Find the distribution of $X$.
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$U sim mathrm Unif (0,1)$. Let $alpha > 0$. Then find the density function of $X=U^-frac 1 alpha$. I have found that if $F$ is the cumulative distribution function of the random variable $X$ then
$$
F(x) = left{
beginarrayll
0 & quad frac 1 x^alpha < 0 \
1 - x^-alpha & quad 0 < frac 1 x^alpha < 1 \
1 & quad frac 1 x^alpha > 1
endarray
right.
$$
Can it be simplified more? Please help me in this regard.
Thank you very much.
probability probability-distributions uniform-distribution density-function
asked Sep 7 at 7:56
Dbchatto67
34313
add a comment |Â
up vote
0
down vote
favorite
$U sim mathrm Unif (0,1)$. Let $alpha > 0$. Then find the density function of $X=U^-frac 1 alpha$. I have found that if $F$ is the cumulative distribution function of the random variable $X$ then
$$
F(x) = left{
beginarrayll
0 & quad frac 1 x^alpha < 0 \
1 - x^-alpha & quad 0 < frac 1 x^alpha < 1 \
1 & quad frac 1 x^alpha > 1
endarray
right.
$$
Can it be simplified more? Please help me in this regard.
Thank you very much.
probability probability-distributions uniform-distribution density-function
asked Sep 7 at 7:56
Dbchatto67
34313
But the problem is that $F$ is not defined at $0$.
– Dbchatto67
Sep 7 at 8:06
@Dbchatto67 CDF are right-continuous.
– Yuta
Sep 7 at 8:11
@drhab The random variable is greater than $1$ so the expressions obtained by OP are wrong.
– Kavi Rama Murthy
Sep 7 at 8:17
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$U sim mathrm Unif (0,1)$. Let $alpha > 0$. Then find the density function of $X=U^-frac 1 alpha$. I have found that if $F$ is the cumulative distribution function of the random variable $X$ then
$$
F(x) = left{
beginarrayll
0 & quad frac 1 x^alpha < 0 \
1 - x^-alpha & quad 0 < frac 1 x^alpha < 1 \
1 & quad frac 1 x^alpha > 1
endarray
right.
$$
Can it be simplified more? Please help me in this regard.
Thank you very much.
probability probability-distributions uniform-distribution density-function
asked Sep 7 at 7:56
Dbchatto67
34313
$U sim mathrm Unif (0,1)$. Let $alpha > 0$. Then find the density function of $X=U^-frac 1 alpha$. I have found that if $F$ is the cumulative distribution function of the random variable $X$ then
$$
F(x) = left{
beginarrayll
0 & quad frac 1 x^alpha < 0 \
1 - x^-alpha & quad 0 < frac 1 x^alpha < 1 \
1 & quad frac 1 x^alpha > 1
endarray
right.
$$
Can it be simplified more? Please help me in this regard.
Thank you very much.
probability probability-distributions uniform-distribution density-function
probability probability-distributions uniform-distribution density-function
asked Sep 7 at 7:56
Dbchatto67
34313
asked Sep 7 at 7:56
Dbchatto67
34313
asked Sep 7 at 7:56
Dbchatto67
34313
asked Sep 7 at 7:56
Dbchatto67
34313
asked Sep 7 at 7:56
Dbchatto67
34313
34313
But the problem is that $F$ is not defined at $0$.
– Dbchatto67
Sep 7 at 8:06
@Dbchatto67 CDF are right-continuous.
– Yuta
Sep 7 at 8:11
@drhab The random variable is greater than $1$ so the expressions obtained by OP are wrong.
– Kavi Rama Murthy
Sep 7 at 8:17
add a comment |Â
But the problem is that $F$ is not defined at $0$.
– Dbchatto67
Sep 7 at 8:06
@Dbchatto67 CDF are right-continuous.
– Yuta
Sep 7 at 8:11
@drhab The random variable is greater than $1$ so the expressions obtained by OP are wrong.
– Kavi Rama Murthy
Sep 7 at 8:17
But the problem is that $F$ is not defined at $0$.
– Dbchatto67
Sep 7 at 8:06
But the problem is that $F$ is not defined at $0$.
– Dbchatto67
Sep 7 at 8:06
@Dbchatto67 CDF are right-continuous.
– Yuta
Sep 7 at 8:11
@Dbchatto67 CDF are right-continuous.
– Yuta
Sep 7 at 8:11
@drhab The random variable is greater than $1$ so the expressions obtained by OP are wrong.
– Kavi Rama Murthy
Sep 7 at 8:17
@drhab The random variable is greater than $1$ so the expressions obtained by OP are wrong.
– Kavi Rama Murthy
Sep 7 at 8:17
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
The correct answer is this: $F(x)=0$ for $x<1$ and $F(x)=1-x^-alpha$ for $x geq 1$. The density is given by $f(x)=0$ for $x<1$ and $f(x)=alpha x^-alpha -1$ for $x >1$
answered Sep 7 at 8:14
Kavi Rama Murthy
26.7k31438
sir that means $X sim mathrm Pareto (alpha)$. Right?
– Dbchatto67
Sep 7 at 8:16
You could equally say $F(x)=0$ for $xle 1$ and $F(x)=1-x^-alpha$ for $x gt 1$
– Henry
Sep 7 at 8:17
@Henry Of course! I don't know what point you are trying to make because your answer and mine are exactly the same.
– Kavi Rama Murthy
Sep 7 at 8:19
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The correct answer is this: $F(x)=0$ for $x<1$ and $F(x)=1-x^-alpha$ for $x geq 1$. The density is given by $f(x)=0$ for $x<1$ and $f(x)=alpha x^-alpha -1$ for $x >1$
answered Sep 7 at 8:14
Kavi Rama Murthy
26.7k31438
sir that means $X sim mathrm Pareto (alpha)$. Right?
– Dbchatto67
Sep 7 at 8:16
You could equally say $F(x)=0$ for $xle 1$ and $F(x)=1-x^-alpha$ for $x gt 1$
– Henry
Sep 7 at 8:17
@Henry Of course! I don't know what point you are trying to make because your answer and mine are exactly the same.
– Kavi Rama Murthy
Sep 7 at 8:19
add a comment |Â
up vote
3
down vote
accepted
The correct answer is this: $F(x)=0$ for $x<1$ and $F(x)=1-x^-alpha$ for $x geq 1$. The density is given by $f(x)=0$ for $x<1$ and $f(x)=alpha x^-alpha -1$ for $x >1$
answered Sep 7 at 8:14
Kavi Rama Murthy
26.7k31438
sir that means $X sim mathrm Pareto (alpha)$. Right?
– Dbchatto67
Sep 7 at 8:16
You could equally say $F(x)=0$ for $xle 1$ and $F(x)=1-x^-alpha$ for $x gt 1$
– Henry
Sep 7 at 8:17
@Henry Of course! I don't know what point you are trying to make because your answer and mine are exactly the same.
– Kavi Rama Murthy
Sep 7 at 8:19
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The correct answer is this: $F(x)=0$ for $x<1$ and $F(x)=1-x^-alpha$ for $x geq 1$. The density is given by $f(x)=0$ for $x<1$ and $f(x)=alpha x^-alpha -1$ for $x >1$
answered Sep 7 at 8:14
Kavi Rama Murthy
26.7k31438
The correct answer is this: $F(x)=0$ for $x<1$ and $F(x)=1-x^-alpha$ for $x geq 1$. The density is given by $f(x)=0$ for $x<1$ and $f(x)=alpha x^-alpha -1$ for $x >1$
answered Sep 7 at 8:14
Kavi Rama Murthy
26.7k31438
answered Sep 7 at 8:14
Kavi Rama Murthy
26.7k31438
answered Sep 7 at 8:14
Kavi Rama Murthy
26.7k31438
answered Sep 7 at 8:14
Kavi Rama Murthy
26.7k31438
26.7k31438
sir that means $X sim mathrm Pareto (alpha)$. Right?
– Dbchatto67
Sep 7 at 8:16
You could equally say $F(x)=0$ for $xle 1$ and $F(x)=1-x^-alpha$ for $x gt 1$
– Henry
Sep 7 at 8:17
@Henry Of course! I don't know what point you are trying to make because your answer and mine are exactly the same.
– Kavi Rama Murthy
Sep 7 at 8:19
add a comment |Â
sir that means $X sim mathrm Pareto (alpha)$. Right?
– Dbchatto67
Sep 7 at 8:16
You could equally say $F(x)=0$ for $xle 1$ and $F(x)=1-x^-alpha$ for $x gt 1$
– Henry
Sep 7 at 8:17
@Henry Of course! I don't know what point you are trying to make because your answer and mine are exactly the same.
– Kavi Rama Murthy
Sep 7 at 8:19
sir that means $X sim mathrm Pareto (alpha)$. Right?
– Dbchatto67
Sep 7 at 8:16
sir that means $X sim mathrm Pareto (alpha)$. Right?
– Dbchatto67
Sep 7 at 8:16
You could equally say $F(x)=0$ for $xle 1$ and $F(x)=1-x^-alpha$ for $x gt 1$
– Henry
Sep 7 at 8:17
You could equally say $F(x)=0$ for $xle 1$ and $F(x)=1-x^-alpha$ for $x gt 1$
– Henry
Sep 7 at 8:17
@Henry Of course! I don't know what point you are trying to make because your answer and mine are exactly the same.
– Kavi Rama Murthy
Sep 7 at 8:19
@Henry Of course! I don't know what point you are trying to make because your answer and mine are exactly the same.
– Kavi Rama Murthy
Sep 7 at 8:19
add a comment |Â
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But the problem is that $F$ is not defined at $0$.
– Dbchatto67
Sep 7 at 8:06
@Dbchatto67 CDF are right-continuous.
– Yuta
Sep 7 at 8:11
@drhab The random variable is greater than $1$ so the expressions obtained by OP are wrong.
– Kavi Rama Murthy
Sep 7 at 8:17