Question: enter image description here

I was solving test papers where I found this one.

My doubt:

I know to work with conditional probabilities and Jaccobian Transformation and part A and B can be done applying the above..But my problem is that, here Y | Z~ N(1+Z, 1). What I cannot understand is how the mean of the given normal distribution is (1+Z) as Z is itself a random variable ? Or does this denote something different ?

Also I cannot understand how to solve part "C".

Please help

Thank you.


1 Answer 1


Consider in this way: You and I will go to draw random numbers from normal distribution. The process is I draw one random number $Z_1$ from $N(0,1)$ first, then based on what I get $z_1$, you draw a random number $Y_1$ from $N(z_1+1,1)$. Repeat this process $N$ times, we get $(Z_i, Y_i)$, $i=1,...,N$, where $Z_i\sim N(0,1)$ and $Y_i|Z_i\sim N(Z_i+1,1)$.

C: From B, we know that $U = 1+Z$. So $U=1.7 ==> Z = 0.7$. $$E(Y|U=1.7) = E(Y|Z=0.7)$$ $$Y|Z \sim N(1+Z,1) ==> E(Y|Z) = 1+Z $$

Combine two equations above, you should get the answer.

  • $\begingroup$ Thank you..I got your point..Bt still something is not clear..Is the pdf structure of f(y|z)=1/sqrt(pi) exp[(x-(1+z))^2/2] ? i.e the mean(1+z) in the pdf is a variable quantity and not a constant as in general ? $\endgroup$
    – P db
    Commented Nov 24, 2018 at 23:03
  • $\begingroup$ Conditional on $z$, then $z$ should be constant. $\endgroup$
    – user158565
    Commented Nov 24, 2018 at 23:06

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