How is the subtraction of a uniform (0, k) and its entire part distributed?
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Let X be a random variable distributed as $U[0, K]$ for an integer K. Find the density function of $Y = f (x) = x- [x]$, where [x] denotes the integer part of the real number x.
I think that [X] represents a discrete uniform but this by definition would be with values at x = 1,2, ... k, and its density would be 1/k which is the same as that of the continuous uniform.
Now, I understand that the distribution [X] is dependent on X, then I could not assume independence to use some kind of transformation because I do not know the joint.
I appreciate if you can give me some other way that I have not considered, thank you very much.
statistics uniform-distribution
asked Dec 30 '18 at 23:16
Cristian PerdomoCristian Perdomo
61
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add a comment |
$begingroup$
Let X be a random variable distributed as $U[0, K]$ for an integer K. Find the density function of $Y = f (x) = x- [x]$, where [x] denotes the integer part of the real number x.
I think that [X] represents a discrete uniform but this by definition would be with values at x = 1,2, ... k, and its density would be 1/k which is the same as that of the continuous uniform.
Now, I understand that the distribution [X] is dependent on X, then I could not assume independence to use some kind of transformation because I do not know the joint.
I appreciate if you can give me some other way that I have not considered, thank you very much.
statistics uniform-distribution
asked Dec 30 '18 at 23:16
Cristian PerdomoCristian Perdomo
61
$endgroup$
add a comment |
$begingroup$
Let X be a random variable distributed as $U[0, K]$ for an integer K. Find the density function of $Y = f (x) = x- [x]$, where [x] denotes the integer part of the real number x.
I think that [X] represents a discrete uniform but this by definition would be with values at x = 1,2, ... k, and its density would be 1/k which is the same as that of the continuous uniform.
Now, I understand that the distribution [X] is dependent on X, then I could not assume independence to use some kind of transformation because I do not know the joint.
I appreciate if you can give me some other way that I have not considered, thank you very much.
statistics uniform-distribution
asked Dec 30 '18 at 23:16
Cristian PerdomoCristian Perdomo
61
$endgroup$
Let X be a random variable distributed as $U[0, K]$ for an integer K. Find the density function of $Y = f (x) = x- [x]$, where [x] denotes the integer part of the real number x.
I think that [X] represents a discrete uniform but this by definition would be with values at x = 1,2, ... k, and its density would be 1/k which is the same as that of the continuous uniform.
Now, I understand that the distribution [X] is dependent on X, then I could not assume independence to use some kind of transformation because I do not know the joint.
I appreciate if you can give me some other way that I have not considered, thank you very much.
statistics uniform-distribution
statistics uniform-distribution
asked Dec 30 '18 at 23:16
Cristian PerdomoCristian Perdomo
61
asked Dec 30 '18 at 23:16
Cristian PerdomoCristian Perdomo
61
asked Dec 30 '18 at 23:16
Cristian PerdomoCristian Perdomo
61
asked Dec 30 '18 at 23:16
Cristian PerdomoCristian Perdomo
61
asked Dec 30 '18 at 23:16
Cristian PerdomoCristian Perdomo
61
61
add a comment |
add a comment |
1 Answer
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So, set ${x}=x-lfloor xrfloor$ to be the fractional part of $x$. Check that ${x} in [0,1)$. Hence, $f_X(x)$ is defined for $xin [0,1)$. Let's compute the CDF. Fix a $cin[0,1)$, and study $mathbb{P}({x}leq c)$. Observe that,
$$
{{x} leq c}=bigcup_{k=0}^{K-1}{xin [k,k+c)},
$$
hence $mathbb{P}({x}leq c)=Kcdot frac{c}{K}=c$. Hence, it turns out that, ${x}$ is uniform on $[0,1)$ (and also, $lfloor xrfloor$ is uniform on ${0,dots,K-1}$.
answered Dec 30 '18 at 23:21
AaronAaron
1,947415
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$begingroup$
Thanks, I understood.
$endgroup$
– Cristian Perdomo
Dec 31 '18 at 0:26
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
So, set ${x}=x-lfloor xrfloor$ to be the fractional part of $x$. Check that ${x} in [0,1)$. Hence, $f_X(x)$ is defined for $xin [0,1)$. Let's compute the CDF. Fix a $cin[0,1)$, and study $mathbb{P}({x}leq c)$. Observe that,
$$
{{x} leq c}=bigcup_{k=0}^{K-1}{xin [k,k+c)},
$$
hence $mathbb{P}({x}leq c)=Kcdot frac{c}{K}=c$. Hence, it turns out that, ${x}$ is uniform on $[0,1)$ (and also, $lfloor xrfloor$ is uniform on ${0,dots,K-1}$.
answered Dec 30 '18 at 23:21
AaronAaron
1,947415
$endgroup$
$begingroup$
Thanks, I understood.
$endgroup$
– Cristian Perdomo
Dec 31 '18 at 0:26
add a comment |
$begingroup$
So, set ${x}=x-lfloor xrfloor$ to be the fractional part of $x$. Check that ${x} in [0,1)$. Hence, $f_X(x)$ is defined for $xin [0,1)$. Let's compute the CDF. Fix a $cin[0,1)$, and study $mathbb{P}({x}leq c)$. Observe that,
$$
{{x} leq c}=bigcup_{k=0}^{K-1}{xin [k,k+c)},
$$
hence $mathbb{P}({x}leq c)=Kcdot frac{c}{K}=c$. Hence, it turns out that, ${x}$ is uniform on $[0,1)$ (and also, $lfloor xrfloor$ is uniform on ${0,dots,K-1}$.
answered Dec 30 '18 at 23:21
AaronAaron
1,947415
$endgroup$
$begingroup$
Thanks, I understood.
$endgroup$
– Cristian Perdomo
Dec 31 '18 at 0:26
add a comment |
$begingroup$
So, set ${x}=x-lfloor xrfloor$ to be the fractional part of $x$. Check that ${x} in [0,1)$. Hence, $f_X(x)$ is defined for $xin [0,1)$. Let's compute the CDF. Fix a $cin[0,1)$, and study $mathbb{P}({x}leq c)$. Observe that,
$$
{{x} leq c}=bigcup_{k=0}^{K-1}{xin [k,k+c)},
$$
hence $mathbb{P}({x}leq c)=Kcdot frac{c}{K}=c$. Hence, it turns out that, ${x}$ is uniform on $[0,1)$ (and also, $lfloor xrfloor$ is uniform on ${0,dots,K-1}$.
answered Dec 30 '18 at 23:21
AaronAaron
1,947415
$endgroup$
So, set ${x}=x-lfloor xrfloor$ to be the fractional part of $x$. Check that ${x} in [0,1)$. Hence, $f_X(x)$ is defined for $xin [0,1)$. Let's compute the CDF. Fix a $cin[0,1)$, and study $mathbb{P}({x}leq c)$. Observe that,
$$
{{x} leq c}=bigcup_{k=0}^{K-1}{xin [k,k+c)},
$$
hence $mathbb{P}({x}leq c)=Kcdot frac{c}{K}=c$. Hence, it turns out that, ${x}$ is uniform on $[0,1)$ (and also, $lfloor xrfloor$ is uniform on ${0,dots,K-1}$.
answered Dec 30 '18 at 23:21
AaronAaron
1,947415
answered Dec 30 '18 at 23:21
AaronAaron
1,947415
answered Dec 30 '18 at 23:21
AaronAaron
1,947415
answered Dec 30 '18 at 23:21
AaronAaron
1,947415
1,947415
$begingroup$
Thanks, I understood.
$endgroup$
– Cristian Perdomo
Dec 31 '18 at 0:26
add a comment |
$begingroup$
Thanks, I understood.
$endgroup$
– Cristian Perdomo
Dec 31 '18 at 0:26
$begingroup$
Thanks, I understood.
$endgroup$
– Cristian Perdomo
Dec 31 '18 at 0:26
$begingroup$
Thanks, I understood.
$endgroup$
– Cristian Perdomo
Dec 31 '18 at 0:26
add a comment |
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