What Is the Correct Calculation for the Conditional Expected Value of an Even X?

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Homework Help Overview

The discussion revolves around calculating the conditional expected value of a discrete random variable X, which takes values from 1 to 6 with a specified probability density function. The focus is on determining the expected value given that X is even.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of the expected value conditioned on X being even, questioning the correct application of probability rules and the interpretation of the probability density function.

Discussion Status

Participants are actively discussing the correct approach to calculating the conditional probabilities and expected values. Some have suggested using the product and sum rules for probabilities, while others are verifying the calculations and definitions involved. There is a recognition of mistakes and clarifications being made regarding the formulas used.

Contextual Notes

There appears to be confusion regarding the calculation of P(X even) and how it affects the expected value computation. Participants are also checking the accuracy of their individual calculations against the provided probability density values.

ParisSpart
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we have a variable that takes values 1,...,6 with density:
n 1 2 3 4 5 6
f(n) 0.1 0.2 0.1 0.3 0,176 0,124

What is the average price (expected value) of X under the condition that X is even?

E(X/X=even)=k*P(X=k/X=even)=0.2*2+4*0.3+0.124*6
i am doing this but its says its not correct what i am doing wrong in the type?
 
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Use the product rule:

P(X=k|X even) = P(X=k) P(X even)
 
P(X even) here is 1/2? or 1/3?
 
P(X even) = P(X=2 or X=4 or X=6)

Use the sum rule for mutually exclusive events:

P(A or B) = P(A) + P(B) if A and B are mutually exclusive.
 
P(X=2)=0.2? from the table?
 
Yes...
 
but its says that its not correct my answer... i find this 1.462656 in the final answer
 
What did you calculate then?
 
first i found this P(X=2 OR X=4 OR X=6)=0.624
AND E(X/X=even)=k*P(X=k/X=even)=0.2*2*0.624+0.624*4*0.3+0.124*6*0.624...=1.4626
 
  • #10
Oops. My mistake.
The rule is: P(X=k|X even) = P(X=k) / P(X even)
 
  • #11
k*(P(X=k)/P(X even)) like this?
 
  • #12
Yes.
 

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