Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a question if the following relations on conditional probabilities hold for independent random variables?

$$P_{X \mid Y, G(Y)}(x_1)=P_{X \mid \{Y\}}(x_2)$$ where $G$ is not necessarily invertible.

Also can we say $$P_{X \mid Y, H(Z)}(x_3)=P_{X \mid Y, \{Z\}}(x_4)$$ where $H$ is not necessarily invertible.

Finally can we say $$P_{X \mid Y, U}(x_5)=P_{X \mid Y,\{V\}}(x_6)$$ where $U=g(V)$ is not necessarily invertible.

The values $x_1,x_2 $ can be equal or related through some function. same goes to $x_3, x_4$ and $x_5,x_6$.If the mappings $H$, $G$ are invertible does the answer change? Please explain the answer or provide a reference which I can find online.

My guess is they hold with $G$, $H$, $g$ being invertible or not.

Example: For continuous and independent random variables $X,N$ I think we can derive $$Y=X+N$$ $$F_{Y\mid X}(y)=F_N(y-x)$$ where $F$ is the distribution function. So the above relations are a generalization.

share|improve this question
1  
Why give $Y$ and $G(Y)$ in the first case? The second two cases look identical. What's with the curly braces on one side of each equation? How is your example a special case of the above when it uses distribution functions for two different variables whilst the above always is $F\mid_X$-based? –  Sharkos Aug 18 '13 at 16:17
    
@Sharkos The curly braces meant to imply a subset of that random variable that is inside. I used it because I thought since the mappings are not assumed to be invertible given the function value we may have multiple values of the random variable which satisfy it. Example may be wrong. Thank you –  triomphe Aug 19 '13 at 2:58
add comment

1 Answer 1

This seems awfully related to your previous question, no?

First question: the conditional distribution of $X$ conditionally on $(Y,G(Y))=(y,G(y))$ is the same as the conditional distribution of $X$ conditionally on $Y=y$, whether $G$ is injective or not.

Second question: the conditional distribution of $X$ conditionally on $(Y,H(Z))=(y,H(z))$ is not necessarily the same as the conditional distribution of $X$ conditionally on $(Y,Z)=(y,z)$, except if $H$ is injective.

Third question: this is a duplicate of the second question.

share|improve this answer
1  
It becomes obvious that you are lacking a proper definition of conditional distribution. Which textbook are you using? –  Did Aug 18 '13 at 19:38
    
Thank you. Yes it is related to the previous question. I do not have a text book. I am self studying reading online lecture notes. Could you please recommend me a book ? –  triomphe Aug 19 '13 at 2:45
    
David Williams, Probability with martingales. –  Did Aug 19 '13 at 5:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.