Range for continuos distribution in Julia
I am trying to calculate the density function of a continuos random variable in range in Julia using Distributions, but I am not able to define the range. I used Truncator constructor to construct the distribution, but I have no idea how to define the range. By density function I mean P(a
Would appreciate any help. The distribution I'm using is Gamma btw!
Thanks
julia-lang probability distribution julia-jump
asked Nov 22 '18 at 22:10
Chris Martin
183
add a comment |
I am trying to calculate the density function of a continuos random variable in range in Julia using Distributions, but I am not able to define the range. I used Truncator constructor to construct the distribution, but I have no idea how to define the range. By density function I mean P(a
Would appreciate any help. The distribution I'm using is Gamma btw!
Thanks
julia-lang probability distribution julia-jump
asked Nov 22 '18 at 22:10
Chris Martin
183
add a comment |
I am trying to calculate the density function of a continuos random variable in range in Julia using Distributions, but I am not able to define the range. I used Truncator constructor to construct the distribution, but I have no idea how to define the range. By density function I mean P(a
Would appreciate any help. The distribution I'm using is Gamma btw!
Thanks
julia-lang probability distribution julia-jump
asked Nov 22 '18 at 22:10
Chris Martin
183
I am trying to calculate the density function of a continuos random variable in range in Julia using Distributions, but I am not able to define the range. I used Truncator constructor to construct the distribution, but I have no idea how to define the range. By density function I mean P(a
Would appreciate any help. The distribution I'm using is Gamma btw!
Thanks
julia-lang probability distribution julia-jump
julia-lang probability distribution julia-jump
asked Nov 22 '18 at 22:10
Chris Martin
183
asked Nov 22 '18 at 22:10
Chris Martin
183
edited Nov 23 '18 at 4:05
asked Nov 22 '18 at 22:10
Chris Martin
183
asked Nov 22 '18 at 22:10
Chris Martin
183
asked Nov 22 '18 at 22:10
Chris Martin
183
183
add a comment |
add a comment |
2 Answers
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To get the maximum and minimum of the support of distribution d just write maximum(d) and minimum(d) respectively. Note that for some distributions this might be infinity, e.g. maximum(Normal()) is Inf.
answered Nov 22 '18 at 22:23
Bogumił Kamiński
11.9k11120
My questing is about continuos random variables not discrete and this works fine for discrete, but of course it wont work for continuos that's why I'm using truncation
– Chris Martin
Nov 22 '18 at 23:09
This works both for continuous and discrete random variables. But I see that you have changed your question now - you have the answer to it in the comment by 张实唯 below.
– Bogumił Kamiński
Nov 23 '18 at 7:42
add a comment |
What version of Julia and Distributions du you use? In Distribution v0.16.4, it can be easily defined with the second and third arguments of Truncated.
julia> a = Gamma()
Gamma{Float64}(α=1.0, θ=1.0)
julia> b = Truncated(a, 2, 3)
Truncated(Gamma{Float64}(α=1.0, θ=1.0), range=(2.0, 3.0))
julia> p = rand(b, 1000);
julia> extrema(p)
(2.0007680527633305, 2.99864177354943)
You can see the document of Truncated by typing ?Truncated in REPL and enter.
answered Nov 23 '18 at 2:02
张实唯
1,644821
Thanks for the response! But I don't think my question was understood properly (I edited it btw). What I am trying to calculate is the probability of x being between two values. So for example in your example, I want something like P(2<x<3). I think what you did is just defining x in that range but that's not the probability. I'm using Julia 0.6.2
– Chris Martin
Nov 23 '18 at 4:06
2
Then, why not simplycdf(x, 3) - cdf(x, 2).
– 张实唯
Nov 23 '18 at 6:56
1
Exactly. With one small note thatcdfevaluatesP(X<= a)notP(X<a)(which in case of continuous distributions is the same, but not in general).
– Bogumił Kamiński
Nov 23 '18 at 7:40
I have no idea why this didn't cross my mind! But yeah you're right, it works well now
– Chris Martin
Nov 23 '18 at 16:50
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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To get the maximum and minimum of the support of distribution d just write maximum(d) and minimum(d) respectively. Note that for some distributions this might be infinity, e.g. maximum(Normal()) is Inf.
answered Nov 22 '18 at 22:23
Bogumił Kamiński
11.9k11120
My questing is about continuos random variables not discrete and this works fine for discrete, but of course it wont work for continuos that's why I'm using truncation
– Chris Martin
Nov 22 '18 at 23:09
This works both for continuous and discrete random variables. But I see that you have changed your question now - you have the answer to it in the comment by 张实唯 below.
– Bogumił Kamiński
Nov 23 '18 at 7:42
add a comment |
To get the maximum and minimum of the support of distribution d just write maximum(d) and minimum(d) respectively. Note that for some distributions this might be infinity, e.g. maximum(Normal()) is Inf.
answered Nov 22 '18 at 22:23
Bogumił Kamiński
11.9k11120
My questing is about continuos random variables not discrete and this works fine for discrete, but of course it wont work for continuos that's why I'm using truncation
– Chris Martin
Nov 22 '18 at 23:09
This works both for continuous and discrete random variables. But I see that you have changed your question now - you have the answer to it in the comment by 张实唯 below.
– Bogumił Kamiński
Nov 23 '18 at 7:42
add a comment |
To get the maximum and minimum of the support of distribution d just write maximum(d) and minimum(d) respectively. Note that for some distributions this might be infinity, e.g. maximum(Normal()) is Inf.
answered Nov 22 '18 at 22:23
Bogumił Kamiński
11.9k11120
To get the maximum and minimum of the support of distribution d just write maximum(d) and minimum(d) respectively. Note that for some distributions this might be infinity, e.g. maximum(Normal()) is Inf.
answered Nov 22 '18 at 22:23
Bogumił Kamiński
11.9k11120
answered Nov 22 '18 at 22:23
Bogumił Kamiński
11.9k11120
answered Nov 22 '18 at 22:23
Bogumił Kamiński
11.9k11120
answered Nov 22 '18 at 22:23
Bogumił Kamiński
11.9k11120
11.9k11120
My questing is about continuos random variables not discrete and this works fine for discrete, but of course it wont work for continuos that's why I'm using truncation
– Chris Martin
Nov 22 '18 at 23:09
This works both for continuous and discrete random variables. But I see that you have changed your question now - you have the answer to it in the comment by 张实唯 below.
– Bogumił Kamiński
Nov 23 '18 at 7:42
add a comment |
My questing is about continuos random variables not discrete and this works fine for discrete, but of course it wont work for continuos that's why I'm using truncation
– Chris Martin
Nov 22 '18 at 23:09
This works both for continuous and discrete random variables. But I see that you have changed your question now - you have the answer to it in the comment by 张实唯 below.
– Bogumił Kamiński
Nov 23 '18 at 7:42
My questing is about continuos random variables not discrete and this works fine for discrete, but of course it wont work for continuos that's why I'm using truncation
– Chris Martin
Nov 22 '18 at 23:09
My questing is about continuos random variables not discrete and this works fine for discrete, but of course it wont work for continuos that's why I'm using truncation
– Chris Martin
Nov 22 '18 at 23:09
This works both for continuous and discrete random variables. But I see that you have changed your question now - you have the answer to it in the comment by 张实唯 below.
– Bogumił Kamiński
Nov 23 '18 at 7:42
This works both for continuous and discrete random variables. But I see that you have changed your question now - you have the answer to it in the comment by 张实唯 below.
– Bogumił Kamiński
Nov 23 '18 at 7:42
add a comment |
What version of Julia and Distributions du you use? In Distribution v0.16.4, it can be easily defined with the second and third arguments of Truncated.
julia> a = Gamma()
Gamma{Float64}(α=1.0, θ=1.0)
julia> b = Truncated(a, 2, 3)
Truncated(Gamma{Float64}(α=1.0, θ=1.0), range=(2.0, 3.0))
julia> p = rand(b, 1000);
julia> extrema(p)
(2.0007680527633305, 2.99864177354943)
You can see the document of Truncated by typing ?Truncated in REPL and enter.
answered Nov 23 '18 at 2:02
张实唯
1,644821
Thanks for the response! But I don't think my question was understood properly (I edited it btw). What I am trying to calculate is the probability of x being between two values. So for example in your example, I want something like P(2<x<3). I think what you did is just defining x in that range but that's not the probability. I'm using Julia 0.6.2
– Chris Martin
Nov 23 '18 at 4:06
2
Then, why not simplycdf(x, 3) - cdf(x, 2).
– 张实唯
Nov 23 '18 at 6:56
1
Exactly. With one small note thatcdfevaluatesP(X<= a)notP(X<a)(which in case of continuous distributions is the same, but not in general).
– Bogumił Kamiński
Nov 23 '18 at 7:40
I have no idea why this didn't cross my mind! But yeah you're right, it works well now
– Chris Martin
Nov 23 '18 at 16:50
add a comment |
What version of Julia and Distributions du you use? In Distribution v0.16.4, it can be easily defined with the second and third arguments of Truncated.
julia> a = Gamma()
Gamma{Float64}(α=1.0, θ=1.0)
julia> b = Truncated(a, 2, 3)
Truncated(Gamma{Float64}(α=1.0, θ=1.0), range=(2.0, 3.0))
julia> p = rand(b, 1000);
julia> extrema(p)
(2.0007680527633305, 2.99864177354943)
You can see the document of Truncated by typing ?Truncated in REPL and enter.
answered Nov 23 '18 at 2:02
张实唯
1,644821
Thanks for the response! But I don't think my question was understood properly (I edited it btw). What I am trying to calculate is the probability of x being between two values. So for example in your example, I want something like P(2<x<3). I think what you did is just defining x in that range but that's not the probability. I'm using Julia 0.6.2
– Chris Martin
Nov 23 '18 at 4:06
2
Then, why not simplycdf(x, 3) - cdf(x, 2).
– 张实唯
Nov 23 '18 at 6:56
1
Exactly. With one small note thatcdfevaluatesP(X<= a)notP(X<a)(which in case of continuous distributions is the same, but not in general).
– Bogumił Kamiński
Nov 23 '18 at 7:40
I have no idea why this didn't cross my mind! But yeah you're right, it works well now
– Chris Martin
Nov 23 '18 at 16:50
add a comment |
What version of Julia and Distributions du you use? In Distribution v0.16.4, it can be easily defined with the second and third arguments of Truncated.
julia> a = Gamma()
Gamma{Float64}(α=1.0, θ=1.0)
julia> b = Truncated(a, 2, 3)
Truncated(Gamma{Float64}(α=1.0, θ=1.0), range=(2.0, 3.0))
julia> p = rand(b, 1000);
julia> extrema(p)
(2.0007680527633305, 2.99864177354943)
You can see the document of Truncated by typing ?Truncated in REPL and enter.
answered Nov 23 '18 at 2:02
张实唯
1,644821
What version of Julia and Distributions du you use? In Distribution v0.16.4, it can be easily defined with the second and third arguments of Truncated.
julia> a = Gamma()
Gamma{Float64}(α=1.0, θ=1.0)
julia> b = Truncated(a, 2, 3)
Truncated(Gamma{Float64}(α=1.0, θ=1.0), range=(2.0, 3.0))
julia> p = rand(b, 1000);
julia> extrema(p)
(2.0007680527633305, 2.99864177354943)
You can see the document of Truncated by typing ?Truncated in REPL and enter.
answered Nov 23 '18 at 2:02
张实唯
1,644821
answered Nov 23 '18 at 2:02
张实唯
1,644821
answered Nov 23 '18 at 2:02
张实唯
1,644821
answered Nov 23 '18 at 2:02
张实唯
1,644821
1,644821
Thanks for the response! But I don't think my question was understood properly (I edited it btw). What I am trying to calculate is the probability of x being between two values. So for example in your example, I want something like P(2<x<3). I think what you did is just defining x in that range but that's not the probability. I'm using Julia 0.6.2
– Chris Martin
Nov 23 '18 at 4:06
2
Then, why not simplycdf(x, 3) - cdf(x, 2).
– 张实唯
Nov 23 '18 at 6:56
1
Exactly. With one small note thatcdfevaluatesP(X<= a)notP(X<a)(which in case of continuous distributions is the same, but not in general).
– Bogumił Kamiński
Nov 23 '18 at 7:40
I have no idea why this didn't cross my mind! But yeah you're right, it works well now
– Chris Martin
Nov 23 '18 at 16:50
add a comment |
Thanks for the response! But I don't think my question was understood properly (I edited it btw). What I am trying to calculate is the probability of x being between two values. So for example in your example, I want something like P(2<x<3). I think what you did is just defining x in that range but that's not the probability. I'm using Julia 0.6.2
– Chris Martin
Nov 23 '18 at 4:06
2
Then, why not simplycdf(x, 3) - cdf(x, 2).
– 张实唯
Nov 23 '18 at 6:56
1
Exactly. With one small note thatcdfevaluatesP(X<= a)notP(X<a)(which in case of continuous distributions is the same, but not in general).
– Bogumił Kamiński
Nov 23 '18 at 7:40
I have no idea why this didn't cross my mind! But yeah you're right, it works well now
– Chris Martin
Nov 23 '18 at 16:50
Thanks for the response! But I don't think my question was understood properly (I edited it btw). What I am trying to calculate is the probability of x being between two values. So for example in your example, I want something like P(2<x<3). I think what you did is just defining x in that range but that's not the probability. I'm using Julia 0.6.2
– Chris Martin
Nov 23 '18 at 4:06
Thanks for the response! But I don't think my question was understood properly (I edited it btw). What I am trying to calculate is the probability of x being between two values. So for example in your example, I want something like P(2<x<3). I think what you did is just defining x in that range but that's not the probability. I'm using Julia 0.6.2
– Chris Martin
Nov 23 '18 at 4:06
2
2
Then, why not simply
cdf(x, 3) - cdf(x, 2).– 张实唯
Nov 23 '18 at 6:56
Then, why not simply
cdf(x, 3) - cdf(x, 2).– 张实唯
Nov 23 '18 at 6:56
1
1
Exactly. With one small note that
cdf evaluates P(X<= a) not P(X<a) (which in case of continuous distributions is the same, but not in general).– Bogumił Kamiński
Nov 23 '18 at 7:40
Exactly. With one small note that
cdf evaluates P(X<= a) not P(X<a) (which in case of continuous distributions is the same, but not in general).– Bogumił Kamiński
Nov 23 '18 at 7:40
I have no idea why this didn't cross my mind! But yeah you're right, it works well now
– Chris Martin
Nov 23 '18 at 16:50
I have no idea why this didn't cross my mind! But yeah you're right, it works well now
– Chris Martin
Nov 23 '18 at 16:50
add a comment |
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