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What is the name of the function $D(a,x) = fracx^a e^-xGamma(a+1)$?
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Does the function $dfracx^a e^-xGamma(a+1)$ have its own specific name?
Temme [1] introduced the function in (3.1)
$$D(a,x) = fracx^a e^-xGamma(a+1).$$ It is the dominant part in many representations of the incomplete gamma functions (his corresponding Algol function is named dax).
Earlier DiDonato and Morris [2] called it $R(a,x)$.
A closely related function is called regularised_gamma_prefix in the Boost library.
I did not find a 'name' for this function. Are there more known references to it?
[1] N.M. Temme, A Set of Algorithms for the Incomplete Gamma Functions, Probability
in the Engineering and Informational Sciences, 8 (1994), pp. 291-307. Available as
https://ir.cwi.nl/pub/10080/10080D.pdf
[2] A.R. DiDonato, A.H. Morris, Computation of the Incomplete Gamma Function
Ratios and their Inverse. ACM TOMS, Vol 12, No 4, Dec 1986, pp. 377-393.
functions notation special-functions gamma-function
asked Jul 27 at 15:57
gammatester
15.7k21429
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up vote
5
down vote
favorite
Does the function $dfracx^a e^-xGamma(a+1)$ have its own specific name?
Temme [1] introduced the function in (3.1)
$$D(a,x) = fracx^a e^-xGamma(a+1).$$ It is the dominant part in many representations of the incomplete gamma functions (his corresponding Algol function is named dax).
Earlier DiDonato and Morris [2] called it $R(a,x)$.
A closely related function is called regularised_gamma_prefix in the Boost library.
I did not find a 'name' for this function. Are there more known references to it?
[1] N.M. Temme, A Set of Algorithms for the Incomplete Gamma Functions, Probability
in the Engineering and Informational Sciences, 8 (1994), pp. 291-307. Available as
https://ir.cwi.nl/pub/10080/10080D.pdf
[2] A.R. DiDonato, A.H. Morris, Computation of the Incomplete Gamma Function
Ratios and their Inverse. ACM TOMS, Vol 12, No 4, Dec 1986, pp. 377-393.
functions notation special-functions gamma-function
asked Jul 27 at 15:57
gammatester
15.7k21429
1
Gamma PDF with parameters $(a+1,1)$.
– Did
Jul 27 at 16:24
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Does the function $dfracx^a e^-xGamma(a+1)$ have its own specific name?
Temme [1] introduced the function in (3.1)
$$D(a,x) = fracx^a e^-xGamma(a+1).$$ It is the dominant part in many representations of the incomplete gamma functions (his corresponding Algol function is named dax).
Earlier DiDonato and Morris [2] called it $R(a,x)$.
A closely related function is called regularised_gamma_prefix in the Boost library.
I did not find a 'name' for this function. Are there more known references to it?
[1] N.M. Temme, A Set of Algorithms for the Incomplete Gamma Functions, Probability
in the Engineering and Informational Sciences, 8 (1994), pp. 291-307. Available as
https://ir.cwi.nl/pub/10080/10080D.pdf
[2] A.R. DiDonato, A.H. Morris, Computation of the Incomplete Gamma Function
Ratios and their Inverse. ACM TOMS, Vol 12, No 4, Dec 1986, pp. 377-393.
functions notation special-functions gamma-function
asked Jul 27 at 15:57
gammatester
15.7k21429
Does the function $dfracx^a e^-xGamma(a+1)$ have its own specific name?
Temme [1] introduced the function in (3.1)
$$D(a,x) = fracx^a e^-xGamma(a+1).$$ It is the dominant part in many representations of the incomplete gamma functions (his corresponding Algol function is named dax).
Earlier DiDonato and Morris [2] called it $R(a,x)$.
A closely related function is called regularised_gamma_prefix in the Boost library.
I did not find a 'name' for this function. Are there more known references to it?
[1] N.M. Temme, A Set of Algorithms for the Incomplete Gamma Functions, Probability
in the Engineering and Informational Sciences, 8 (1994), pp. 291-307. Available as
https://ir.cwi.nl/pub/10080/10080D.pdf
[2] A.R. DiDonato, A.H. Morris, Computation of the Incomplete Gamma Function
Ratios and their Inverse. ACM TOMS, Vol 12, No 4, Dec 1986, pp. 377-393.
functions notation special-functions gamma-function
asked Jul 27 at 15:57
gammatester
15.7k21429
edited Jul 27 at 16:06
asked Jul 27 at 15:57
gammatester
15.7k21429
asked Jul 27 at 15:57
gammatester
15.7k21429
asked Jul 27 at 15:57
gammatester
15.7k21429
15.7k21429
1
Gamma PDF with parameters $(a+1,1)$.
– Did
Jul 27 at 16:24
add a comment |Â
1
Gamma PDF with parameters $(a+1,1)$.
– Did
Jul 27 at 16:24
1
1
Gamma PDF with parameters $(a+1,1)$.
– Did
Jul 27 at 16:24
Gamma PDF with parameters $(a+1,1)$.
– Did
Jul 27 at 16:24
add a comment |Â
1 Answer
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You may see it called the probability density function of a gamma distribution. That is the restriction of this function to the interval $[0,+infty).$ This distribution is that of the sum of $a+1$ independent random variables each of which is exponentially distributed, so that the probability of its being greater than $x$ is $e^-x$ for $xge0.$
One more often sees it written as $dfracx^alpha-1 e^-xGamma(alpha).$ That way $alpha$ is the number of exponentially distributed random variables that are added.
Among those probability distributions called gamma distributions one also finds instances in which a scale parameter is present, thus:
$$
frac left( x/mu right)^alpha-1 e^-x/mu Gamma(alpha) left( frac dx muright) text for xge0. tag$mu$
$$
Sometimes it is parametrized using the reciprocal of the scale parameter, $lambda= dfrac 1 mu:$
$$
frac(lambda x)^alpha-1 e^-lambda xGamma(alpha) (lambda , dx). tag$lambda$
$$
It may be that $alpha$ is not a positive integer, in which case the statement about the sum of $alpha$ random variables may seem not to make sense. But suppose $X_1,X_2$ are independent random variables with
$$
Pr(X_iin A) = int_A fracx^alpha_i-1 e^-xGamma(alpha_i) , dx quad textfor measurable A subseteq [0,+infty), quad i=1,2.
$$
In that case, one has
$$
Pr(X_1+X_2in A) = int_A fracx^alpha_1+alpha_2-1 e^-xGamma(alpha_1+alpha_2) , dx quad textfor measurable A subseteq [0,+infty).
$$
This last probability density function is the convolution of those corresponding to $alpha_1$ and $alpha_2.$ Thus putting $alpha-1$ in the exponent rather than $alpha$ makes this a convolution semigroup in which convolution of these functions corresponds to addition of $alpha$s.
When it's written in the $(lambda)$ form rather than the $mu$ form and $x$ is time in a Poisson process, then $lambda x$ is the average number of arrivals in a time interval of length $x.$ When it's written in the $(mu)$ form, then $mu$ is the average time until the next arrival.
answered Jul 27 at 16:12
Michael Hardy
204k23186461
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
You may see it called the probability density function of a gamma distribution. That is the restriction of this function to the interval $[0,+infty).$ This distribution is that of the sum of $a+1$ independent random variables each of which is exponentially distributed, so that the probability of its being greater than $x$ is $e^-x$ for $xge0.$
One more often sees it written as $dfracx^alpha-1 e^-xGamma(alpha).$ That way $alpha$ is the number of exponentially distributed random variables that are added.
Among those probability distributions called gamma distributions one also finds instances in which a scale parameter is present, thus:
$$
frac left( x/mu right)^alpha-1 e^-x/mu Gamma(alpha) left( frac dx muright) text for xge0. tag$mu$
$$
Sometimes it is parametrized using the reciprocal of the scale parameter, $lambda= dfrac 1 mu:$
$$
frac(lambda x)^alpha-1 e^-lambda xGamma(alpha) (lambda , dx). tag$lambda$
$$
It may be that $alpha$ is not a positive integer, in which case the statement about the sum of $alpha$ random variables may seem not to make sense. But suppose $X_1,X_2$ are independent random variables with
$$
Pr(X_iin A) = int_A fracx^alpha_i-1 e^-xGamma(alpha_i) , dx quad textfor measurable A subseteq [0,+infty), quad i=1,2.
$$
In that case, one has
$$
Pr(X_1+X_2in A) = int_A fracx^alpha_1+alpha_2-1 e^-xGamma(alpha_1+alpha_2) , dx quad textfor measurable A subseteq [0,+infty).
$$
This last probability density function is the convolution of those corresponding to $alpha_1$ and $alpha_2.$ Thus putting $alpha-1$ in the exponent rather than $alpha$ makes this a convolution semigroup in which convolution of these functions corresponds to addition of $alpha$s.
When it's written in the $(lambda)$ form rather than the $mu$ form and $x$ is time in a Poisson process, then $lambda x$ is the average number of arrivals in a time interval of length $x.$ When it's written in the $(mu)$ form, then $mu$ is the average time until the next arrival.
answered Jul 27 at 16:12
Michael Hardy
204k23186461
add a comment |Â
up vote
5
down vote
accepted
You may see it called the probability density function of a gamma distribution. That is the restriction of this function to the interval $[0,+infty).$ This distribution is that of the sum of $a+1$ independent random variables each of which is exponentially distributed, so that the probability of its being greater than $x$ is $e^-x$ for $xge0.$
One more often sees it written as $dfracx^alpha-1 e^-xGamma(alpha).$ That way $alpha$ is the number of exponentially distributed random variables that are added.
Among those probability distributions called gamma distributions one also finds instances in which a scale parameter is present, thus:
$$
frac left( x/mu right)^alpha-1 e^-x/mu Gamma(alpha) left( frac dx muright) text for xge0. tag$mu$
$$
Sometimes it is parametrized using the reciprocal of the scale parameter, $lambda= dfrac 1 mu:$
$$
frac(lambda x)^alpha-1 e^-lambda xGamma(alpha) (lambda , dx). tag$lambda$
$$
It may be that $alpha$ is not a positive integer, in which case the statement about the sum of $alpha$ random variables may seem not to make sense. But suppose $X_1,X_2$ are independent random variables with
$$
Pr(X_iin A) = int_A fracx^alpha_i-1 e^-xGamma(alpha_i) , dx quad textfor measurable A subseteq [0,+infty), quad i=1,2.
$$
In that case, one has
$$
Pr(X_1+X_2in A) = int_A fracx^alpha_1+alpha_2-1 e^-xGamma(alpha_1+alpha_2) , dx quad textfor measurable A subseteq [0,+infty).
$$
This last probability density function is the convolution of those corresponding to $alpha_1$ and $alpha_2.$ Thus putting $alpha-1$ in the exponent rather than $alpha$ makes this a convolution semigroup in which convolution of these functions corresponds to addition of $alpha$s.
When it's written in the $(lambda)$ form rather than the $mu$ form and $x$ is time in a Poisson process, then $lambda x$ is the average number of arrivals in a time interval of length $x.$ When it's written in the $(mu)$ form, then $mu$ is the average time until the next arrival.
answered Jul 27 at 16:12
Michael Hardy
204k23186461
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
You may see it called the probability density function of a gamma distribution. That is the restriction of this function to the interval $[0,+infty).$ This distribution is that of the sum of $a+1$ independent random variables each of which is exponentially distributed, so that the probability of its being greater than $x$ is $e^-x$ for $xge0.$
One more often sees it written as $dfracx^alpha-1 e^-xGamma(alpha).$ That way $alpha$ is the number of exponentially distributed random variables that are added.
Among those probability distributions called gamma distributions one also finds instances in which a scale parameter is present, thus:
$$
frac left( x/mu right)^alpha-1 e^-x/mu Gamma(alpha) left( frac dx muright) text for xge0. tag$mu$
$$
Sometimes it is parametrized using the reciprocal of the scale parameter, $lambda= dfrac 1 mu:$
$$
frac(lambda x)^alpha-1 e^-lambda xGamma(alpha) (lambda , dx). tag$lambda$
$$
It may be that $alpha$ is not a positive integer, in which case the statement about the sum of $alpha$ random variables may seem not to make sense. But suppose $X_1,X_2$ are independent random variables with
$$
Pr(X_iin A) = int_A fracx^alpha_i-1 e^-xGamma(alpha_i) , dx quad textfor measurable A subseteq [0,+infty), quad i=1,2.
$$
In that case, one has
$$
Pr(X_1+X_2in A) = int_A fracx^alpha_1+alpha_2-1 e^-xGamma(alpha_1+alpha_2) , dx quad textfor measurable A subseteq [0,+infty).
$$
This last probability density function is the convolution of those corresponding to $alpha_1$ and $alpha_2.$ Thus putting $alpha-1$ in the exponent rather than $alpha$ makes this a convolution semigroup in which convolution of these functions corresponds to addition of $alpha$s.
When it's written in the $(lambda)$ form rather than the $mu$ form and $x$ is time in a Poisson process, then $lambda x$ is the average number of arrivals in a time interval of length $x.$ When it's written in the $(mu)$ form, then $mu$ is the average time until the next arrival.
answered Jul 27 at 16:12
Michael Hardy
204k23186461
You may see it called the probability density function of a gamma distribution. That is the restriction of this function to the interval $[0,+infty).$ This distribution is that of the sum of $a+1$ independent random variables each of which is exponentially distributed, so that the probability of its being greater than $x$ is $e^-x$ for $xge0.$
One more often sees it written as $dfracx^alpha-1 e^-xGamma(alpha).$ That way $alpha$ is the number of exponentially distributed random variables that are added.
Among those probability distributions called gamma distributions one also finds instances in which a scale parameter is present, thus:
$$
frac left( x/mu right)^alpha-1 e^-x/mu Gamma(alpha) left( frac dx muright) text for xge0. tag$mu$
$$
Sometimes it is parametrized using the reciprocal of the scale parameter, $lambda= dfrac 1 mu:$
$$
frac(lambda x)^alpha-1 e^-lambda xGamma(alpha) (lambda , dx). tag$lambda$
$$
It may be that $alpha$ is not a positive integer, in which case the statement about the sum of $alpha$ random variables may seem not to make sense. But suppose $X_1,X_2$ are independent random variables with
$$
Pr(X_iin A) = int_A fracx^alpha_i-1 e^-xGamma(alpha_i) , dx quad textfor measurable A subseteq [0,+infty), quad i=1,2.
$$
In that case, one has
$$
Pr(X_1+X_2in A) = int_A fracx^alpha_1+alpha_2-1 e^-xGamma(alpha_1+alpha_2) , dx quad textfor measurable A subseteq [0,+infty).
$$
This last probability density function is the convolution of those corresponding to $alpha_1$ and $alpha_2.$ Thus putting $alpha-1$ in the exponent rather than $alpha$ makes this a convolution semigroup in which convolution of these functions corresponds to addition of $alpha$s.
When it's written in the $(lambda)$ form rather than the $mu$ form and $x$ is time in a Poisson process, then $lambda x$ is the average number of arrivals in a time interval of length $x.$ When it's written in the $(mu)$ form, then $mu$ is the average time until the next arrival.
answered Jul 27 at 16:12
Michael Hardy
204k23186461
edited Jul 27 at 16:25
answered Jul 27 at 16:12
Michael Hardy
204k23186461
answered Jul 27 at 16:12
Michael Hardy
204k23186461
answered Jul 27 at 16:12
Michael Hardy
204k23186461
204k23186461
add a comment |Â
add a comment |Â
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1
Gamma PDF with parameters $(a+1,1)$.
– Did
Jul 27 at 16:24