Find PMF of X^2 if X~Dunif(0,1,…,n)
$begingroup$
Follow up on this: Find PMF of $X^2$ if $X$~Dunif
(I do not have enough "reputation points" to comment, so if this is an inappropriate way to ask for a follow up, please let me know)
Is this a correct way to solve:
$Y=X^2$
X~DUnif(0,1,...,n)
- Find the PMF relating X and y:
$F_Y(y) = P(Y<y) = P(X^2<y) = P(X<sqrt{y})$
- Find CDF given X is DUnif:
$F_X(sqrt{(y)}) = int_0^{sqrt{y}}frac{1}{n}dx = frac{sqrt{y}}{n}$
- Take derivative to find PDF:
$frac{1}{2nsqrt{y}}$
for y in {0,1,...,n}
I think I am missing something due to the squared/square root. Any help?
random-variables uniform-distribution
asked Dec 8 '18 at 22:19
user603569user603569
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add a comment |
$begingroup$
Follow up on this: Find PMF of $X^2$ if $X$~Dunif
(I do not have enough "reputation points" to comment, so if this is an inappropriate way to ask for a follow up, please let me know)
Is this a correct way to solve:
$Y=X^2$
X~DUnif(0,1,...,n)
- Find the PMF relating X and y:
$F_Y(y) = P(Y<y) = P(X^2<y) = P(X<sqrt{y})$
- Find CDF given X is DUnif:
$F_X(sqrt{(y)}) = int_0^{sqrt{y}}frac{1}{n}dx = frac{sqrt{y}}{n}$
- Take derivative to find PDF:
$frac{1}{2nsqrt{y}}$
for y in {0,1,...,n}
I think I am missing something due to the squared/square root. Any help?
random-variables uniform-distribution
asked Dec 8 '18 at 22:19
user603569user603569
728
$endgroup$
add a comment |
$begingroup$
Follow up on this: Find PMF of $X^2$ if $X$~Dunif
(I do not have enough "reputation points" to comment, so if this is an inappropriate way to ask for a follow up, please let me know)
Is this a correct way to solve:
$Y=X^2$
X~DUnif(0,1,...,n)
- Find the PMF relating X and y:
$F_Y(y) = P(Y<y) = P(X^2<y) = P(X<sqrt{y})$
- Find CDF given X is DUnif:
$F_X(sqrt{(y)}) = int_0^{sqrt{y}}frac{1}{n}dx = frac{sqrt{y}}{n}$
- Take derivative to find PDF:
$frac{1}{2nsqrt{y}}$
for y in {0,1,...,n}
I think I am missing something due to the squared/square root. Any help?
random-variables uniform-distribution
asked Dec 8 '18 at 22:19
user603569user603569
728
$endgroup$
Follow up on this: Find PMF of $X^2$ if $X$~Dunif
(I do not have enough "reputation points" to comment, so if this is an inappropriate way to ask for a follow up, please let me know)
Is this a correct way to solve:
$Y=X^2$
X~DUnif(0,1,...,n)
- Find the PMF relating X and y:
$F_Y(y) = P(Y<y) = P(X^2<y) = P(X<sqrt{y})$
- Find CDF given X is DUnif:
$F_X(sqrt{(y)}) = int_0^{sqrt{y}}frac{1}{n}dx = frac{sqrt{y}}{n}$
- Take derivative to find PDF:
$frac{1}{2nsqrt{y}}$
for y in {0,1,...,n}
I think I am missing something due to the squared/square root. Any help?
random-variables uniform-distribution
random-variables uniform-distribution
asked Dec 8 '18 at 22:19
user603569user603569
728
asked Dec 8 '18 at 22:19
user603569user603569
728
asked Dec 8 '18 at 22:19
user603569user603569
728
asked Dec 8 '18 at 22:19
user603569user603569
728
asked Dec 8 '18 at 22:19
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1 Answer
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$begingroup$
If $X$ is uniformly distributed over ${0,1,ldots,n}$, then the distribution of $Y=X^2$ is straightforward to determine:
begin{align}
mathbb P(Y=1) &= mathbb P(X=0)+mathbb P(X=1)=frac 2{n+1}\
mathbb P(Y=k^2) &= mathbb P(X=k) = frac1{n+1}, k=2,ldots,n.
end{align}
The distribution function of $Y$ is thus
$$
F_Y(y) = frac2{n+1}mathsf 1_{[1,infty)}(y) + sum_{k=2}^nfrac1{n+1}mathsf 1_{[k^2,infty)}(y).
$$
Taking a derivative does not make sense here since the random variables are not continuous.
answered Dec 8 '18 at 22:29
Math1000Math1000
19.1k31745
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1 Answer
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1 Answer
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active
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$begingroup$
If $X$ is uniformly distributed over ${0,1,ldots,n}$, then the distribution of $Y=X^2$ is straightforward to determine:
begin{align}
mathbb P(Y=1) &= mathbb P(X=0)+mathbb P(X=1)=frac 2{n+1}\
mathbb P(Y=k^2) &= mathbb P(X=k) = frac1{n+1}, k=2,ldots,n.
end{align}
The distribution function of $Y$ is thus
$$
F_Y(y) = frac2{n+1}mathsf 1_{[1,infty)}(y) + sum_{k=2}^nfrac1{n+1}mathsf 1_{[k^2,infty)}(y).
$$
Taking a derivative does not make sense here since the random variables are not continuous.
answered Dec 8 '18 at 22:29
Math1000Math1000
19.1k31745
$endgroup$
add a comment |
$begingroup$
If $X$ is uniformly distributed over ${0,1,ldots,n}$, then the distribution of $Y=X^2$ is straightforward to determine:
begin{align}
mathbb P(Y=1) &= mathbb P(X=0)+mathbb P(X=1)=frac 2{n+1}\
mathbb P(Y=k^2) &= mathbb P(X=k) = frac1{n+1}, k=2,ldots,n.
end{align}
The distribution function of $Y$ is thus
$$
F_Y(y) = frac2{n+1}mathsf 1_{[1,infty)}(y) + sum_{k=2}^nfrac1{n+1}mathsf 1_{[k^2,infty)}(y).
$$
Taking a derivative does not make sense here since the random variables are not continuous.
answered Dec 8 '18 at 22:29
Math1000Math1000
19.1k31745
$endgroup$
add a comment |
$begingroup$
If $X$ is uniformly distributed over ${0,1,ldots,n}$, then the distribution of $Y=X^2$ is straightforward to determine:
begin{align}
mathbb P(Y=1) &= mathbb P(X=0)+mathbb P(X=1)=frac 2{n+1}\
mathbb P(Y=k^2) &= mathbb P(X=k) = frac1{n+1}, k=2,ldots,n.
end{align}
The distribution function of $Y$ is thus
$$
F_Y(y) = frac2{n+1}mathsf 1_{[1,infty)}(y) + sum_{k=2}^nfrac1{n+1}mathsf 1_{[k^2,infty)}(y).
$$
Taking a derivative does not make sense here since the random variables are not continuous.
answered Dec 8 '18 at 22:29
Math1000Math1000
19.1k31745
$endgroup$
If $X$ is uniformly distributed over ${0,1,ldots,n}$, then the distribution of $Y=X^2$ is straightforward to determine:
begin{align}
mathbb P(Y=1) &= mathbb P(X=0)+mathbb P(X=1)=frac 2{n+1}\
mathbb P(Y=k^2) &= mathbb P(X=k) = frac1{n+1}, k=2,ldots,n.
end{align}
The distribution function of $Y$ is thus
$$
F_Y(y) = frac2{n+1}mathsf 1_{[1,infty)}(y) + sum_{k=2}^nfrac1{n+1}mathsf 1_{[k^2,infty)}(y).
$$
Taking a derivative does not make sense here since the random variables are not continuous.
answered Dec 8 '18 at 22:29
Math1000Math1000
19.1k31745
answered Dec 8 '18 at 22:29
Math1000Math1000
19.1k31745
answered Dec 8 '18 at 22:29
Math1000Math1000
19.1k31745
answered Dec 8 '18 at 22:29
Math1000Math1000
19.1k31745
19.1k31745
add a comment |
add a comment |
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