Is My Formula for Conditional Expectation Correct?

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SUMMARY

The discussion confirms the correctness of the formula for the conditional expectation of a random variable Y given that Y is less than another independent random variable Z. The formula is expressed as E(Y | Y < z) = ∫₀ᶻ y · f(y) dy / F(z), where f(y) is the probability density function (pdf) of Y and F(z) is the cumulative distribution function (cdf) of Z. The verification process involves understanding truncated distributions and the relationship between conditional expectations.

PREREQUISITES
  • Understanding of independent and identically distributed (I.I.D) random variables
  • Knowledge of probability density functions (pdf) and cumulative distribution functions (cdf)
  • Familiarity with conditional expectation concepts
  • Basic calculus for evaluating integrals
NEXT STEPS
  • Study the properties of truncated distributions in probability theory
  • Learn about the derivation of conditional expectations in joint distributions
  • Explore advanced topics in probability, such as the Law of Total Expectation
  • Review examples of conditional expectations in statistical analysis
USEFUL FOR

Students and professionals in statistics, data science, and probability theory who are working with conditional expectations and random variables.

JamesF
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This result isn't in our book, but it is in my notes and I want to make sure it's correct. Please verify if you can.

Homework Statement


I have two I.I.D random variables. I want the conditional expectation of Y given Y is less than some other independent random variable Z.

[tex]E(Y \, \vert \, Y < z) = \dfrac{\int_0^{z} y \cdot f(y) \, dy}{F(z)}[/tex]

Where f(y) is the pdf of Y and F(z) is the cdf for Z

The Attempt at a Solution


I've searched the book and the web, but all I find is the formula for conditional expectation for [tex]E(X | Y = y)[/tex] for joint distributions and the like. Is my formula correct?
 
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You know that [tex]\mathbb{E}[X|Y]=\frac{\mathbb{E}[X \mathbf{1}_Y]}{\mathbb{P}(Y)}[/tex] so your formula looks correct.
 
Last edited:
Think this way: if you know [tex]Y \le z[/tex], then the truncated distribution has density

[tex] g(y \mid Y \le z) = \frac{f(y)}{F(z)}[/tex]

so the expectation is

[tex] \int_0^z y g(y \mid Y \le z) \, dy = \frac{\int_0^z y f(y) \, dy}{F(z)}[/tex]

exactly as you have it.
 

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