Mixing
Finding the median and modes of a discrete uniform distribution
$begingroup$
Question Let $X$ be Discrete Uniform on ${1, 2, ldots, n}.$ Find all medians and modes of $X$.
So far, I know that $P(x)=frac{1}{n}$. By intuition, the median is $frac{n+1}{2}$ for odd numbers. I have no idea how to put this in a story proof sort of way, and how to calculate the median and means for both odd and even cases. This is a problem given in the textbook that I came across during practice.
probability statistics probability-distributions
edited Nov 29 '17 at 4:28
Joel
14.4k12340
$endgroup$
add a comment |
$begingroup$
Question Let $X$ be Discrete Uniform on ${1, 2, ldots, n}.$ Find all medians and modes of $X$.
So far, I know that $P(x)=frac{1}{n}$. By intuition, the median is $frac{n+1}{2}$ for odd numbers. I have no idea how to put this in a story proof sort of way, and how to calculate the median and means for both odd and even cases. This is a problem given in the textbook that I came across during practice.
probability statistics probability-distributions
edited Nov 29 '17 at 4:28
Joel
14.4k12340
$endgroup$
$begingroup$
I believe the question is asking about the median and modes of a distribution, not a sample. For $n = 6,$ it would be the 'fair die' distribution that puts probability $1/6$ at each of the values $1,2,3,4,5,6.$ So the median of this distribution is $eta = 3.5$ and (depending on definition) there is no mode or there multiple modes at each of the six equally likely values.
$endgroup$
– BruceET
Nov 6 '17 at 7:43
$begingroup$
@BruceET If you accept multiple modes, then you should be able to accept multiple medians, so for even $n$ either $left[ frac n2, frac n 2 +1 right]$ or $left{ frac n2, frac n 2 +1 right}$ depending on whether the median should be part of the support
$endgroup$
– Henry
Nov 6 '17 at 18:36
$begingroup$
I view the issues with medians and modes to be quite different, but I don't disagree with your suggestion for medians. [In general, are there multiple modes only if all have the same max density/rel.freq., or does any local maximum qualify as a (perhaps) 'minor' mode?]
$endgroup$
– BruceET
Nov 6 '17 at 20:06
add a comment |
$begingroup$
Question Let $X$ be Discrete Uniform on ${1, 2, ldots, n}.$ Find all medians and modes of $X$.
So far, I know that $P(x)=frac{1}{n}$. By intuition, the median is $frac{n+1}{2}$ for odd numbers. I have no idea how to put this in a story proof sort of way, and how to calculate the median and means for both odd and even cases. This is a problem given in the textbook that I came across during practice.
probability statistics probability-distributions
edited Nov 29 '17 at 4:28
Joel
14.4k12340
$endgroup$
Question Let $X$ be Discrete Uniform on ${1, 2, ldots, n}.$ Find all medians and modes of $X$.
So far, I know that $P(x)=frac{1}{n}$. By intuition, the median is $frac{n+1}{2}$ for odd numbers. I have no idea how to put this in a story proof sort of way, and how to calculate the median and means for both odd and even cases. This is a problem given in the textbook that I came across during practice.
probability statistics probability-distributions
probability statistics probability-distributions
edited Nov 29 '17 at 4:28
Joel
14.4k12340
edited Nov 29 '17 at 4:28
Joel
14.4k12340
edited Nov 29 '17 at 4:28
Joel
14.4k12340
edited Nov 29 '17 at 4:28
Joel
14.4k12340
edited Nov 29 '17 at 4:28
Joel
14.4k12340
14.4k12340
asked Nov 6 '17 at 4:42
user490867
$begingroup$
I believe the question is asking about the median and modes of a distribution, not a sample. For $n = 6,$ it would be the 'fair die' distribution that puts probability $1/6$ at each of the values $1,2,3,4,5,6.$ So the median of this distribution is $eta = 3.5$ and (depending on definition) there is no mode or there multiple modes at each of the six equally likely values.
$endgroup$
– BruceET
Nov 6 '17 at 7:43
$begingroup$
@BruceET If you accept multiple modes, then you should be able to accept multiple medians, so for even $n$ either $left[ frac n2, frac n 2 +1 right]$ or $left{ frac n2, frac n 2 +1 right}$ depending on whether the median should be part of the support
$endgroup$
– Henry
Nov 6 '17 at 18:36
$begingroup$
I view the issues with medians and modes to be quite different, but I don't disagree with your suggestion for medians. [In general, are there multiple modes only if all have the same max density/rel.freq., or does any local maximum qualify as a (perhaps) 'minor' mode?]
$endgroup$
– BruceET
Nov 6 '17 at 20:06
add a comment |
$begingroup$
I believe the question is asking about the median and modes of a distribution, not a sample. For $n = 6,$ it would be the 'fair die' distribution that puts probability $1/6$ at each of the values $1,2,3,4,5,6.$ So the median of this distribution is $eta = 3.5$ and (depending on definition) there is no mode or there multiple modes at each of the six equally likely values.
$endgroup$
– BruceET
Nov 6 '17 at 7:43
$begingroup$
@BruceET If you accept multiple modes, then you should be able to accept multiple medians, so for even $n$ either $left[ frac n2, frac n 2 +1 right]$ or $left{ frac n2, frac n 2 +1 right}$ depending on whether the median should be part of the support
$endgroup$
– Henry
Nov 6 '17 at 18:36
$begingroup$
I view the issues with medians and modes to be quite different, but I don't disagree with your suggestion for medians. [In general, are there multiple modes only if all have the same max density/rel.freq., or does any local maximum qualify as a (perhaps) 'minor' mode?]
$endgroup$
– BruceET
Nov 6 '17 at 20:06
$begingroup$
I believe the question is asking about the median and modes of a distribution, not a sample. For $n = 6,$ it would be the 'fair die' distribution that puts probability $1/6$ at each of the values $1,2,3,4,5,6.$ So the median of this distribution is $eta = 3.5$ and (depending on definition) there is no mode or there multiple modes at each of the six equally likely values.
$endgroup$
– BruceET
Nov 6 '17 at 7:43
$begingroup$
I believe the question is asking about the median and modes of a distribution, not a sample. For $n = 6,$ it would be the 'fair die' distribution that puts probability $1/6$ at each of the values $1,2,3,4,5,6.$ So the median of this distribution is $eta = 3.5$ and (depending on definition) there is no mode or there multiple modes at each of the six equally likely values.
$endgroup$
– BruceET
Nov 6 '17 at 7:43
$begingroup$
@BruceET If you accept multiple modes, then you should be able to accept multiple medians, so for even $n$ either $left[ frac n2, frac n 2 +1 right]$ or $left{ frac n2, frac n 2 +1 right}$ depending on whether the median should be part of the support
$endgroup$
– Henry
Nov 6 '17 at 18:36
$begingroup$
@BruceET If you accept multiple modes, then you should be able to accept multiple medians, so for even $n$ either $left[ frac n2, frac n 2 +1 right]$ or $left{ frac n2, frac n 2 +1 right}$ depending on whether the median should be part of the support
$endgroup$
– Henry
Nov 6 '17 at 18:36
$begingroup$
I view the issues with medians and modes to be quite different, but I don't disagree with your suggestion for medians. [In general, are there multiple modes only if all have the same max density/rel.freq., or does any local maximum qualify as a (perhaps) 'minor' mode?]
$endgroup$
– BruceET
Nov 6 '17 at 20:06
$begingroup$
I view the issues with medians and modes to be quite different, but I don't disagree with your suggestion for medians. [In general, are there multiple modes only if all have the same max density/rel.freq., or does any local maximum qualify as a (perhaps) 'minor' mode?]
$endgroup$
– BruceET
Nov 6 '17 at 20:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If the cumulative distribution function is strictly increasing, you can define the median $M$ unambiguously as
$$M=F_X^{-1}(0.5)$$
in which $F_X$ is the cumulative distribution function of your variable $X$. But this definition does not make sense for a discrete distribution, since the cumulative distribution is not injective, and hence non-invertible.
A possible solution is to define a pseudo-inverse $F_X^{-}$ as follows
$$F_X^{-}(y)=inf{x in mathbb{R}|F_X(x)geq y}$$
You can now define the median as being $F_X^-(0.5)$. If you apply this definition on a uniform discrete distribution on the numbers ${1,2,3,4,5}$, you get $3$ as a result. While applying it on the dice ($n=6$) will get you $3$. Maybe you're not very satisfied with that last result. So let's try another definition.
We now define $F_X^{+}$ as
$$F_X^{+}(y)=sup{x in mathbb{R}|F_X(x)leq y}$$
The median now becomes $F_X^{+}(0.5)$. Let's see what that gives us for our previous distributions. For the case of the uniform discrete distribution on ${1,2,3,4,5}$, we still get $3$, but for the dice we now get $4$. Still not happy I suppose?
But wait, can't we just take the mean of those two? In fact, you can do even more you can define
$$F_X^{alpha}(y)=alpha F_X^{-}(y)+(1-alpha)F_X^{+}(y)$$
for any $alpha in [0,1]$. It is now clear that if we pick $alpha=0.5$, we will get exactly what we wanted! In the even case, the median will be $frac{n+1}{2}$.
answered Nov 6 '17 at 18:24
RaskolnikovRaskolnikov
12.6k23571
$endgroup$
$begingroup$
For continuous distributions as well, there can be several medians.
$endgroup$
– Did
Nov 6 '17 at 19:40
$begingroup$
True, I didn't think more deeply about it, but it is easy to construct continuous distributions which are non-injective and that therefore have the same ambiguity problem in defining the median as discrete distributions.
$endgroup$
– Raskolnikov
Nov 6 '17 at 19:50
add a comment |
$begingroup$
Modes are local maximums of the PDF, therefore there are $n$ modes: 1, 2,...,$n$. In this case you can see that talking about a mode does not make much sense, or at least is not really enlightening (but then again, you already know that the random variable is uniform...)
The median depends on whether $n$ is even of odd.
- If $n$ is odd, then there is a single median, ${n+1}over2$.
- If $n$ is even then any value in the (closed) interval [${nover2},
{nover2}+1$] qualifies as a median.
This with the following definition of a median : any value $m$ such that
$$
P(X<m)leq .5 quad text{and } P(X>m)leq .5.
$$
Of course the default median is ${n+1}over2$ regardless of $n$.
answered Nov 6 '17 at 18:34
A.G.A.G.
2,020711
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
oldest
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oldest
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$begingroup$
If the cumulative distribution function is strictly increasing, you can define the median $M$ unambiguously as
$$M=F_X^{-1}(0.5)$$
in which $F_X$ is the cumulative distribution function of your variable $X$. But this definition does not make sense for a discrete distribution, since the cumulative distribution is not injective, and hence non-invertible.
A possible solution is to define a pseudo-inverse $F_X^{-}$ as follows
$$F_X^{-}(y)=inf{x in mathbb{R}|F_X(x)geq y}$$
You can now define the median as being $F_X^-(0.5)$. If you apply this definition on a uniform discrete distribution on the numbers ${1,2,3,4,5}$, you get $3$ as a result. While applying it on the dice ($n=6$) will get you $3$. Maybe you're not very satisfied with that last result. So let's try another definition.
We now define $F_X^{+}$ as
$$F_X^{+}(y)=sup{x in mathbb{R}|F_X(x)leq y}$$
The median now becomes $F_X^{+}(0.5)$. Let's see what that gives us for our previous distributions. For the case of the uniform discrete distribution on ${1,2,3,4,5}$, we still get $3$, but for the dice we now get $4$. Still not happy I suppose?
But wait, can't we just take the mean of those two? In fact, you can do even more you can define
$$F_X^{alpha}(y)=alpha F_X^{-}(y)+(1-alpha)F_X^{+}(y)$$
for any $alpha in [0,1]$. It is now clear that if we pick $alpha=0.5$, we will get exactly what we wanted! In the even case, the median will be $frac{n+1}{2}$.
answered Nov 6 '17 at 18:24
RaskolnikovRaskolnikov
12.6k23571
$endgroup$
$begingroup$
For continuous distributions as well, there can be several medians.
$endgroup$
– Did
Nov 6 '17 at 19:40
$begingroup$
True, I didn't think more deeply about it, but it is easy to construct continuous distributions which are non-injective and that therefore have the same ambiguity problem in defining the median as discrete distributions.
$endgroup$
– Raskolnikov
Nov 6 '17 at 19:50
add a comment |
$begingroup$
If the cumulative distribution function is strictly increasing, you can define the median $M$ unambiguously as
$$M=F_X^{-1}(0.5)$$
in which $F_X$ is the cumulative distribution function of your variable $X$. But this definition does not make sense for a discrete distribution, since the cumulative distribution is not injective, and hence non-invertible.
A possible solution is to define a pseudo-inverse $F_X^{-}$ as follows
$$F_X^{-}(y)=inf{x in mathbb{R}|F_X(x)geq y}$$
You can now define the median as being $F_X^-(0.5)$. If you apply this definition on a uniform discrete distribution on the numbers ${1,2,3,4,5}$, you get $3$ as a result. While applying it on the dice ($n=6$) will get you $3$. Maybe you're not very satisfied with that last result. So let's try another definition.
We now define $F_X^{+}$ as
$$F_X^{+}(y)=sup{x in mathbb{R}|F_X(x)leq y}$$
The median now becomes $F_X^{+}(0.5)$. Let's see what that gives us for our previous distributions. For the case of the uniform discrete distribution on ${1,2,3,4,5}$, we still get $3$, but for the dice we now get $4$. Still not happy I suppose?
But wait, can't we just take the mean of those two? In fact, you can do even more you can define
$$F_X^{alpha}(y)=alpha F_X^{-}(y)+(1-alpha)F_X^{+}(y)$$
for any $alpha in [0,1]$. It is now clear that if we pick $alpha=0.5$, we will get exactly what we wanted! In the even case, the median will be $frac{n+1}{2}$.
answered Nov 6 '17 at 18:24
RaskolnikovRaskolnikov
12.6k23571
$endgroup$
$begingroup$
For continuous distributions as well, there can be several medians.
$endgroup$
– Did
Nov 6 '17 at 19:40
$begingroup$
True, I didn't think more deeply about it, but it is easy to construct continuous distributions which are non-injective and that therefore have the same ambiguity problem in defining the median as discrete distributions.
$endgroup$
– Raskolnikov
Nov 6 '17 at 19:50
add a comment |
$begingroup$
If the cumulative distribution function is strictly increasing, you can define the median $M$ unambiguously as
$$M=F_X^{-1}(0.5)$$
in which $F_X$ is the cumulative distribution function of your variable $X$. But this definition does not make sense for a discrete distribution, since the cumulative distribution is not injective, and hence non-invertible.
A possible solution is to define a pseudo-inverse $F_X^{-}$ as follows
$$F_X^{-}(y)=inf{x in mathbb{R}|F_X(x)geq y}$$
You can now define the median as being $F_X^-(0.5)$. If you apply this definition on a uniform discrete distribution on the numbers ${1,2,3,4,5}$, you get $3$ as a result. While applying it on the dice ($n=6$) will get you $3$. Maybe you're not very satisfied with that last result. So let's try another definition.
We now define $F_X^{+}$ as
$$F_X^{+}(y)=sup{x in mathbb{R}|F_X(x)leq y}$$
The median now becomes $F_X^{+}(0.5)$. Let's see what that gives us for our previous distributions. For the case of the uniform discrete distribution on ${1,2,3,4,5}$, we still get $3$, but for the dice we now get $4$. Still not happy I suppose?
But wait, can't we just take the mean of those two? In fact, you can do even more you can define
$$F_X^{alpha}(y)=alpha F_X^{-}(y)+(1-alpha)F_X^{+}(y)$$
for any $alpha in [0,1]$. It is now clear that if we pick $alpha=0.5$, we will get exactly what we wanted! In the even case, the median will be $frac{n+1}{2}$.
answered Nov 6 '17 at 18:24
RaskolnikovRaskolnikov
12.6k23571
$endgroup$
If the cumulative distribution function is strictly increasing, you can define the median $M$ unambiguously as
$$M=F_X^{-1}(0.5)$$
in which $F_X$ is the cumulative distribution function of your variable $X$. But this definition does not make sense for a discrete distribution, since the cumulative distribution is not injective, and hence non-invertible.
A possible solution is to define a pseudo-inverse $F_X^{-}$ as follows
$$F_X^{-}(y)=inf{x in mathbb{R}|F_X(x)geq y}$$
You can now define the median as being $F_X^-(0.5)$. If you apply this definition on a uniform discrete distribution on the numbers ${1,2,3,4,5}$, you get $3$ as a result. While applying it on the dice ($n=6$) will get you $3$. Maybe you're not very satisfied with that last result. So let's try another definition.
We now define $F_X^{+}$ as
$$F_X^{+}(y)=sup{x in mathbb{R}|F_X(x)leq y}$$
The median now becomes $F_X^{+}(0.5)$. Let's see what that gives us for our previous distributions. For the case of the uniform discrete distribution on ${1,2,3,4,5}$, we still get $3$, but for the dice we now get $4$. Still not happy I suppose?
But wait, can't we just take the mean of those two? In fact, you can do even more you can define
$$F_X^{alpha}(y)=alpha F_X^{-}(y)+(1-alpha)F_X^{+}(y)$$
for any $alpha in [0,1]$. It is now clear that if we pick $alpha=0.5$, we will get exactly what we wanted! In the even case, the median will be $frac{n+1}{2}$.
answered Nov 6 '17 at 18:24
RaskolnikovRaskolnikov
12.6k23571
edited Nov 6 '17 at 19:52
answered Nov 6 '17 at 18:24
RaskolnikovRaskolnikov
12.6k23571
answered Nov 6 '17 at 18:24
RaskolnikovRaskolnikov
12.6k23571
answered Nov 6 '17 at 18:24
RaskolnikovRaskolnikov
12.6k23571
12.6k23571
$begingroup$
For continuous distributions as well, there can be several medians.
$endgroup$
– Did
Nov 6 '17 at 19:40
$begingroup$
True, I didn't think more deeply about it, but it is easy to construct continuous distributions which are non-injective and that therefore have the same ambiguity problem in defining the median as discrete distributions.
$endgroup$
– Raskolnikov
Nov 6 '17 at 19:50
add a comment |
$begingroup$
For continuous distributions as well, there can be several medians.
$endgroup$
– Did
Nov 6 '17 at 19:40
$begingroup$
True, I didn't think more deeply about it, but it is easy to construct continuous distributions which are non-injective and that therefore have the same ambiguity problem in defining the median as discrete distributions.
$endgroup$
– Raskolnikov
Nov 6 '17 at 19:50
$begingroup$
For continuous distributions as well, there can be several medians.
$endgroup$
– Did
Nov 6 '17 at 19:40
$begingroup$
For continuous distributions as well, there can be several medians.
$endgroup$
– Did
Nov 6 '17 at 19:40
$begingroup$
True, I didn't think more deeply about it, but it is easy to construct continuous distributions which are non-injective and that therefore have the same ambiguity problem in defining the median as discrete distributions.
$endgroup$
– Raskolnikov
Nov 6 '17 at 19:50
$begingroup$
True, I didn't think more deeply about it, but it is easy to construct continuous distributions which are non-injective and that therefore have the same ambiguity problem in defining the median as discrete distributions.
$endgroup$
– Raskolnikov
Nov 6 '17 at 19:50
add a comment |
$begingroup$
Modes are local maximums of the PDF, therefore there are $n$ modes: 1, 2,...,$n$. In this case you can see that talking about a mode does not make much sense, or at least is not really enlightening (but then again, you already know that the random variable is uniform...)
The median depends on whether $n$ is even of odd.
- If $n$ is odd, then there is a single median, ${n+1}over2$.
- If $n$ is even then any value in the (closed) interval [${nover2},
{nover2}+1$] qualifies as a median.
This with the following definition of a median : any value $m$ such that
$$
P(X<m)leq .5 quad text{and } P(X>m)leq .5.
$$
Of course the default median is ${n+1}over2$ regardless of $n$.
answered Nov 6 '17 at 18:34
A.G.A.G.
2,020711
$endgroup$
add a comment |
$begingroup$
Modes are local maximums of the PDF, therefore there are $n$ modes: 1, 2,...,$n$. In this case you can see that talking about a mode does not make much sense, or at least is not really enlightening (but then again, you already know that the random variable is uniform...)
The median depends on whether $n$ is even of odd.
- If $n$ is odd, then there is a single median, ${n+1}over2$.
- If $n$ is even then any value in the (closed) interval [${nover2},
{nover2}+1$] qualifies as a median.
This with the following definition of a median : any value $m$ such that
$$
P(X<m)leq .5 quad text{and } P(X>m)leq .5.
$$
Of course the default median is ${n+1}over2$ regardless of $n$.
answered Nov 6 '17 at 18:34
A.G.A.G.
2,020711
$endgroup$
add a comment |
$begingroup$
Modes are local maximums of the PDF, therefore there are $n$ modes: 1, 2,...,$n$. In this case you can see that talking about a mode does not make much sense, or at least is not really enlightening (but then again, you already know that the random variable is uniform...)
The median depends on whether $n$ is even of odd.
- If $n$ is odd, then there is a single median, ${n+1}over2$.
- If $n$ is even then any value in the (closed) interval [${nover2},
{nover2}+1$] qualifies as a median.
This with the following definition of a median : any value $m$ such that
$$
P(X<m)leq .5 quad text{and } P(X>m)leq .5.
$$
Of course the default median is ${n+1}over2$ regardless of $n$.
answered Nov 6 '17 at 18:34
A.G.A.G.
2,020711
$endgroup$
Modes are local maximums of the PDF, therefore there are $n$ modes: 1, 2,...,$n$. In this case you can see that talking about a mode does not make much sense, or at least is not really enlightening (but then again, you already know that the random variable is uniform...)
The median depends on whether $n$ is even of odd.
- If $n$ is odd, then there is a single median, ${n+1}over2$.
- If $n$ is even then any value in the (closed) interval [${nover2},
{nover2}+1$] qualifies as a median.
This with the following definition of a median : any value $m$ such that
$$
P(X<m)leq .5 quad text{and } P(X>m)leq .5.
$$
Of course the default median is ${n+1}over2$ regardless of $n$.
answered Nov 6 '17 at 18:34
A.G.A.G.
2,020711
edited Nov 6 '17 at 21:32
answered Nov 6 '17 at 18:34
A.G.A.G.
2,020711
answered Nov 6 '17 at 18:34
A.G.A.G.
2,020711
answered Nov 6 '17 at 18:34
A.G.A.G.
2,020711
2,020711
add a comment |
add a comment |
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I believe the question is asking about the median and modes of a distribution, not a sample. For $n = 6,$ it would be the 'fair die' distribution that puts probability $1/6$ at each of the values $1,2,3,4,5,6.$ So the median of this distribution is $eta = 3.5$ and (depending on definition) there is no mode or there multiple modes at each of the six equally likely values.
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– BruceET
Nov 6 '17 at 7:43
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@BruceET If you accept multiple modes, then you should be able to accept multiple medians, so for even $n$ either $left[ frac n2, frac n 2 +1 right]$ or $left{ frac n2, frac n 2 +1 right}$ depending on whether the median should be part of the support
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– Henry
Nov 6 '17 at 18:36
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I view the issues with medians and modes to be quite different, but I don't disagree with your suggestion for medians. [In general, are there multiple modes only if all have the same max density/rel.freq., or does any local maximum qualify as a (perhaps) 'minor' mode?]
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– BruceET
Nov 6 '17 at 20:06