$ hence $$F_Y(y)=P(Yleqslant y)=tfrac12(y+sqrt{y}).$$
answered May 4 '14 at 19:43
DidDid
248k23223460
edited May 5 '14 at 6:54
answered May 4 '14 at 19:43
DidDid
248k23223460
answered May 4 '14 at 19:43
DidDid
248k23223460
answered May 4 '14 at 19:43
DidDid
248k23223460
248k23223460
$begingroup$
How can you understand that $[−y<X<sqrt(y)]$
$endgroup$
– Someone
May 4 '14 at 19:56
1
$begingroup$
I think your interval should be -sqrt(y) < X < y, since y^2 < y on (-1,1), but the final answer is correct.
$endgroup$
– Alex Zorn
May 4 '14 at 19:58
1
$begingroup$
Indeed: $Ylt y$ means that $Xlt y$ and $X^2lt y$, that is, $Xlt y$ and $-sqrt{y}lt Xltsqrt{y}$, and $sqrt{y}gt y$ hence this is $-sqrt{y}lt Xlt y$. Thanks for notifying me.
$endgroup$
– Did
May 4 '14 at 20:00
add a comment |
$begingroup$
How can you understand that $[−y<X<sqrt(y)]$
$endgroup$
– Someone
May 4 '14 at 19:56
1
$begingroup$
I think your interval should be -sqrt(y) < X < y, since y^2 < y on (-1,1), but the final answer is correct.
$endgroup$
– Alex Zorn
May 4 '14 at 19:58
1
$begingroup$
Indeed: $Ylt y$ means that $Xlt y$ and $X^2lt y$, that is, $Xlt y$ and $-sqrt{y}lt Xltsqrt{y}$, and $sqrt{y}gt y$ hence this is $-sqrt{y}lt Xlt y$. Thanks for notifying me.
$endgroup$
– Did
May 4 '14 at 20:00
$begingroup$
How can you understand that $[−y<X<sqrt(y)]$
$endgroup$
– Someone
May 4 '14 at 19:56
$begingroup$
How can you understand that $[−y<X<sqrt(y)]$
$endgroup$
– Someone
May 4 '14 at 19:56
1
1
$begingroup$
I think your interval should be -sqrt(y) < X < y, since y^2 < y on (-1,1), but the final answer is correct.
$endgroup$
– Alex Zorn
May 4 '14 at 19:58
$begingroup$
I think your interval should be -sqrt(y) < X < y, since y^2 < y on (-1,1), but the final answer is correct.
$endgroup$
– Alex Zorn
May 4 '14 at 19:58
1
1
$begingroup$
Indeed: $Ylt y$ means that $Xlt y$ and $X^2lt y$, that is, $Xlt y$ and $-sqrt{y}lt Xltsqrt{y}$, and $sqrt{y}gt y$ hence this is $-sqrt{y}lt Xlt y$. Thanks for notifying me.
$endgroup$
– Did
May 4 '14 at 20:00
$begingroup$
Indeed: $Ylt y$ means that $Xlt y$ and $X^2lt y$, that is, $Xlt y$ and $-sqrt{y}lt Xltsqrt{y}$, and $sqrt{y}gt y$ hence this is $-sqrt{y}lt Xlt y$. Thanks for notifying me.
$endgroup$
– Did
May 4 '14 at 20:00
add a comment |
$begingroup$
Using the joint cdf may not be the easiest way.
If $Xle 0$, then $Y=X^2$. If $Xgt 0$, then $Y=X$.
Let us find the cdf of $Y$. The only interesting part is for $0le yle 1$.
With probability $frac{1}{2}$, we have $Xle 0$. Conditional on $Xle 0$, the distribution of $X$ is uniform on $[-1,0]$. Thus
$$Pr(Yle y|Xle 0)=Pr(X^2le y|Xle 0)=Pr(-sqrt{y}le Xle 0|Xle 0)=sqrt{y}.$$
With probability $frac{1}{2}$ we have $Xgt 0$. Conditional on $Xgt 0$, the distribution of $X$ is uniform on $[0,1]$. Thus
$$Pr(Yle y|Xgt 0)=Pr(Xle y|Xgt 0)=y.$$
We conclude that
$$F_Y(y)=frac{1}{2}(sqrt{y})+frac{1}{2}y$$
for $0le yle 1$.
answered May 4 '14 at 19:54
André NicolasAndré Nicolas
453k36425810
$endgroup$
1
$begingroup$
I don't think that your solution is right. Because when we put $y=1$ the result is $1/2$. But it must be $1$ I think. @Andre Nicolas
$endgroup$
– Someone
May 4 '14 at 20:00
$begingroup$
@Someone: Thanks, got tangled up with $1-(1-dots)$. Fixed.
$endgroup$
– André Nicolas
May 4 '14 at 20:07
add a comment |
$begingroup$
Using the joint cdf may not be the easiest way.
If $Xle 0$, then $Y=X^2$. If $Xgt 0$, then $Y=X$.
Let us find the cdf of $Y$. The only interesting part is for $0le yle 1$.
With probability $frac{1}{2}$, we have $Xle 0$. Conditional on $Xle 0$, the distribution of $X$ is uniform on $[-1,0]$. Thus
$$Pr(Yle y|Xle 0)=Pr(X^2le y|Xle 0)=Pr(-sqrt{y}le Xle 0|Xle 0)=sqrt{y}.$$
With probability $frac{1}{2}$ we have $Xgt 0$. Conditional on $Xgt 0$, the distribution of $X$ is uniform on $[0,1]$. Thus
$$Pr(Yle y|Xgt 0)=Pr(Xle y|Xgt 0)=y.$$
We conclude that
$$F_Y(y)=frac{1}{2}(sqrt{y})+frac{1}{2}y$$
for $0le yle 1$.
answered May 4 '14 at 19:54
André NicolasAndré Nicolas
453k36425810
$endgroup$
1
$begingroup$
I don't think that your solution is right. Because when we put $y=1$ the result is $1/2$. But it must be $1$ I think. @Andre Nicolas
$endgroup$
– Someone
May 4 '14 at 20:00
$begingroup$
@Someone: Thanks, got tangled up with $1-(1-dots)$. Fixed.
$endgroup$
– André Nicolas
May 4 '14 at 20:07
add a comment |
$begingroup$
Using the joint cdf may not be the easiest way.
If $Xle 0$, then $Y=X^2$. If $Xgt 0$, then $Y=X$.
Let us find the cdf of $Y$. The only interesting part is for $0le yle 1$.
With probability $frac{1}{2}$, we have $Xle 0$. Conditional on $Xle 0$, the distribution of $X$ is uniform on $[-1,0]$. Thus
$$Pr(Yle y|Xle 0)=Pr(X^2le y|Xle 0)=Pr(-sqrt{y}le Xle 0|Xle 0)=sqrt{y}.$$
With probability $frac{1}{2}$ we have $Xgt 0$. Conditional on $Xgt 0$, the distribution of $X$ is uniform on $[0,1]$. Thus
$$Pr(Yle y|Xgt 0)=Pr(Xle y|Xgt 0)=y.$$
We conclude that
$$F_Y(y)=frac{1}{2}(sqrt{y})+frac{1}{2}y$$
for $0le yle 1$.
answered May 4 '14 at 19:54
André NicolasAndré Nicolas
453k36425810
$endgroup$
Using the joint cdf may not be the easiest way.
If $Xle 0$, then $Y=X^2$. If $Xgt 0$, then $Y=X$.
Let us find the cdf of $Y$. The only interesting part is for $0le yle 1$.
With probability $frac{1}{2}$, we have $Xle 0$. Conditional on $Xle 0$, the distribution of $X$ is uniform on $[-1,0]$. Thus
$$Pr(Yle y|Xle 0)=Pr(X^2le y|Xle 0)=Pr(-sqrt{y}le Xle 0|Xle 0)=sqrt{y}.$$
With probability $frac{1}{2}$ we have $Xgt 0$. Conditional on $Xgt 0$, the distribution of $X$ is uniform on $[0,1]$. Thus
$$Pr(Yle y|Xgt 0)=Pr(Xle y|Xgt 0)=y.$$
We conclude that
$$F_Y(y)=frac{1}{2}(sqrt{y})+frac{1}{2}y$$
for $0le yle 1$.
answered May 4 '14 at 19:54
André NicolasAndré Nicolas
453k36425810
edited May 4 '14 at 20:06
answered May 4 '14 at 19:54
André NicolasAndré Nicolas
453k36425810
answered May 4 '14 at 19:54
André NicolasAndré Nicolas
453k36425810
answered May 4 '14 at 19:54
André NicolasAndré Nicolas
453k36425810
453k36425810
1
$begingroup$
I don't think that your solution is right. Because when we put $y=1$ the result is $1/2$. But it must be $1$ I think. @Andre Nicolas
$endgroup$
– Someone
May 4 '14 at 20:00
$begingroup$
@Someone: Thanks, got tangled up with $1-(1-dots)$. Fixed.
$endgroup$
– André Nicolas
May 4 '14 at 20:07
add a comment |
1
$begingroup$
I don't think that your solution is right. Because when we put $y=1$ the result is $1/2$. But it must be $1$ I think. @Andre Nicolas
$endgroup$
– Someone
May 4 '14 at 20:00
$begingroup$
@Someone: Thanks, got tangled up with $1-(1-dots)$. Fixed.
$endgroup$
– André Nicolas
May 4 '14 at 20:07
1
1
$begingroup$
I don't think that your solution is right. Because when we put $y=1$ the result is $1/2$. But it must be $1$ I think. @Andre Nicolas
$endgroup$
– Someone
May 4 '14 at 20:00
$begingroup$
I don't think that your solution is right. Because when we put $y=1$ the result is $1/2$. But it must be $1$ I think. @Andre Nicolas
$endgroup$
– Someone
May 4 '14 at 20:00
$begingroup$
@Someone: Thanks, got tangled up with $1-(1-dots)$. Fixed.
$endgroup$
– André Nicolas
May 4 '14 at 20:07
$begingroup$
@Someone: Thanks, got tangled up with $1-(1-dots)$. Fixed.
$endgroup$
– André Nicolas
May 4 '14 at 20:07
add a comment |
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