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I have the following random variables $X$ ann $Y$ following respectively $N(0,b)$ and $N(0,c)$

$Z=\dfrac{(X+a)}{Y}$ with $a$ a real number.

What's the expectation of Z, i.e $E(Z)$ ?

UPDATE : sorry for this missing element : in my case,$X$ and $Y$ are 2 joint Gaussian distributed with a correlation coefficient $\rho$.

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    This may be helpful: https://en.wikipedia.org/wiki/Ratio_distribution#Uncorrelated_noncentral_normal_ratio – Stephan Kolassa Dec 06 '22 at 09:32
  • The expectation of the ratio of two Normal independent variables is the ratio of the expectations, that is $a/0$. You can bootstrap it with a software, say R. – Luke Dec 06 '22 at 09:52
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    @Luke The statement in your first sentence is untrue in general. Consider, for example, something simple like $\mu_X=\mu_Y=1$ and $\sigma_X=\sigma_Y=1$; $E(X/Y)\neq 1$ – Glen_b Dec 06 '22 at 10:29
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    Note that we could try to go from $E(X/Y) = E(X \times \frac{1}{Y}) = E(X) E(\frac{1}{Y})$, so if $E(\frac{1}{Y})$ were finite we could use the fact that $E(\frac{1}{Y})\neq 1/E(Y)$ in general, but for the normal $E(\frac{1}{Y})$ is not finite. – Glen_b Dec 06 '22 at 10:39
  • The analysis at https://stats.stackexchange.com/questions/299722 applies here, because you can re-express $X$ as a linear combination of $Y$ and another variable $Z$ independent of $X,$ with $(Y,Z)$ Binormal and independent. – whuber Dec 08 '22 at 15:06
  • @whuber . Thanks for your quick answer, could you give me please the expression of this linear combination of $Y$ and $Z$, I have difficulties to make the link with my case and your link. Regards –  Dec 08 '22 at 15:49
  • Because linear combinations of multivariate Normals are Normal and Normal distributions are determined by their means and covariance matrices, just equate the covariance matrix of $(\alpha Y + \beta Z, Y)$ with the covariance matrix of $(X,Y)$ and solve for the coefficients $(\alpha,\beta).$ – whuber Dec 08 '22 at 18:04

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