0
$\begingroup$

I have a bivariate normal distribution$$(X, Y)\sim N(\mu_{x}, \mu_{y}, \sigma_{x}^2, \sigma_{y}^2, \rho)$$ My question is : when $X > k$ ($k$ is a constant),how to get the distribution of $Y$? Can anyone tell me how to solve it? For exaple, let $$(X, Y) \sim N(0, 0, 1, 1, 0.7)$$ when $X > 1$, the distribution of $Y$?

$\endgroup$
1
$\begingroup$

Using usual notation, the conditional (truncated) distribution $Y\mid X>k$ for some fixed $k$ is given by

\begin{align} f_{Y\mid X>k}(y)&=\int_k^\infty\frac{f_{X,Y}(x,y)}{P(X>k)}\,dx \\\\&=\frac{1}{P(X>k)}\int_k^\infty f_{Y\mid X=x}(y\mid x)f_X(x)\,dx\qquad,\,y\in\mathbb R \end{align}

You can now find this density explicitly given any joint distribution $(X,Y)$.

$\endgroup$
  • $\begingroup$ I'm sorry I can't understand well about the expression you showed. Can you recomment relevant materials? Please $\endgroup$ – riskingitall Nov 21 '18 at 11:47
  • $\begingroup$ @riskingitall $f_{X,Y},f_X,f_{Y\mid X}$ are respectively the joint density of $(X,Y)$, the marginal density of $X$ and the conditional density of $Y\mid X$. Which expression is unclear? $\endgroup$ – StubbornAtom Nov 21 '18 at 12:13
  • $\begingroup$ wow, I got it! You're a genius.Thank you very very much.This question has been bothering me for a long time. $\endgroup$ – riskingitall Nov 21 '18 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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