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I want to know how to compute joint pdf of (x,y) where x follows exponential distribution with parameter $\lambda$ and $y=x+\epsilon $, $\epsilon$ follows standard normal distribution. x and $\epsilon$ are independent.

and how to compute pdf of y.

the question I met is to show $f(x|y)=\frac{\phi(x-(y-\lambda))}{\Phi(y-\lambda)}$, where $\phi,\Phi$ are pdf and cdf of standard normal. I guess this equation is originated from $f(x|y) = \frac{f(x,y)}{f(y)}$, so I need to compute the joint pdf and the pdf of y. I tried but can't get the same form as shown above.

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  • $\begingroup$ The PDF of $y$ conditioned on $x$ is the PDF of a standard normal distribution with mean $x$, i.e., $f(y|x)=\phi(y-x)$. $\endgroup$ – Bernhard Sep 15 '15 at 12:16
  • $\begingroup$ @Bernhard .yes and $f(y|x)f(x)=f(x,y)$. how do we get $f(y)$? Is it $\int f(x,y)dx$? I don't know how to get $\Phi(y-\lambda)$ $\endgroup$ – Luyi Tian Sep 15 '15 at 12:32
  • $\begingroup$ Exactly, you have to integrate the joint distribution over $x$. I have not worked through it yet, but I expect you will be able to include the exponent of the exponential PDF into the exponent of the normal distribution. $\endgroup$ – Bernhard Sep 15 '15 at 13:51
  • $\begingroup$ @Bernhard yes I got $f(x,y)$ and in the integration I have a quadratic expression of x like $e^{x^2+x+1}$. How do I transform this to the $\Phi(y-\lambda)$ $\endgroup$ – Luyi Tian Sep 16 '15 at 13:00
  • $\begingroup$ Try to transform $e^{x^2+bx+c}$ into $e^{(x+a)^2+d}=C\cdot e^{(x+a)^2}$. The integral over the latter should then be $\Phi(y-\lambda)$. $\endgroup$ – Bernhard Sep 16 '15 at 14:24

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