What Is E{x|x+y+z=1} for Independent Standard Normal Variables?

  • Context: Graduate 
  • Thread starter Thread starter purplebird
  • Start date Start date
  • Tags Tags
    Expectation
Click For Summary

Discussion Overview

The discussion centers around the calculation of the conditional expectation E{x|x+y+z=1} for independent standard normal variables x, y, and z. Participants explore the implications of symmetry and independence in the context of this problem.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant states the problem of finding E{x|x+y+z=1} given that x, y, and z are independent standard normal variables.
  • Another participant highlights the symmetry among x, y, and z, suggesting that interchanging these variables does not change the value of the conditional expectation.
  • A third participant expresses confusion about the symmetry argument and requests clarification.
  • A later reply explains that the identical distribution of x, y, and z is crucial, and that the condition x+y+z=1 remains unchanged under permutations of the variables, leading to the conclusion that E{x|x+y+z=1} equals E{y|x+y+z=1} and E{z|x+y+z=1}.
  • One participant emphasizes the importance of the statistical independence of x, y, and z, noting that dependence among the variables could disrupt the symmetry and equality of their expected values.

Areas of Agreement / Disagreement

Participants generally agree on the role of symmetry and identical distribution but express differing levels of understanding regarding these concepts. The discussion does not reach a consensus on the calculation of the conditional expectation.

Contextual Notes

Participants note that the problem does not require complex integrals, suggesting a simpler approach may be possible. However, the implications of statistical independence and the exact nature of the distributions are not fully resolved.

purplebird
Messages
14
Reaction score
0
Given x,y and z are standard normal distributions with mean 0 and standard deviation 1. x,y and z are also statistically independent.

Find E{x|x+y+z=1}.
 
Physics news on Phys.org
symmetry of x, y, and z. For example, interchanging x and y in E[x| x+y+z=1] to get E[y|y+x+z = 1] doesn't change its value.
 
Adeimantus said:
symmetry of x, y, and z. For example, interchanging x and y in E[x| x+y+z=1] to get E[y|y+x+z = 1] doesn't change its value.

I am not able to understand what you are referring to. Could you please explain. Thanks.
 
Sure thing... If I understand the problem correctly, it doesn't really matter what the exact distributions of x, y, and z are. It only matters that they are identically distributed. Also, note that the condition x+y+z = 1 is unchanged by permuting the letters x,y,z. This, together with their being identically distributed means that

E{x|x+y+z=1} = E{y|x+y+z=1} = E{z|x+y+z=1}


Also, think about what E{x+y+z|x+y+z=1} would be, and remember the additive property of the expected value.

edit: think simple. no hard integrals needed, which was the first thing that came to my mind when I read the problem.
 
Last edited:
I left something out...it is also important that x,y,z are statistically independent. If, for example, x and y were statistically dependent, but x and z were not, then that would create an asymmetry and you could no longer conclude that the expected values of x, y, and z were equal.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K