Mathematica 中的符号条件期望 [英] Symbolic Conditional Expectation in Mathematica

问题描述

我想实现条件期望运算符.我将使用大写的 epsilon E 来表示运算符.我希望至少有以下输入(下划线表示下标)

I want to implement the conditional expectation operator. I will use the capital epsilon E to denote the operator. I expect at least the following input (underscore means subscript)

E_2[a]
E_2[x_1]
E_2[x_1 + y_5]
E_1[3 a + b - 4 + 2 x_]
E_6[x_5 x_7]
E_t[x_t]
E_t[3 x_{t - 1} x_{t + 2}]

产生以下输出

a
x_1
E_2[y_5] + x_1
-4 + 3 a + b + 2 E_2[x_5]
E_6[x_7] x_5
x_t
3 E_t[x_{t + 2}] x_{t - 1}

上面的示例并不是我需要生成的唯一输入/输出对，而是作为我喜欢的语法的测试和说明.

The examples above are not the only input/output pairs I need to produce, but rather serve as a test and an illustration for the syntax I prefer.

我已经到了这一步.ce 表示 Conditional Expectation，它的第三个组成部分是期望传播"是否最终确定(否则在乘积规则中会出现无限递归)，mv 表示 Measurable Variable.

I've got this far. ce means Conditional Expectation, its third component is whether the "expectation propagation" is finalized or not (otherwise the infinite recursion occurs in the product rule), mv stands for Measurable Variable.

Notation[Subscript[E, t_][y_]  ==> ce[y_, t_, False]];
Notation[Subscript[E, t_][y_] <==  ce[y_, t_, _]];

Notation[Subscript[x_, t_] <==> mv[x_, t_]];

(* Atomic Elements and Measurable Variables *)
ce[x_, t_, _] := x /; (AtomQ[x] || Head[x] === mv && 0 <= t - x[[2]]);

(* Distribution over Addition *)
ce[x_ + y__, t_, s_] := ce[x, t, s] + Plus @@ (ce[#, t, s] & /@ {y});

(* Distribution over Product *)
ce[x__Times, t_, False] := Module[{v, m, n},

   (* All Variables in the Product *)
   v = List @@ x;

   (* Measurable Among Them *)
   m = Select[v, AtomQ[#] || Head[#] === mv && 0 <= t - #[[2]] &];

   (* The Rest is not Measurable *)
   n = Complement[v, m];

   Times[Times @@ m, ce[Times @@ n, t, True]]

];       

推荐答案

我想我可以让你接近你想要的；不过，我不会全部完成，因为这可能很棘手，但我会为您指明正确的方向.

I think I can get you close to what you want; I'm not going to do it all, though, as it can be tricky, but I'll point you in the right direction.

首先，在 Mathematica 中使用下标来表示不同的变量是很棘手的，因为它会将 E0 解释为 Subscript[E,0] 和 E 和 Subscript 都是保留的.(正如 Sjoerd 所说，E = 2.718....)让 Mathematica 识别 作为一个独特的符号，您需要通过 <<Notations` 加载 Notations 包.然后使用符号调色板，Symbolize Subscript[E,0].(请注意，不要在不使用调色板正确设置代码的情况下尝试这样做，否则可能无法正常工作.)

First of all, using subscripts to denote different variables is tricky in Mathematica as it interprets E0 as Subscript[E,0] and both E and Subscript are reserved. (As Sjoerd said, E = 2.718....) To get Mathematica to recognize <anything><something> as a distinct symbol, you need to load the Notations package via <<Notations`. Then using the Notations Palette, Symbolize Subscript[E,0]. (As a word of caution, don't try to do that without using the palette to set up the code correctly, otherwise it may not work.)

根据需要对所有变量进行符号化后，您需要设置适当的转换规则.前两个最简单，回车

Once all of your variables are symbolized, as needed, you need to set up the appropriate transformation rules. The first two are simplest, enter

E_0[a] = a
E_0[x_0] = x_0

规则 3 和 4:

E_0[x_Plus]:=Distribute[E_0[x]]
E_0[x_Times]:=Distribute[E_0[x], Times]

那些是简单的，接下来的三个需要不同类型的关联，都不是 Set 也不是 SetDelayed 将在这里工作，因为被评估的外部符号是 Dt，并且您不能将新规则与其关联，因为它是 Protected.但是，有两种方法可以将此类表达式与内部符号相关联: UpSet (^=)(或 UpSetDelayed (^:=)) 或 TagSet (/:).我更喜欢使用 TagSet 因为它更明确，但两者都应该有效.

Those were the easy ones, the next three require a different kind of association, neither Set nor SetDelayed will work here as the outer symbol being evaluated is Dt, and you can't associate new rules with it as it is Protected. However, there are two methods of associating such expressions with an internal symbol: UpSet (^=) (or UpSetDelayed (^:=)) or TagSet (/:). I prefer to use TagSet as it is more explicit, but either should work.

规则 5 和 6:

E_0 /: Dt[ E_0[ x_ ], y_ ] := E_0[ Dt[x,y] ]

这也将使您接近规则 7，但将其与规则 3 和 4 一起添加会导致递归限制错误，因为它会反复尝试找出如何评估它.相反，将规则 3 和 4 替换为

This will also get you close to rule 7, but adding this alongside rules 3 and 4 causes a Recursion Limit error as it bounces back and forth trying to figure out how to evaluate it. Instead, replace rule 3 and 4 with

E_0[x_ + y__]:= E_0[x] + Plus@@( E_0 /@ {y} )
E_0[x_ y__ ] := E_0[x] Times@@( E_0 /@ {y} )

这对递归有明确的限制.就规则 7 而言，你明白了

Which puts definite limits on the recursion. As far as rule 7 is concerned, you get this

E_0[D[x_1[t_1,q_0], t_1]] E_0[Dt[t_1, y_0]] 
+ E_0[D[x_1[t_1,q_0], q_0]] E_0[Dt[q_0,y]]

这是 Dt 规则和规则 4 的结果.为了让 E_0 不分布在 D 和 Dt 留作练习.

which is a consequence of the Dt rule and rule 4. To get E_0 not to distribute over D and Dt is left as an exercise.

编辑:我想对您提供的解决方案代码发表一些评论.首先，巧妙地使用布尔值来停止递归，它与您的 Notation 配合得很好.不过，我建议您对产品分发进行一些更改.首先，我会使用 x__Times 而不是条件 (/; Head[x] == Times) 因为它更容易阅读，我相信(但没有t 测试)它可能更快，即处理它的开销更少.其次，将Table 的使用替换为List @@ x，其中@@，称为Apply，将 Times 替换为 <代码>列表，它再次更容易阅读和编写.对于 n 的定义，请考虑使用 补码;我不知道它是否更快，但我倾向于为这种类型的事物设置理论结构.最后，除非您需要一个变量在使用时重新评估，否则不要使用 SetDelayed (:=)，使用 Set (=).通过使用 :=，m 被评估两次，v 被评估 3 次！

Edit: I'd like to make a few comments on the solution code you've provided. First, clever use of a Boolean to stop the recursion, and it works well with your Notation. I'd suggest several changes to you product distribution, though. First, I'd use x__Times instead of the condition (/; Head[x] == Times) as it is easier to read, and I believe (but haven't tested) it may be faster, i.e. less overhead to process it. Second, replace you use of Table with List @@ x, where @@, called Apply, replaces Times with List, and it is again easier to read and write. For your definition of n, consider using Complement; I don't know if it is faster, but I tend to prefer set theoretic constructs for this type of thing. Lastly, unless you need a variable to be reevaluated whenever it is used, do not use SetDelayed (:=), use Set (=). By using :=, m is evaluated twice, and v is evaluated 3 times!

优点和缺点:这样做的主要原因是易用性和可读性.通过定义您自己的对象及其行为方式，您可以为自己提供很大的灵活性并简化您的代码.仅此一点就值得.但是，我过去在这样做时遇到了困难，而且这样的设置可能很挑剔，我建议进行彻底的测试.其次，通过添加这些额外的层，您可能会减慢代码速度，因此如果将其用于关键任务应用程序，请务必小心.此外，每次使用时都必须包含 Notation，而调色板在某些时候会变得烦人.虽然，可以通过在加载 Notation 包之前设置 AutoLoadNotationPalette = False 来处理调色板.

Pros and Cons: The chief reasons to do this is ease of use and readability. By defining your own objects and how they behave, you give yourself a lot of flexibility and simplify your code. That alone makes it worth it. However, I've had difficulty doing this in the past, and such a setup can be persnickety, and I'd recommend thorough testing. Secondly, by adding these extra layers, you may slow down your code, so be careful if this is to be used in mission critical applications. Additionally, you have to include Notation every time you use it, and the palette will become annoying at some point. Although, the palette can be dealt with by setting AutoLoadNotationPalette = False prior to loading the Notation package.

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