Overloading functions like Mean for distributions
Clash Royale CLAN TAG#URR8PPP
I'd like to understand how functions like Mean are overloaded to provide the right behavior for the distributions in Mathematica (like NormalDistribution or PoissonDistribution).
I originally assumed it was through the use of UpValues, but now I'm not so sure...
So, if I wanted to implement a function or distribution and define a behavior for it when another function like Mean is applied to it, how would I go about it? I know it's not the following:
f[x_]:=2+x
f /: Mean[f[x_]] := 3 x
Desired behavior:
f[3]
Mean[f[3]]
(*
==> 5
==> 9
*)
upvalues
asked Dec 22 '18 at 3:16
hmode
32728
add a comment |
I'd like to understand how functions like Mean are overloaded to provide the right behavior for the distributions in Mathematica (like NormalDistribution or PoissonDistribution).
I originally assumed it was through the use of UpValues, but now I'm not so sure...
So, if I wanted to implement a function or distribution and define a behavior for it when another function like Mean is applied to it, how would I go about it? I know it's not the following:
f[x_]:=2+x
f /: Mean[f[x_]] := 3 x
Desired behavior:
f[3]
Mean[f[3]]
(*
==> 5
==> 9
*)
upvalues
asked Dec 22 '18 at 3:16
hmode
32728
1
Have you seenProbabilityDistribution?
– Edmund
Dec 22 '18 at 21:54
Nope - also really useful. Thanks!
– hmode
Dec 23 '18 at 1:39
add a comment |
I'd like to understand how functions like Mean are overloaded to provide the right behavior for the distributions in Mathematica (like NormalDistribution or PoissonDistribution).
I originally assumed it was through the use of UpValues, but now I'm not so sure...
So, if I wanted to implement a function or distribution and define a behavior for it when another function like Mean is applied to it, how would I go about it? I know it's not the following:
f[x_]:=2+x
f /: Mean[f[x_]] := 3 x
Desired behavior:
f[3]
Mean[f[3]]
(*
==> 5
==> 9
*)
upvalues
asked Dec 22 '18 at 3:16
hmode
32728
I'd like to understand how functions like Mean are overloaded to provide the right behavior for the distributions in Mathematica (like NormalDistribution or PoissonDistribution).
I originally assumed it was through the use of UpValues, but now I'm not so sure...
So, if I wanted to implement a function or distribution and define a behavior for it when another function like Mean is applied to it, how would I go about it? I know it's not the following:
f[x_]:=2+x
f /: Mean[f[x_]] := 3 x
Desired behavior:
f[3]
Mean[f[3]]
(*
==> 5
==> 9
*)
upvalues
upvalues
asked Dec 22 '18 at 3:16
hmode
32728
asked Dec 22 '18 at 3:16
hmode
32728
asked Dec 22 '18 at 3:16
hmode
32728
asked Dec 22 '18 at 3:16
hmode
32728
asked Dec 22 '18 at 3:16
hmode
32728
32728
1
Have you seenProbabilityDistribution?
– Edmund
Dec 22 '18 at 21:54
Nope - also really useful. Thanks!
– hmode
Dec 23 '18 at 1:39
add a comment |
1
Have you seenProbabilityDistribution?
– Edmund
Dec 22 '18 at 21:54
Nope - also really useful. Thanks!
– hmode
Dec 23 '18 at 1:39
1
1
Have you seen
ProbabilityDistribution?– Edmund
Dec 22 '18 at 21:54
Have you seen
ProbabilityDistribution?– Edmund
Dec 22 '18 at 21:54
Nope - also really useful. Thanks!
– hmode
Dec 23 '18 at 1:39
Nope - also really useful. Thanks!
– hmode
Dec 23 '18 at 1:39
add a comment |
2 Answers
2
active
oldest
votes
Symbolic(!) distributions are recognized by their head, not by their PDF:
Mean[PDF[NormalDistribution[0, 1], x]]
Mean[E^(-(x^2/2))/Sqrt[2 π]]
So instead of messing around with Mean, I would rather suggest
distro /: PDF[distro[μ_, σ_], x_] := E^(-((x - μ)^2/(2 σ^2)))/(Sqrt[2 π] σ);
distro /: Mean[distro[μ_, σ_]] := μ;
distro /: Variance[distro[μ_, σ_]] := σ^2
Mean[distro[0, 1]]
Variance[distro[0, 1]]
0
1
answered Dec 22 '18 at 11:49
Henrik Schumacher
49k467139
add a comment |
This works:
Unprotect[Mean];
SetAttributes[Mean, HoldFirst];
Protect[Mean];
f[x_] := 2 + x
f /: Mean[f[x_]] := 3 x
Since using Unprotect is not reccomended here's another way.
mean[x_] := Mean[x]
SetAttributes[mean, HoldFirst]
f[x_] := 2 + x
f /: mean[f[x_]] := 3 x
answered Dec 22 '18 at 5:57
Andrew
1,9011115
2
It should be mentioned thatSetAttributes[Mean, HoldFirst]is likely to break many internal functions that depend onMean
– Jason B.
Dec 22 '18 at 15:05
Thanks for this solution - I did learn something from your answer but I was trying to avoid setting DownValues.
– hmode
Dec 23 '18 at 1:35
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Symbolic(!) distributions are recognized by their head, not by their PDF:
Mean[PDF[NormalDistribution[0, 1], x]]
Mean[E^(-(x^2/2))/Sqrt[2 π]]
So instead of messing around with Mean, I would rather suggest
distro /: PDF[distro[μ_, σ_], x_] := E^(-((x - μ)^2/(2 σ^2)))/(Sqrt[2 π] σ);
distro /: Mean[distro[μ_, σ_]] := μ;
distro /: Variance[distro[μ_, σ_]] := σ^2
Mean[distro[0, 1]]
Variance[distro[0, 1]]
0
1
answered Dec 22 '18 at 11:49
Henrik Schumacher
49k467139
add a comment |
Symbolic(!) distributions are recognized by their head, not by their PDF:
Mean[PDF[NormalDistribution[0, 1], x]]
Mean[E^(-(x^2/2))/Sqrt[2 π]]
So instead of messing around with Mean, I would rather suggest
distro /: PDF[distro[μ_, σ_], x_] := E^(-((x - μ)^2/(2 σ^2)))/(Sqrt[2 π] σ);
distro /: Mean[distro[μ_, σ_]] := μ;
distro /: Variance[distro[μ_, σ_]] := σ^2
Mean[distro[0, 1]]
Variance[distro[0, 1]]
0
1
answered Dec 22 '18 at 11:49
Henrik Schumacher
49k467139
add a comment |
Symbolic(!) distributions are recognized by their head, not by their PDF:
Mean[PDF[NormalDistribution[0, 1], x]]
Mean[E^(-(x^2/2))/Sqrt[2 π]]
So instead of messing around with Mean, I would rather suggest
distro /: PDF[distro[μ_, σ_], x_] := E^(-((x - μ)^2/(2 σ^2)))/(Sqrt[2 π] σ);
distro /: Mean[distro[μ_, σ_]] := μ;
distro /: Variance[distro[μ_, σ_]] := σ^2
Mean[distro[0, 1]]
Variance[distro[0, 1]]
0
1
answered Dec 22 '18 at 11:49
Henrik Schumacher
49k467139
Symbolic(!) distributions are recognized by their head, not by their PDF:
Mean[PDF[NormalDistribution[0, 1], x]]
Mean[E^(-(x^2/2))/Sqrt[2 π]]
So instead of messing around with Mean, I would rather suggest
distro /: PDF[distro[μ_, σ_], x_] := E^(-((x - μ)^2/(2 σ^2)))/(Sqrt[2 π] σ);
distro /: Mean[distro[μ_, σ_]] := μ;
distro /: Variance[distro[μ_, σ_]] := σ^2
Mean[distro[0, 1]]
Variance[distro[0, 1]]
0
1
answered Dec 22 '18 at 11:49
Henrik Schumacher
49k467139
edited Dec 23 '18 at 9:08
answered Dec 22 '18 at 11:49
Henrik Schumacher
49k467139
answered Dec 22 '18 at 11:49
Henrik Schumacher
49k467139
answered Dec 22 '18 at 11:49
Henrik Schumacher
49k467139
49k467139
add a comment |
add a comment |
This works:
Unprotect[Mean];
SetAttributes[Mean, HoldFirst];
Protect[Mean];
f[x_] := 2 + x
f /: Mean[f[x_]] := 3 x
Since using Unprotect is not reccomended here's another way.
mean[x_] := Mean[x]
SetAttributes[mean, HoldFirst]
f[x_] := 2 + x
f /: mean[f[x_]] := 3 x
answered Dec 22 '18 at 5:57
Andrew
1,9011115
2
It should be mentioned thatSetAttributes[Mean, HoldFirst]is likely to break many internal functions that depend onMean
– Jason B.
Dec 22 '18 at 15:05
Thanks for this solution - I did learn something from your answer but I was trying to avoid setting DownValues.
– hmode
Dec 23 '18 at 1:35
add a comment |
This works:
Unprotect[Mean];
SetAttributes[Mean, HoldFirst];
Protect[Mean];
f[x_] := 2 + x
f /: Mean[f[x_]] := 3 x
Since using Unprotect is not reccomended here's another way.
mean[x_] := Mean[x]
SetAttributes[mean, HoldFirst]
f[x_] := 2 + x
f /: mean[f[x_]] := 3 x
answered Dec 22 '18 at 5:57
Andrew
1,9011115
2
It should be mentioned thatSetAttributes[Mean, HoldFirst]is likely to break many internal functions that depend onMean
– Jason B.
Dec 22 '18 at 15:05
Thanks for this solution - I did learn something from your answer but I was trying to avoid setting DownValues.
– hmode
Dec 23 '18 at 1:35
add a comment |
This works:
Unprotect[Mean];
SetAttributes[Mean, HoldFirst];
Protect[Mean];
f[x_] := 2 + x
f /: Mean[f[x_]] := 3 x
Since using Unprotect is not reccomended here's another way.
mean[x_] := Mean[x]
SetAttributes[mean, HoldFirst]
f[x_] := 2 + x
f /: mean[f[x_]] := 3 x
answered Dec 22 '18 at 5:57
Andrew
1,9011115
This works:
Unprotect[Mean];
SetAttributes[Mean, HoldFirst];
Protect[Mean];
f[x_] := 2 + x
f /: Mean[f[x_]] := 3 x
Since using Unprotect is not reccomended here's another way.
mean[x_] := Mean[x]
SetAttributes[mean, HoldFirst]
f[x_] := 2 + x
f /: mean[f[x_]] := 3 x
answered Dec 22 '18 at 5:57
Andrew
1,9011115
answered Dec 22 '18 at 5:57
Andrew
1,9011115
answered Dec 22 '18 at 5:57
Andrew
1,9011115
answered Dec 22 '18 at 5:57
Andrew
1,9011115
1,9011115
2
It should be mentioned thatSetAttributes[Mean, HoldFirst]is likely to break many internal functions that depend onMean
– Jason B.
Dec 22 '18 at 15:05
Thanks for this solution - I did learn something from your answer but I was trying to avoid setting DownValues.
– hmode
Dec 23 '18 at 1:35
add a comment |
2
It should be mentioned thatSetAttributes[Mean, HoldFirst]is likely to break many internal functions that depend onMean
– Jason B.
Dec 22 '18 at 15:05
Thanks for this solution - I did learn something from your answer but I was trying to avoid setting DownValues.
– hmode
Dec 23 '18 at 1:35
2
2
It should be mentioned that
SetAttributes[Mean, HoldFirst] is likely to break many internal functions that depend on Mean– Jason B.
Dec 22 '18 at 15:05
It should be mentioned that
SetAttributes[Mean, HoldFirst] is likely to break many internal functions that depend on Mean– Jason B.
Dec 22 '18 at 15:05
Thanks for this solution - I did learn something from your answer but I was trying to avoid setting DownValues.
– hmode
Dec 23 '18 at 1:35
Thanks for this solution - I did learn something from your answer but I was trying to avoid setting DownValues.
– hmode
Dec 23 '18 at 1:35
add a comment |
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1
Have you seen
ProbabilityDistribution?– Edmund
Dec 22 '18 at 21:54
Nope - also really useful. Thanks!
– hmode
Dec 23 '18 at 1:39