Mixing
How to get the value of 'scaled' binomial distribution?
$begingroup$
People kindly told me that there is not a equivalent popular distribution for $aX$ when $X$ is distributed as Binomial, but it is just a 'scaled' distribution. Here, $a$ is a positive constant.
Question
How to get the value of 'scaled' binomial distribution? Do I need to use $Gamma$ function? A simpler way is appreciated.
To be more precise, I want to calculate $$g(x) = frac{1}{|a|}f(frac xa) = {n choose x/a} p^{frac xa} (1-p)^{(n-frac xa)},$$ where $f(x)$ is $B(n,p)$ and $a$ is positive constant.
What is the simplest way to calculate $g(x)$?
Background
I'm considering this question under the same condition with this question, where $X$ is distributed as $Binomial$ and I want to know the distribution of $aX$. $a$ is a real and positive number.
I know
$aX sim frac{1}{|a|}f(frac xa)$ ($f$ means the distribution of $X$),so I tried to calculate the probability of $aX$ by $frac{1}{|a|}f(frac xa) = {n choose x/a} p^{frac ax} (1-p)^{(n-frac xa)}$. But, this was not easy for me because $x/a$ is not a integer.
I think with $Gamma$ function $nCr$ can be calculated for any real number $r (< n)$. But is there any straightforward way to calculate 'scaled' binomial distribution?
probability probability-distributions
edited Apr 13 '17 at 12:19
Community♦
1
asked Apr 27 '15 at 10:11
rkjt50r983rkjt50r983
259210
$endgroup$
add a comment |
$begingroup$
People kindly told me that there is not a equivalent popular distribution for $aX$ when $X$ is distributed as Binomial, but it is just a 'scaled' distribution. Here, $a$ is a positive constant.
Question
How to get the value of 'scaled' binomial distribution? Do I need to use $Gamma$ function? A simpler way is appreciated.
To be more precise, I want to calculate $$g(x) = frac{1}{|a|}f(frac xa) = {n choose x/a} p^{frac xa} (1-p)^{(n-frac xa)},$$ where $f(x)$ is $B(n,p)$ and $a$ is positive constant.
What is the simplest way to calculate $g(x)$?
Background
I'm considering this question under the same condition with this question, where $X$ is distributed as $Binomial$ and I want to know the distribution of $aX$. $a$ is a real and positive number.
I know
$aX sim frac{1}{|a|}f(frac xa)$ ($f$ means the distribution of $X$),so I tried to calculate the probability of $aX$ by $frac{1}{|a|}f(frac xa) = {n choose x/a} p^{frac ax} (1-p)^{(n-frac xa)}$. But, this was not easy for me because $x/a$ is not a integer.
I think with $Gamma$ function $nCr$ can be calculated for any real number $r (< n)$. But is there any straightforward way to calculate 'scaled' binomial distribution?
probability probability-distributions
edited Apr 13 '17 at 12:19
Community♦
1
asked Apr 27 '15 at 10:11
rkjt50r983rkjt50r983
259210
$endgroup$
$begingroup$
You may want to look at the beta function—en.wikipedia.org/wiki/Beta_function—as it provides a way to extend the binomial distribution to real values. (It's indicated near the end of the "Properties" section.)
$endgroup$
– Brian Tung
Apr 27 '15 at 16:54
$begingroup$
So, yes, you will need to use the gamma function. I suspect there's no way around it for your application.
$endgroup$
– Brian Tung
Apr 27 '15 at 16:55
$begingroup$
@BrianTung thank you. This is what I'm looking for. If you replace your comment to an answer, I would give a check mark :)
$endgroup$
– rkjt50r983
Apr 28 '15 at 2:15
$begingroup$
Thanks for your kind offer! But I'd probably only move it to an answer if you're looking to give out a checkmark. (I don't remember if you get any points for doing that.) Otherwise, if my contribution is merely a link to another page, I generally leave it in a comment. If it helped you out, that's good enough for me.
$endgroup$
– Brian Tung
Apr 28 '15 at 16:23
$begingroup$
@rkjt50r983, your scale function is not a density function because it does not normalise to 1. I think you consider that as a critical issue as well.
$endgroup$
– Chamberlain Foncha
Nov 12 '17 at 15:01
add a comment |
$begingroup$
People kindly told me that there is not a equivalent popular distribution for $aX$ when $X$ is distributed as Binomial, but it is just a 'scaled' distribution. Here, $a$ is a positive constant.
Question
How to get the value of 'scaled' binomial distribution? Do I need to use $Gamma$ function? A simpler way is appreciated.
To be more precise, I want to calculate $$g(x) = frac{1}{|a|}f(frac xa) = {n choose x/a} p^{frac xa} (1-p)^{(n-frac xa)},$$ where $f(x)$ is $B(n,p)$ and $a$ is positive constant.
What is the simplest way to calculate $g(x)$?
Background
I'm considering this question under the same condition with this question, where $X$ is distributed as $Binomial$ and I want to know the distribution of $aX$. $a$ is a real and positive number.
I know
$aX sim frac{1}{|a|}f(frac xa)$ ($f$ means the distribution of $X$),so I tried to calculate the probability of $aX$ by $frac{1}{|a|}f(frac xa) = {n choose x/a} p^{frac ax} (1-p)^{(n-frac xa)}$. But, this was not easy for me because $x/a$ is not a integer.
I think with $Gamma$ function $nCr$ can be calculated for any real number $r (< n)$. But is there any straightforward way to calculate 'scaled' binomial distribution?
probability probability-distributions
edited Apr 13 '17 at 12:19
Community♦
1
asked Apr 27 '15 at 10:11
rkjt50r983rkjt50r983
259210
$endgroup$
People kindly told me that there is not a equivalent popular distribution for $aX$ when $X$ is distributed as Binomial, but it is just a 'scaled' distribution. Here, $a$ is a positive constant.
Question
How to get the value of 'scaled' binomial distribution? Do I need to use $Gamma$ function? A simpler way is appreciated.
To be more precise, I want to calculate $$g(x) = frac{1}{|a|}f(frac xa) = {n choose x/a} p^{frac xa} (1-p)^{(n-frac xa)},$$ where $f(x)$ is $B(n,p)$ and $a$ is positive constant.
What is the simplest way to calculate $g(x)$?
Background
I'm considering this question under the same condition with this question, where $X$ is distributed as $Binomial$ and I want to know the distribution of $aX$. $a$ is a real and positive number.
I know
$aX sim frac{1}{|a|}f(frac xa)$ ($f$ means the distribution of $X$),so I tried to calculate the probability of $aX$ by $frac{1}{|a|}f(frac xa) = {n choose x/a} p^{frac ax} (1-p)^{(n-frac xa)}$. But, this was not easy for me because $x/a$ is not a integer.
I think with $Gamma$ function $nCr$ can be calculated for any real number $r (< n)$. But is there any straightforward way to calculate 'scaled' binomial distribution?
probability probability-distributions
probability probability-distributions
edited Apr 13 '17 at 12:19
Community♦
1
asked Apr 27 '15 at 10:11
rkjt50r983rkjt50r983
259210
edited Apr 13 '17 at 12:19
Community♦
1
asked Apr 27 '15 at 10:11
rkjt50r983rkjt50r983
259210
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
asked Apr 27 '15 at 10:11
rkjt50r983rkjt50r983
259210
asked Apr 27 '15 at 10:11
rkjt50r983rkjt50r983
259210
asked Apr 27 '15 at 10:11
rkjt50r983rkjt50r983
259210
259210
$begingroup$
You may want to look at the beta function—en.wikipedia.org/wiki/Beta_function—as it provides a way to extend the binomial distribution to real values. (It's indicated near the end of the "Properties" section.)
$endgroup$
– Brian Tung
Apr 27 '15 at 16:54
$begingroup$
So, yes, you will need to use the gamma function. I suspect there's no way around it for your application.
$endgroup$
– Brian Tung
Apr 27 '15 at 16:55
$begingroup$
@BrianTung thank you. This is what I'm looking for. If you replace your comment to an answer, I would give a check mark :)
$endgroup$
– rkjt50r983
Apr 28 '15 at 2:15
$begingroup$
Thanks for your kind offer! But I'd probably only move it to an answer if you're looking to give out a checkmark. (I don't remember if you get any points for doing that.) Otherwise, if my contribution is merely a link to another page, I generally leave it in a comment. If it helped you out, that's good enough for me.
$endgroup$
– Brian Tung
Apr 28 '15 at 16:23
$begingroup$
@rkjt50r983, your scale function is not a density function because it does not normalise to 1. I think you consider that as a critical issue as well.
$endgroup$
– Chamberlain Foncha
Nov 12 '17 at 15:01
add a comment |
$begingroup$
You may want to look at the beta function—en.wikipedia.org/wiki/Beta_function—as it provides a way to extend the binomial distribution to real values. (It's indicated near the end of the "Properties" section.)
$endgroup$
– Brian Tung
Apr 27 '15 at 16:54
$begingroup$
So, yes, you will need to use the gamma function. I suspect there's no way around it for your application.
$endgroup$
– Brian Tung
Apr 27 '15 at 16:55
$begingroup$
@BrianTung thank you. This is what I'm looking for. If you replace your comment to an answer, I would give a check mark :)
$endgroup$
– rkjt50r983
Apr 28 '15 at 2:15
$begingroup$
Thanks for your kind offer! But I'd probably only move it to an answer if you're looking to give out a checkmark. (I don't remember if you get any points for doing that.) Otherwise, if my contribution is merely a link to another page, I generally leave it in a comment. If it helped you out, that's good enough for me.
$endgroup$
– Brian Tung
Apr 28 '15 at 16:23
$begingroup$
@rkjt50r983, your scale function is not a density function because it does not normalise to 1. I think you consider that as a critical issue as well.
$endgroup$
– Chamberlain Foncha
Nov 12 '17 at 15:01
$begingroup$
You may want to look at the beta function—en.wikipedia.org/wiki/Beta_function—as it provides a way to extend the binomial distribution to real values. (It's indicated near the end of the "Properties" section.)
$endgroup$
– Brian Tung
Apr 27 '15 at 16:54
$begingroup$
You may want to look at the beta function—en.wikipedia.org/wiki/Beta_function—as it provides a way to extend the binomial distribution to real values. (It's indicated near the end of the "Properties" section.)
$endgroup$
– Brian Tung
Apr 27 '15 at 16:54
$begingroup$
So, yes, you will need to use the gamma function. I suspect there's no way around it for your application.
$endgroup$
– Brian Tung
Apr 27 '15 at 16:55
$begingroup$
So, yes, you will need to use the gamma function. I suspect there's no way around it for your application.
$endgroup$
– Brian Tung
Apr 27 '15 at 16:55
$begingroup$
@BrianTung thank you. This is what I'm looking for. If you replace your comment to an answer, I would give a check mark :)
$endgroup$
– rkjt50r983
Apr 28 '15 at 2:15
$begingroup$
@BrianTung thank you. This is what I'm looking for. If you replace your comment to an answer, I would give a check mark :)
$endgroup$
– rkjt50r983
Apr 28 '15 at 2:15
$begingroup$
Thanks for your kind offer! But I'd probably only move it to an answer if you're looking to give out a checkmark. (I don't remember if you get any points for doing that.) Otherwise, if my contribution is merely a link to another page, I generally leave it in a comment. If it helped you out, that's good enough for me.
$endgroup$
– Brian Tung
Apr 28 '15 at 16:23
$begingroup$
Thanks for your kind offer! But I'd probably only move it to an answer if you're looking to give out a checkmark. (I don't remember if you get any points for doing that.) Otherwise, if my contribution is merely a link to another page, I generally leave it in a comment. If it helped you out, that's good enough for me.
$endgroup$
– Brian Tung
Apr 28 '15 at 16:23
$begingroup$
@rkjt50r983, your scale function is not a density function because it does not normalise to 1. I think you consider that as a critical issue as well.
$endgroup$
– Chamberlain Foncha
Nov 12 '17 at 15:01
$begingroup$
@rkjt50r983, your scale function is not a density function because it does not normalise to 1. I think you consider that as a critical issue as well.
$endgroup$
– Chamberlain Foncha
Nov 12 '17 at 15:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For a random variable $X$ of binomial distribution of parameters $n,p$ the possible values taken are $0,1,2, cdots, n$ with probabilities$$P(X=k)={n choose k}p^k(1-p)^{n-k}.$$
If $a$ is a real constant and $Y=aX$ then the valueas taken by $Y$ are $1a,2a,cdots , na$ with probabilities $$P(Y=ak)=P(X=k).$$
I don't know if this answers the question.
EDITED
Perhaps, the reason of confusion is that there is no pdf. The cdf is
$$F_Y(x)=P(aX<x)=Pleft(X<frac{x}{a}right)=F_Xleft(frac{x}{a}right).$$
If the derivative existed then one could get a formula similar to what you quoted for the Gaussian distribution.
answered Apr 27 '15 at 10:35
zolizoli
17k41945
$endgroup$
$begingroup$
thank you, but I think it is necessary to calculate ${n choose k/a}p^{k/a}(1-p)^{n-k/a}.$. More specifically, how do I define and calculate ${n choose k/a}$?
$endgroup$
– rkjt50r983
Apr 27 '15 at 10:47
$begingroup$
@rkjt50r938, The probabilities that $Y=ai$ are given. What else would want anybody know? The is no pdf in this case. Perhaps you want see the form of the cdf??
$endgroup$
– zoli
Apr 27 '15 at 10:55
$begingroup$
@rkjt50r938: I've edited my answer.
$endgroup$
– zoli
Apr 27 '15 at 11:39
add a comment |
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$begingroup$
For a random variable $X$ of binomial distribution of parameters $n,p$ the possible values taken are $0,1,2, cdots, n$ with probabilities$$P(X=k)={n choose k}p^k(1-p)^{n-k}.$$
If $a$ is a real constant and $Y=aX$ then the valueas taken by $Y$ are $1a,2a,cdots , na$ with probabilities $$P(Y=ak)=P(X=k).$$
I don't know if this answers the question.
EDITED
Perhaps, the reason of confusion is that there is no pdf. The cdf is
$$F_Y(x)=P(aX<x)=Pleft(X<frac{x}{a}right)=F_Xleft(frac{x}{a}right).$$
If the derivative existed then one could get a formula similar to what you quoted for the Gaussian distribution.
answered Apr 27 '15 at 10:35
zolizoli
17k41945
$endgroup$
$begingroup$
thank you, but I think it is necessary to calculate ${n choose k/a}p^{k/a}(1-p)^{n-k/a}.$. More specifically, how do I define and calculate ${n choose k/a}$?
$endgroup$
– rkjt50r983
Apr 27 '15 at 10:47
$begingroup$
@rkjt50r938, The probabilities that $Y=ai$ are given. What else would want anybody know? The is no pdf in this case. Perhaps you want see the form of the cdf??
$endgroup$
– zoli
Apr 27 '15 at 10:55
$begingroup$
@rkjt50r938: I've edited my answer.
$endgroup$
– zoli
Apr 27 '15 at 11:39
add a comment |
$begingroup$
For a random variable $X$ of binomial distribution of parameters $n,p$ the possible values taken are $0,1,2, cdots, n$ with probabilities$$P(X=k)={n choose k}p^k(1-p)^{n-k}.$$
If $a$ is a real constant and $Y=aX$ then the valueas taken by $Y$ are $1a,2a,cdots , na$ with probabilities $$P(Y=ak)=P(X=k).$$
I don't know if this answers the question.
EDITED
Perhaps, the reason of confusion is that there is no pdf. The cdf is
$$F_Y(x)=P(aX<x)=Pleft(X<frac{x}{a}right)=F_Xleft(frac{x}{a}right).$$
If the derivative existed then one could get a formula similar to what you quoted for the Gaussian distribution.
answered Apr 27 '15 at 10:35
zolizoli
17k41945
$endgroup$
$begingroup$
thank you, but I think it is necessary to calculate ${n choose k/a}p^{k/a}(1-p)^{n-k/a}.$. More specifically, how do I define and calculate ${n choose k/a}$?
$endgroup$
– rkjt50r983
Apr 27 '15 at 10:47
$begingroup$
@rkjt50r938, The probabilities that $Y=ai$ are given. What else would want anybody know? The is no pdf in this case. Perhaps you want see the form of the cdf??
$endgroup$
– zoli
Apr 27 '15 at 10:55
$begingroup$
@rkjt50r938: I've edited my answer.
$endgroup$
– zoli
Apr 27 '15 at 11:39
add a comment |
$begingroup$
For a random variable $X$ of binomial distribution of parameters $n,p$ the possible values taken are $0,1,2, cdots, n$ with probabilities$$P(X=k)={n choose k}p^k(1-p)^{n-k}.$$
If $a$ is a real constant and $Y=aX$ then the valueas taken by $Y$ are $1a,2a,cdots , na$ with probabilities $$P(Y=ak)=P(X=k).$$
I don't know if this answers the question.
EDITED
Perhaps, the reason of confusion is that there is no pdf. The cdf is
$$F_Y(x)=P(aX<x)=Pleft(X<frac{x}{a}right)=F_Xleft(frac{x}{a}right).$$
If the derivative existed then one could get a formula similar to what you quoted for the Gaussian distribution.
answered Apr 27 '15 at 10:35
zolizoli
17k41945
$endgroup$
For a random variable $X$ of binomial distribution of parameters $n,p$ the possible values taken are $0,1,2, cdots, n$ with probabilities$$P(X=k)={n choose k}p^k(1-p)^{n-k}.$$
If $a$ is a real constant and $Y=aX$ then the valueas taken by $Y$ are $1a,2a,cdots , na$ with probabilities $$P(Y=ak)=P(X=k).$$
I don't know if this answers the question.
EDITED
Perhaps, the reason of confusion is that there is no pdf. The cdf is
$$F_Y(x)=P(aX<x)=Pleft(X<frac{x}{a}right)=F_Xleft(frac{x}{a}right).$$
If the derivative existed then one could get a formula similar to what you quoted for the Gaussian distribution.
answered Apr 27 '15 at 10:35
zolizoli
17k41945
edited Apr 27 '15 at 18:16
answered Apr 27 '15 at 10:35
zolizoli
17k41945
answered Apr 27 '15 at 10:35
zolizoli
17k41945
answered Apr 27 '15 at 10:35
zolizoli
17k41945
17k41945
$begingroup$
thank you, but I think it is necessary to calculate ${n choose k/a}p^{k/a}(1-p)^{n-k/a}.$. More specifically, how do I define and calculate ${n choose k/a}$?
$endgroup$
– rkjt50r983
Apr 27 '15 at 10:47
$begingroup$
@rkjt50r938, The probabilities that $Y=ai$ are given. What else would want anybody know? The is no pdf in this case. Perhaps you want see the form of the cdf??
$endgroup$
– zoli
Apr 27 '15 at 10:55
$begingroup$
@rkjt50r938: I've edited my answer.
$endgroup$
– zoli
Apr 27 '15 at 11:39
add a comment |
$begingroup$
thank you, but I think it is necessary to calculate ${n choose k/a}p^{k/a}(1-p)^{n-k/a}.$. More specifically, how do I define and calculate ${n choose k/a}$?
$endgroup$
– rkjt50r983
Apr 27 '15 at 10:47
$begingroup$
@rkjt50r938, The probabilities that $Y=ai$ are given. What else would want anybody know? The is no pdf in this case. Perhaps you want see the form of the cdf??
$endgroup$
– zoli
Apr 27 '15 at 10:55
$begingroup$
@rkjt50r938: I've edited my answer.
$endgroup$
– zoli
Apr 27 '15 at 11:39
$begingroup$
thank you, but I think it is necessary to calculate ${n choose k/a}p^{k/a}(1-p)^{n-k/a}.$. More specifically, how do I define and calculate ${n choose k/a}$?
$endgroup$
– rkjt50r983
Apr 27 '15 at 10:47
$begingroup$
thank you, but I think it is necessary to calculate ${n choose k/a}p^{k/a}(1-p)^{n-k/a}.$. More specifically, how do I define and calculate ${n choose k/a}$?
$endgroup$
– rkjt50r983
Apr 27 '15 at 10:47
$begingroup$
@rkjt50r938, The probabilities that $Y=ai$ are given. What else would want anybody know? The is no pdf in this case. Perhaps you want see the form of the cdf??
$endgroup$
– zoli
Apr 27 '15 at 10:55
$begingroup$
@rkjt50r938, The probabilities that $Y=ai$ are given. What else would want anybody know? The is no pdf in this case. Perhaps you want see the form of the cdf??
$endgroup$
– zoli
Apr 27 '15 at 10:55
$begingroup$
@rkjt50r938: I've edited my answer.
$endgroup$
– zoli
Apr 27 '15 at 11:39
$begingroup$
@rkjt50r938: I've edited my answer.
$endgroup$
– zoli
Apr 27 '15 at 11:39
add a comment |
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$begingroup$
You may want to look at the beta function—en.wikipedia.org/wiki/Beta_function—as it provides a way to extend the binomial distribution to real values. (It's indicated near the end of the "Properties" section.)
$endgroup$
– Brian Tung
Apr 27 '15 at 16:54
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So, yes, you will need to use the gamma function. I suspect there's no way around it for your application.
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– Brian Tung
Apr 27 '15 at 16:55
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@BrianTung thank you. This is what I'm looking for. If you replace your comment to an answer, I would give a check mark :)
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– rkjt50r983
Apr 28 '15 at 2:15
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Thanks for your kind offer! But I'd probably only move it to an answer if you're looking to give out a checkmark. (I don't remember if you get any points for doing that.) Otherwise, if my contribution is merely a link to another page, I generally leave it in a comment. If it helped you out, that's good enough for me.
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– Brian Tung
Apr 28 '15 at 16:23
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@rkjt50r983, your scale function is not a density function because it does not normalise to 1. I think you consider that as a critical issue as well.
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– Chamberlain Foncha
Nov 12 '17 at 15:01