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I have a question about the prediction. This question stopped me for a whole afternoon, still do not have an idea on how to solve it. Would you please give me a clue. The question is:

U_1,U_2 be independent standard normal random variables and set Z=U_1^2+U_2^2, Y=U_1. Is Z of any value in prediction Y?

Thank you so much.

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At least intuitively, it definitely should! Assume Z=4, then Y is likely to have a high absolute value. Assume Z is close to zero, now Y is definitely small.

What Z will not tell you is any information about whether Y is positive or negative, but you can think of Y as the square root of Z minus the square of a normal random variable, therefore Z is critical in determining Y

Also, please consider changing the U_1, U_2 notation. It makes a very simple question a bit harder to read than it should

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  • $\begingroup$ "Assume Z=4, then Y is likely to have a high value" -- I don't think there's a good basis for this. U1 and U2 may be either positive or negative and either may be the main contributor to Z, or both might be similar in magnitude. If $Z=4$ then $-2\leq U_1 \leq 2$. Now, there is a greater chance of being in a small interval near $\pm 2$ than in an interval of the same width near 0, and in that sense it's somewhat "predictable" (given Z we can give two intervals one of which is likely to contain Y), but that's a different statement to the one in your answer $\endgroup$
    – Glen_b
    Commented Feb 25, 2019 at 12:08
  • $\begingroup$ You are absolutelly right, but I do not understand the difference between your claim and mine. Maybe I am missing something $\endgroup$
    – David
    Commented Feb 25, 2019 at 13:14
  • $\begingroup$ Thank you all. My idea is that when we solve this kind of question, should we consider the E(Y|Z)? If E(Y|Z)=0 then I can say Z could not predict on Y? $\endgroup$
    – Wang Wang
    Commented Feb 25, 2019 at 15:20
  • $\begingroup$ @asdf you said "Y is likely to have a high value". That's demonstrably not the case; the conditional distribution of Y is symmetric -- Y is exactly as likely to be low as it is high. $\endgroup$
    – Glen_b
    Commented Feb 26, 2019 at 2:29
  • $\begingroup$ @WangWang it depends on what you mean by "predict". If you mean "does E(Y|Z=z) vary with z?" then the answer is "clearly not". If you mean "can I give a (possibly non-simple) prediction interval for the conditional distribution of Y that differs from one I'd give for Y without conditioning on Z?" then the answer is clearly "yes". $\endgroup$
    – Glen_b
    Commented Feb 26, 2019 at 2:31

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