Why uniform discrete pdf does not converge to continuous?
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Why a discrete probability distribution does not converge to a continuous when the number of support points grow to infinity?
Consider a uniform distribution with support in $[0, 1]$, if that is continuous, then pdf in each point in the support would be one, while if the distribution is discrete with (infinitely) many support points the pdf in each point would be zero. Why is that?
Is the concept of pdf inherently different in discrete and continuous cases? I am looking for an intuition about this fundamental question.
probability probability-theory probability-distributions
asked Dec 12 '18 at 21:57
user25004user25004
1,44411637
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add a comment |
$begingroup$
Why a discrete probability distribution does not converge to a continuous when the number of support points grow to infinity?
Consider a uniform distribution with support in $[0, 1]$, if that is continuous, then pdf in each point in the support would be one, while if the distribution is discrete with (infinitely) many support points the pdf in each point would be zero. Why is that?
Is the concept of pdf inherently different in discrete and continuous cases? I am looking for an intuition about this fundamental question.
probability probability-theory probability-distributions
asked Dec 12 '18 at 21:57
user25004user25004
1,44411637
$endgroup$
add a comment |
$begingroup$
Why a discrete probability distribution does not converge to a continuous when the number of support points grow to infinity?
Consider a uniform distribution with support in $[0, 1]$, if that is continuous, then pdf in each point in the support would be one, while if the distribution is discrete with (infinitely) many support points the pdf in each point would be zero. Why is that?
Is the concept of pdf inherently different in discrete and continuous cases? I am looking for an intuition about this fundamental question.
probability probability-theory probability-distributions
asked Dec 12 '18 at 21:57
user25004user25004
1,44411637
$endgroup$
Why a discrete probability distribution does not converge to a continuous when the number of support points grow to infinity?
Consider a uniform distribution with support in $[0, 1]$, if that is continuous, then pdf in each point in the support would be one, while if the distribution is discrete with (infinitely) many support points the pdf in each point would be zero. Why is that?
Is the concept of pdf inherently different in discrete and continuous cases? I am looking for an intuition about this fundamental question.
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Dec 12 '18 at 21:57
user25004user25004
1,44411637
asked Dec 12 '18 at 21:57
user25004user25004
1,44411637
edited Dec 13 '18 at 14:31
user25004
asked Dec 12 '18 at 21:57
user25004user25004
1,44411637
asked Dec 12 '18 at 21:57
user25004user25004
1,44411637
asked Dec 12 '18 at 21:57
user25004user25004
1,44411637
1,44411637
add a comment |
add a comment |
2 Answers
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Let, for every $n geq 1$, $p_n$ is the uniform probability on $left{0,frac{1}{n},ldots,frac{n-1}{n},1right}$.
Then $p_n$ converges in distribution to the uniform probability measure on $[0,1]$ ($mu$), with the following (standard) meaning:
For every continuous function $f:[0,1] rightarrow mathbb{R}$, $int{fdp_n} rightarrow int{fdmu}$.
answered Dec 12 '18 at 22:36
MindlackMindlack
4,900211
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add a comment |
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There are several ways to look at what's going on here - indeed, several ways to define what it would mean for such a convergence to occur. Let's look at a few:
- The cdf of a continuous $U(0,,1)$ distribution is $x$ on $[0,,1]$, so it rises smoothly as a line. The cdf of its discrete counterpart looks more like a staircase. Does the staircase become the line? Here's a similar puzzle: the staircase consists of horizontal and vertical segments respectively summing to $1$ and $1$ so is of length $2$, so why is the straight path only of length $sqrt{2}$?
- The distribution which gives $0,,frac{1}{n},,cdots,,1$ a probability of $frac{1}{n+1}$ each has moment-generating fucntion $frac{1}{n+1}sum_{k=0}^ne^{tk/n}$. Is the $ntoinfty$ limit the expected $frac{e^t-1}{t}$? Sure; the series tends to the integral $int_0^1 e^{tx} dx=frac{e^t-1}{t}$.
- If you want to think of the probability mass functions of discrete distributions as if they were probability density functions, read about the Dirac delta. Our discrete distribution's "density" would be $frac{1}{n+1}sum_{k=0}^ndelta(x-k/n)$. Again, in the $ntoinfty$ limit the sum becomes an integral, $int_0^1delta(x-y)dy=1$ for $xin [0,,1]$.
answered Dec 12 '18 at 22:31
J.G.J.G.
32.6k23250
$endgroup$
$begingroup$
Did you mean $sqrt{2}$ at the end of 1.?
$endgroup$
– Clement C.
Dec 12 '18 at 22:40
$begingroup$
@ClementC. Thanks; fixed.
$endgroup$
– J.G.
Dec 12 '18 at 22:57
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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$begingroup$
Let, for every $n geq 1$, $p_n$ is the uniform probability on $left{0,frac{1}{n},ldots,frac{n-1}{n},1right}$.
Then $p_n$ converges in distribution to the uniform probability measure on $[0,1]$ ($mu$), with the following (standard) meaning:
For every continuous function $f:[0,1] rightarrow mathbb{R}$, $int{fdp_n} rightarrow int{fdmu}$.
answered Dec 12 '18 at 22:36
MindlackMindlack
4,900211
$endgroup$
add a comment |
$begingroup$
Let, for every $n geq 1$, $p_n$ is the uniform probability on $left{0,frac{1}{n},ldots,frac{n-1}{n},1right}$.
Then $p_n$ converges in distribution to the uniform probability measure on $[0,1]$ ($mu$), with the following (standard) meaning:
For every continuous function $f:[0,1] rightarrow mathbb{R}$, $int{fdp_n} rightarrow int{fdmu}$.
answered Dec 12 '18 at 22:36
MindlackMindlack
4,900211
$endgroup$
add a comment |
$begingroup$
Let, for every $n geq 1$, $p_n$ is the uniform probability on $left{0,frac{1}{n},ldots,frac{n-1}{n},1right}$.
Then $p_n$ converges in distribution to the uniform probability measure on $[0,1]$ ($mu$), with the following (standard) meaning:
For every continuous function $f:[0,1] rightarrow mathbb{R}$, $int{fdp_n} rightarrow int{fdmu}$.
answered Dec 12 '18 at 22:36
MindlackMindlack
4,900211
$endgroup$
Let, for every $n geq 1$, $p_n$ is the uniform probability on $left{0,frac{1}{n},ldots,frac{n-1}{n},1right}$.
Then $p_n$ converges in distribution to the uniform probability measure on $[0,1]$ ($mu$), with the following (standard) meaning:
For every continuous function $f:[0,1] rightarrow mathbb{R}$, $int{fdp_n} rightarrow int{fdmu}$.
answered Dec 12 '18 at 22:36
MindlackMindlack
4,900211
answered Dec 12 '18 at 22:36
MindlackMindlack
4,900211
answered Dec 12 '18 at 22:36
MindlackMindlack
4,900211
answered Dec 12 '18 at 22:36
MindlackMindlack
4,900211
4,900211
add a comment |
add a comment |
$begingroup$
There are several ways to look at what's going on here - indeed, several ways to define what it would mean for such a convergence to occur. Let's look at a few:
- The cdf of a continuous $U(0,,1)$ distribution is $x$ on $[0,,1]$, so it rises smoothly as a line. The cdf of its discrete counterpart looks more like a staircase. Does the staircase become the line? Here's a similar puzzle: the staircase consists of horizontal and vertical segments respectively summing to $1$ and $1$ so is of length $2$, so why is the straight path only of length $sqrt{2}$?
- The distribution which gives $0,,frac{1}{n},,cdots,,1$ a probability of $frac{1}{n+1}$ each has moment-generating fucntion $frac{1}{n+1}sum_{k=0}^ne^{tk/n}$. Is the $ntoinfty$ limit the expected $frac{e^t-1}{t}$? Sure; the series tends to the integral $int_0^1 e^{tx} dx=frac{e^t-1}{t}$.
- If you want to think of the probability mass functions of discrete distributions as if they were probability density functions, read about the Dirac delta. Our discrete distribution's "density" would be $frac{1}{n+1}sum_{k=0}^ndelta(x-k/n)$. Again, in the $ntoinfty$ limit the sum becomes an integral, $int_0^1delta(x-y)dy=1$ for $xin [0,,1]$.
answered Dec 12 '18 at 22:31
J.G.J.G.
32.6k23250
$endgroup$
$begingroup$
Did you mean $sqrt{2}$ at the end of 1.?
$endgroup$
– Clement C.
Dec 12 '18 at 22:40
$begingroup$
@ClementC. Thanks; fixed.
$endgroup$
– J.G.
Dec 12 '18 at 22:57
add a comment |
$begingroup$
There are several ways to look at what's going on here - indeed, several ways to define what it would mean for such a convergence to occur. Let's look at a few:
- The cdf of a continuous $U(0,,1)$ distribution is $x$ on $[0,,1]$, so it rises smoothly as a line. The cdf of its discrete counterpart looks more like a staircase. Does the staircase become the line? Here's a similar puzzle: the staircase consists of horizontal and vertical segments respectively summing to $1$ and $1$ so is of length $2$, so why is the straight path only of length $sqrt{2}$?
- The distribution which gives $0,,frac{1}{n},,cdots,,1$ a probability of $frac{1}{n+1}$ each has moment-generating fucntion $frac{1}{n+1}sum_{k=0}^ne^{tk/n}$. Is the $ntoinfty$ limit the expected $frac{e^t-1}{t}$? Sure; the series tends to the integral $int_0^1 e^{tx} dx=frac{e^t-1}{t}$.
- If you want to think of the probability mass functions of discrete distributions as if they were probability density functions, read about the Dirac delta. Our discrete distribution's "density" would be $frac{1}{n+1}sum_{k=0}^ndelta(x-k/n)$. Again, in the $ntoinfty$ limit the sum becomes an integral, $int_0^1delta(x-y)dy=1$ for $xin [0,,1]$.
answered Dec 12 '18 at 22:31
J.G.J.G.
32.6k23250
$endgroup$
$begingroup$
Did you mean $sqrt{2}$ at the end of 1.?
$endgroup$
– Clement C.
Dec 12 '18 at 22:40
$begingroup$
@ClementC. Thanks; fixed.
$endgroup$
– J.G.
Dec 12 '18 at 22:57
add a comment |
$begingroup$
There are several ways to look at what's going on here - indeed, several ways to define what it would mean for such a convergence to occur. Let's look at a few:
- The cdf of a continuous $U(0,,1)$ distribution is $x$ on $[0,,1]$, so it rises smoothly as a line. The cdf of its discrete counterpart looks more like a staircase. Does the staircase become the line? Here's a similar puzzle: the staircase consists of horizontal and vertical segments respectively summing to $1$ and $1$ so is of length $2$, so why is the straight path only of length $sqrt{2}$?
- The distribution which gives $0,,frac{1}{n},,cdots,,1$ a probability of $frac{1}{n+1}$ each has moment-generating fucntion $frac{1}{n+1}sum_{k=0}^ne^{tk/n}$. Is the $ntoinfty$ limit the expected $frac{e^t-1}{t}$? Sure; the series tends to the integral $int_0^1 e^{tx} dx=frac{e^t-1}{t}$.
- If you want to think of the probability mass functions of discrete distributions as if they were probability density functions, read about the Dirac delta. Our discrete distribution's "density" would be $frac{1}{n+1}sum_{k=0}^ndelta(x-k/n)$. Again, in the $ntoinfty$ limit the sum becomes an integral, $int_0^1delta(x-y)dy=1$ for $xin [0,,1]$.
answered Dec 12 '18 at 22:31
J.G.J.G.
32.6k23250
$endgroup$
There are several ways to look at what's going on here - indeed, several ways to define what it would mean for such a convergence to occur. Let's look at a few:
- The cdf of a continuous $U(0,,1)$ distribution is $x$ on $[0,,1]$, so it rises smoothly as a line. The cdf of its discrete counterpart looks more like a staircase. Does the staircase become the line? Here's a similar puzzle: the staircase consists of horizontal and vertical segments respectively summing to $1$ and $1$ so is of length $2$, so why is the straight path only of length $sqrt{2}$?
- The distribution which gives $0,,frac{1}{n},,cdots,,1$ a probability of $frac{1}{n+1}$ each has moment-generating fucntion $frac{1}{n+1}sum_{k=0}^ne^{tk/n}$. Is the $ntoinfty$ limit the expected $frac{e^t-1}{t}$? Sure; the series tends to the integral $int_0^1 e^{tx} dx=frac{e^t-1}{t}$.
- If you want to think of the probability mass functions of discrete distributions as if they were probability density functions, read about the Dirac delta. Our discrete distribution's "density" would be $frac{1}{n+1}sum_{k=0}^ndelta(x-k/n)$. Again, in the $ntoinfty$ limit the sum becomes an integral, $int_0^1delta(x-y)dy=1$ for $xin [0,,1]$.
answered Dec 12 '18 at 22:31
J.G.J.G.
32.6k23250
edited Dec 12 '18 at 22:56
answered Dec 12 '18 at 22:31
J.G.J.G.
32.6k23250
answered Dec 12 '18 at 22:31
J.G.J.G.
32.6k23250
answered Dec 12 '18 at 22:31
J.G.J.G.
32.6k23250
32.6k23250
$begingroup$
Did you mean $sqrt{2}$ at the end of 1.?
$endgroup$
– Clement C.
Dec 12 '18 at 22:40
$begingroup$
@ClementC. Thanks; fixed.
$endgroup$
– J.G.
Dec 12 '18 at 22:57
add a comment |
$begingroup$
Did you mean $sqrt{2}$ at the end of 1.?
$endgroup$
– Clement C.
Dec 12 '18 at 22:40
$begingroup$
@ClementC. Thanks; fixed.
$endgroup$
– J.G.
Dec 12 '18 at 22:57
$begingroup$
Did you mean $sqrt{2}$ at the end of 1.?
$endgroup$
– Clement C.
Dec 12 '18 at 22:40
$begingroup$
Did you mean $sqrt{2}$ at the end of 1.?
$endgroup$
– Clement C.
Dec 12 '18 at 22:40
$begingroup$
@ClementC. Thanks; fixed.
$endgroup$
– J.G.
Dec 12 '18 at 22:57
$begingroup$
@ClementC. Thanks; fixed.
$endgroup$
– J.G.
Dec 12 '18 at 22:57
add a comment |
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