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Random variable with exponential distribution.
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Let $X$ be random variable with exponential distribution $mathcalE(2)$ and let $Y$ be another random variable such that $$Y=maxleft(X^2, fracX+12 right).$$
Find the distribution for random variable $Y$.
Distribution for $X$ is $f_X(x)= 2e^-2x, x>0$ and zero otherwise.
Now, for variable $Y$ we have that it's distribution is zero whenever $y leq frac14$
For $y=t> frac12$ we have the following:
$F_Y(t)=int_0^2t-1 f_X(x)dx= 1- e^2-4t$
Similarly, for $y=t>1$ we have
$F_Y(t)=int_0^sqrtt f_X(x)dx= 1- e^-2sqrtt$
But, i cannot understand what happens in case that $y$ takes random value on interval $(frac14, frac12)$. It's the black line on the graph. How can i handle situations like this? Any help appreciated!
probability probability-distributions exponential-distribution
edited Jul 30 at 8:43
Davide Giraudo
121k15146249
asked Jul 30 at 8:15
cdummie
689412
add a comment |Â
up vote
5
down vote
favorite
Let $X$ be random variable with exponential distribution $mathcalE(2)$ and let $Y$ be another random variable such that $$Y=maxleft(X^2, fracX+12 right).$$
Find the distribution for random variable $Y$.
Distribution for $X$ is $f_X(x)= 2e^-2x, x>0$ and zero otherwise.
Now, for variable $Y$ we have that it's distribution is zero whenever $y leq frac14$
For $y=t> frac12$ we have the following:
$F_Y(t)=int_0^2t-1 f_X(x)dx= 1- e^2-4t$
Similarly, for $y=t>1$ we have
$F_Y(t)=int_0^sqrtt f_X(x)dx= 1- e^-2sqrtt$
But, i cannot understand what happens in case that $y$ takes random value on interval $(frac14, frac12)$. It's the black line on the graph. How can i handle situations like this? Any help appreciated!
probability probability-distributions exponential-distribution
edited Jul 30 at 8:43
Davide Giraudo
121k15146249
asked Jul 30 at 8:15
cdummie
689412
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let $X$ be random variable with exponential distribution $mathcalE(2)$ and let $Y$ be another random variable such that $$Y=maxleft(X^2, fracX+12 right).$$
Find the distribution for random variable $Y$.
Distribution for $X$ is $f_X(x)= 2e^-2x, x>0$ and zero otherwise.
Now, for variable $Y$ we have that it's distribution is zero whenever $y leq frac14$
For $y=t> frac12$ we have the following:
$F_Y(t)=int_0^2t-1 f_X(x)dx= 1- e^2-4t$
Similarly, for $y=t>1$ we have
$F_Y(t)=int_0^sqrtt f_X(x)dx= 1- e^-2sqrtt$
But, i cannot understand what happens in case that $y$ takes random value on interval $(frac14, frac12)$. It's the black line on the graph. How can i handle situations like this? Any help appreciated!
probability probability-distributions exponential-distribution
edited Jul 30 at 8:43
Davide Giraudo
121k15146249
asked Jul 30 at 8:15
cdummie
689412
Let $X$ be random variable with exponential distribution $mathcalE(2)$ and let $Y$ be another random variable such that $$Y=maxleft(X^2, fracX+12 right).$$
Find the distribution for random variable $Y$.
Distribution for $X$ is $f_X(x)= 2e^-2x, x>0$ and zero otherwise.
Now, for variable $Y$ we have that it's distribution is zero whenever $y leq frac14$
For $y=t> frac12$ we have the following:
$F_Y(t)=int_0^2t-1 f_X(x)dx= 1- e^2-4t$
Similarly, for $y=t>1$ we have
$F_Y(t)=int_0^sqrtt f_X(x)dx= 1- e^-2sqrtt$
But, i cannot understand what happens in case that $y$ takes random value on interval $(frac14, frac12)$. It's the black line on the graph. How can i handle situations like this? Any help appreciated!
probability probability-distributions exponential-distribution
edited Jul 30 at 8:43
Davide Giraudo
121k15146249
asked Jul 30 at 8:15
cdummie
689412
edited Jul 30 at 8:43
Davide Giraudo
121k15146249
edited Jul 30 at 8:43
Davide Giraudo
121k15146249
edited Jul 30 at 8:43
Davide Giraudo
121k15146249
121k15146249
asked Jul 30 at 8:15
cdummie
689412
asked Jul 30 at 8:15
cdummie
689412
asked Jul 30 at 8:15
cdummie
689412
689412
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
Let's compute the CDF of $Y$ firstly: $$beginalignP(Y le y) &= P(maxleft(X^2, fracX+12right) le y) \&= P(X^2 le y, fracX+12 le y) \&= P(X^2 le y, X le 2y - 1) \&= P(-sqrty le X le sqrty, X le 2y - 1) \&= begincases P(X le sqrty) , y > 1 \ P(X le 2y - 1), frac12 le y le 1 \ 0, textotherwiseendcasesendalign$$
As we can see $Y$ can't be less than $frac12$ because $max(X^2, fracX+12)$ always $ge frac12$
edited Jul 30 at 16:23
Paul Sinclair
18.6k21439
answered Jul 30 at 8:32
D F
1,0551218
@D F so whenever i conclude that $Y$ can't take any value below (or above) some specified value $a$ then i can say that it's distribution below (or above) that value $a$ must be zero. Is this correct?
– cdummie
Jul 30 at 11:50
1
almost correct. If $Y$ cannot take any value below some value $a$ then the CDF $F_Y(y) = 0 forall y le a$ because the CDF of $Y$ is $P(Y le y)$ hence if $Y$ cannot be below $a$ then $forall y le a to P(Y le y) = F_Y(y) = 0$. However, if $Y$ cannot be above some value i.e. $P(Y ge a) = 0$ then $forall y ge a to F_Y(y) = 1$ since $P(Y ge y) = 1 - P(Y le y) = 1 - F_Y(y)$
– D F
Jul 30 at 12:00
add a comment |Â
up vote
3
down vote
Here from $X>0$ a.s. it follows directly that $Y>frac12$ a.s.
Consequently $F_Y(y)=0$ for $yleqfrac12$.
Observe that this matches with the expression $F_Y(t)=1-e^2-4t$ for $t>frac12$ close to $frac12$.
If there $t$ approaches $frac12$ then the expression approaches $0$.
That is enough already to conclude that $F_Y(frac12)=0$ and if this is the case then automatically $F_Y(t)=0$ for every $tleqfrac12$.
This simply because $F$ is monotonically increasing and only takes values $geq0$ (and $leq1$ of course, but that is not relevant here).
answered Jul 30 at 8:31
drhab
85.9k540118
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Let's compute the CDF of $Y$ firstly: $$beginalignP(Y le y) &= P(maxleft(X^2, fracX+12right) le y) \&= P(X^2 le y, fracX+12 le y) \&= P(X^2 le y, X le 2y - 1) \&= P(-sqrty le X le sqrty, X le 2y - 1) \&= begincases P(X le sqrty) , y > 1 \ P(X le 2y - 1), frac12 le y le 1 \ 0, textotherwiseendcasesendalign$$
As we can see $Y$ can't be less than $frac12$ because $max(X^2, fracX+12)$ always $ge frac12$
edited Jul 30 at 16:23
Paul Sinclair
18.6k21439
answered Jul 30 at 8:32
D F
1,0551218
@D F so whenever i conclude that $Y$ can't take any value below (or above) some specified value $a$ then i can say that it's distribution below (or above) that value $a$ must be zero. Is this correct?
– cdummie
Jul 30 at 11:50
1
almost correct. If $Y$ cannot take any value below some value $a$ then the CDF $F_Y(y) = 0 forall y le a$ because the CDF of $Y$ is $P(Y le y)$ hence if $Y$ cannot be below $a$ then $forall y le a to P(Y le y) = F_Y(y) = 0$. However, if $Y$ cannot be above some value i.e. $P(Y ge a) = 0$ then $forall y ge a to F_Y(y) = 1$ since $P(Y ge y) = 1 - P(Y le y) = 1 - F_Y(y)$
– D F
Jul 30 at 12:00
add a comment |Â
up vote
5
down vote
accepted
Let's compute the CDF of $Y$ firstly: $$beginalignP(Y le y) &= P(maxleft(X^2, fracX+12right) le y) \&= P(X^2 le y, fracX+12 le y) \&= P(X^2 le y, X le 2y - 1) \&= P(-sqrty le X le sqrty, X le 2y - 1) \&= begincases P(X le sqrty) , y > 1 \ P(X le 2y - 1), frac12 le y le 1 \ 0, textotherwiseendcasesendalign$$
As we can see $Y$ can't be less than $frac12$ because $max(X^2, fracX+12)$ always $ge frac12$
edited Jul 30 at 16:23
Paul Sinclair
18.6k21439
answered Jul 30 at 8:32
D F
1,0551218
@D F so whenever i conclude that $Y$ can't take any value below (or above) some specified value $a$ then i can say that it's distribution below (or above) that value $a$ must be zero. Is this correct?
– cdummie
Jul 30 at 11:50
1
almost correct. If $Y$ cannot take any value below some value $a$ then the CDF $F_Y(y) = 0 forall y le a$ because the CDF of $Y$ is $P(Y le y)$ hence if $Y$ cannot be below $a$ then $forall y le a to P(Y le y) = F_Y(y) = 0$. However, if $Y$ cannot be above some value i.e. $P(Y ge a) = 0$ then $forall y ge a to F_Y(y) = 1$ since $P(Y ge y) = 1 - P(Y le y) = 1 - F_Y(y)$
– D F
Jul 30 at 12:00
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Let's compute the CDF of $Y$ firstly: $$beginalignP(Y le y) &= P(maxleft(X^2, fracX+12right) le y) \&= P(X^2 le y, fracX+12 le y) \&= P(X^2 le y, X le 2y - 1) \&= P(-sqrty le X le sqrty, X le 2y - 1) \&= begincases P(X le sqrty) , y > 1 \ P(X le 2y - 1), frac12 le y le 1 \ 0, textotherwiseendcasesendalign$$
As we can see $Y$ can't be less than $frac12$ because $max(X^2, fracX+12)$ always $ge frac12$
edited Jul 30 at 16:23
Paul Sinclair
18.6k21439
answered Jul 30 at 8:32
D F
1,0551218
Let's compute the CDF of $Y$ firstly: $$beginalignP(Y le y) &= P(maxleft(X^2, fracX+12right) le y) \&= P(X^2 le y, fracX+12 le y) \&= P(X^2 le y, X le 2y - 1) \&= P(-sqrty le X le sqrty, X le 2y - 1) \&= begincases P(X le sqrty) , y > 1 \ P(X le 2y - 1), frac12 le y le 1 \ 0, textotherwiseendcasesendalign$$
As we can see $Y$ can't be less than $frac12$ because $max(X^2, fracX+12)$ always $ge frac12$
edited Jul 30 at 16:23
Paul Sinclair
18.6k21439
answered Jul 30 at 8:32
D F
1,0551218
edited Jul 30 at 16:23
Paul Sinclair
18.6k21439
edited Jul 30 at 16:23
Paul Sinclair
18.6k21439
edited Jul 30 at 16:23
Paul Sinclair
18.6k21439
18.6k21439
answered Jul 30 at 8:32
D F
1,0551218
answered Jul 30 at 8:32
D F
1,0551218
answered Jul 30 at 8:32
D F
1,0551218
1,0551218
@D F so whenever i conclude that $Y$ can't take any value below (or above) some specified value $a$ then i can say that it's distribution below (or above) that value $a$ must be zero. Is this correct?
– cdummie
Jul 30 at 11:50
1
almost correct. If $Y$ cannot take any value below some value $a$ then the CDF $F_Y(y) = 0 forall y le a$ because the CDF of $Y$ is $P(Y le y)$ hence if $Y$ cannot be below $a$ then $forall y le a to P(Y le y) = F_Y(y) = 0$. However, if $Y$ cannot be above some value i.e. $P(Y ge a) = 0$ then $forall y ge a to F_Y(y) = 1$ since $P(Y ge y) = 1 - P(Y le y) = 1 - F_Y(y)$
– D F
Jul 30 at 12:00
add a comment |Â
@D F so whenever i conclude that $Y$ can't take any value below (or above) some specified value $a$ then i can say that it's distribution below (or above) that value $a$ must be zero. Is this correct?
– cdummie
Jul 30 at 11:50
1
almost correct. If $Y$ cannot take any value below some value $a$ then the CDF $F_Y(y) = 0 forall y le a$ because the CDF of $Y$ is $P(Y le y)$ hence if $Y$ cannot be below $a$ then $forall y le a to P(Y le y) = F_Y(y) = 0$. However, if $Y$ cannot be above some value i.e. $P(Y ge a) = 0$ then $forall y ge a to F_Y(y) = 1$ since $P(Y ge y) = 1 - P(Y le y) = 1 - F_Y(y)$
– D F
Jul 30 at 12:00
@D F so whenever i conclude that $Y$ can't take any value below (or above) some specified value $a$ then i can say that it's distribution below (or above) that value $a$ must be zero. Is this correct?
– cdummie
Jul 30 at 11:50
@D F so whenever i conclude that $Y$ can't take any value below (or above) some specified value $a$ then i can say that it's distribution below (or above) that value $a$ must be zero. Is this correct?
– cdummie
Jul 30 at 11:50
1
1
almost correct. If $Y$ cannot take any value below some value $a$ then the CDF $F_Y(y) = 0 forall y le a$ because the CDF of $Y$ is $P(Y le y)$ hence if $Y$ cannot be below $a$ then $forall y le a to P(Y le y) = F_Y(y) = 0$. However, if $Y$ cannot be above some value i.e. $P(Y ge a) = 0$ then $forall y ge a to F_Y(y) = 1$ since $P(Y ge y) = 1 - P(Y le y) = 1 - F_Y(y)$
– D F
Jul 30 at 12:00
almost correct. If $Y$ cannot take any value below some value $a$ then the CDF $F_Y(y) = 0 forall y le a$ because the CDF of $Y$ is $P(Y le y)$ hence if $Y$ cannot be below $a$ then $forall y le a to P(Y le y) = F_Y(y) = 0$. However, if $Y$ cannot be above some value i.e. $P(Y ge a) = 0$ then $forall y ge a to F_Y(y) = 1$ since $P(Y ge y) = 1 - P(Y le y) = 1 - F_Y(y)$
– D F
Jul 30 at 12:00
add a comment |Â
up vote
3
down vote
Here from $X>0$ a.s. it follows directly that $Y>frac12$ a.s.
Consequently $F_Y(y)=0$ for $yleqfrac12$.
Observe that this matches with the expression $F_Y(t)=1-e^2-4t$ for $t>frac12$ close to $frac12$.
If there $t$ approaches $frac12$ then the expression approaches $0$.
That is enough already to conclude that $F_Y(frac12)=0$ and if this is the case then automatically $F_Y(t)=0$ for every $tleqfrac12$.
This simply because $F$ is monotonically increasing and only takes values $geq0$ (and $leq1$ of course, but that is not relevant here).
answered Jul 30 at 8:31
drhab
85.9k540118
add a comment |Â
up vote
3
down vote
Here from $X>0$ a.s. it follows directly that $Y>frac12$ a.s.
Consequently $F_Y(y)=0$ for $yleqfrac12$.
Observe that this matches with the expression $F_Y(t)=1-e^2-4t$ for $t>frac12$ close to $frac12$.
If there $t$ approaches $frac12$ then the expression approaches $0$.
That is enough already to conclude that $F_Y(frac12)=0$ and if this is the case then automatically $F_Y(t)=0$ for every $tleqfrac12$.
This simply because $F$ is monotonically increasing and only takes values $geq0$ (and $leq1$ of course, but that is not relevant here).
answered Jul 30 at 8:31
drhab
85.9k540118
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Here from $X>0$ a.s. it follows directly that $Y>frac12$ a.s.
Consequently $F_Y(y)=0$ for $yleqfrac12$.
Observe that this matches with the expression $F_Y(t)=1-e^2-4t$ for $t>frac12$ close to $frac12$.
If there $t$ approaches $frac12$ then the expression approaches $0$.
That is enough already to conclude that $F_Y(frac12)=0$ and if this is the case then automatically $F_Y(t)=0$ for every $tleqfrac12$.
This simply because $F$ is monotonically increasing and only takes values $geq0$ (and $leq1$ of course, but that is not relevant here).
answered Jul 30 at 8:31
drhab
85.9k540118
Here from $X>0$ a.s. it follows directly that $Y>frac12$ a.s.
Consequently $F_Y(y)=0$ for $yleqfrac12$.
Observe that this matches with the expression $F_Y(t)=1-e^2-4t$ for $t>frac12$ close to $frac12$.
If there $t$ approaches $frac12$ then the expression approaches $0$.
That is enough already to conclude that $F_Y(frac12)=0$ and if this is the case then automatically $F_Y(t)=0$ for every $tleqfrac12$.
This simply because $F$ is monotonically increasing and only takes values $geq0$ (and $leq1$ of course, but that is not relevant here).
answered Jul 30 at 8:31
drhab
85.9k540118
edited Jul 30 at 8:37
answered Jul 30 at 8:31
drhab
85.9k540118
answered Jul 30 at 8:31
drhab
85.9k540118
answered Jul 30 at 8:31
drhab
85.9k540118
85.9k540118
add a comment |Â
add a comment |Â
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