Conditional expectation on multiple variables

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SUMMARY

The discussion focuses on computing the conditional expectation E[X|Y1,Y2] for discrete random variables. The formula E[X|Y1,Y2] = ∑_x{x p(x|y1,y2)} is confirmed as correct, where p(x|y1,y2) = p(x ∩ y1 ∩ y2) / p(y1 ∩ y2). When Y1 and Y2 are independent and identically distributed (iid), the joint probability simplifies to p(y1, y2) = p(y1)p(y2). The expected value is calculated using summation for discrete variables and integration for continuous variables, with results dependent on both Y1 and Y2.

PREREQUISITES
  • Understanding of conditional probability and joint distributions
  • Familiarity with discrete and continuous random variables
  • Knowledge of probability mass functions and density functions
  • Basic calculus for integration in continuous cases
NEXT STEPS
  • Study the properties of independent and identically distributed (iid) random variables
  • Learn about joint probability distributions and their applications
  • Explore the concept of marginal probability distributions
  • Investigate the use of the Law of Total Expectation in probability theory
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Statisticians, data scientists, and anyone involved in probability theory or statistical modeling who needs to compute conditional expectations involving multiple variables.

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How to compute [tex]E[X|Y1,Y2][/tex]?
Assume all random variables are discrete.

I tried [tex]E[X|Y1,Y2] = \sum_x{x p(x|y1,y2)[/tex] but I'm not sure how to compute [tex]p(x|y1,y2] = \frac{p(x \cap y1 \cap y2)}{p(y1 \cap y2)}[/tex]

If it is correct, how can I simplify the expression if Y1 and Y2 are iid?
 
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If y1 and y2 are independent then p(y1, y2) = p(y1)p(y2).
 
In general

[tex] p(x \mid y_1, y_2) = \frac{p(x,y_1,y_2)}{p(y_1,y_2)}[/tex]

where the numerator is the joint density (or mass function for discrete case) of all three, and the denominator is the marginal of the two ys. You treat this as a function of [itex]x[/itex] alone. Then, in the discrete case, the expected value is

[tex] \sum x p(x \mid y_1, y_2)[/tex]

and in the continuous case it is

[tex] \int x p(x \mid y_1, y_2) \, dx[/tex]

In each case it is possible for the answer to depend on both [itex]y_1, y_2[/itex].
 

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