1
$\begingroup$

Dear all,

I have recently been breaking my head over this question. The idea is that a certain variable $Y$ is normally distributed with a parameter $X$ in both mean and variance.

$Y|X \sim N(\mu X,X^2)$

This parameter $X$ is assumed to be normally distributed as well with parameters $\alpha$ and $\beta$.

$X\sim N(\alpha, \beta)$

Now I am interested in the distribution of $Y$ (with $X$ marginalized out).

My current progress: Simulation shows me that the distribution of Y seems approximately normal as well. This does not proof anything off course.

The integral $\int_{all x}f_{Y|X}(s;x)f_{X}(s)dx$ seems unsolveable.

Kind regards

Improve this question
$\endgroup$
2
  • $\begingroup$ In what context does this question arise? $\endgroup$
    – an12
    Sep 20, 2012 at 2:53
  • $\begingroup$ Sorry, I cannot go into too much detail on the context. In general, I want to add an element of "risk" to a model. Until now the risk of this element has been unrecognized. $\endgroup$
    – Vincent Lous
    Sep 20, 2012 at 7:57

3 Answers 3

Reset to default
1
$\begingroup$

The characteristic function is $$\eqalign{ E\left[e^{itY}\right] &= E\left[ E\left[ e^{itY}|X \right]\right] \cr &= E \left[ \exp(it\mu X - t^2 X^2/2 \right] \cr &= \frac{\exp \left(\left(-(\alpha^2+\beta \mu^2) t^2 + 2 i \mu \alpha t\right)/\left(2 t^2 \beta + 2\right) \right)}{\sqrt {{t}^{2} \beta+1}}\cr}$$

It is certainly not a normal distribution, but might be approximated by a normal distribution when $\beta$ is small and $\alpha \ne 0$. In fact, by expanding this in a series in powers of $\beta$ and taking the inverse Fourier transform, I get a density

$$f(y) = \frac{\exp(-(y-\mu \alpha)^2/(2 \alpha^2))}{\sqrt{2 \pi |\alpha|}} \left( 1 + \frac { \left( 2\;{\alpha}^{4}+4\;x{\alpha}^{3}\mu+{x}^{2} \left( -5+{\mu}^{2} \right) {\alpha}^{2}-2\;{x}^{3}\alpha\mu+{x}^{4 } \right) }{2{\alpha}^{6}} \beta + \ldots\right) $$

Improve this answer
$\endgroup$
1
$\begingroup$

$$N(\mu X,X^2) \sim XN(\mu,1) \sim \mu X + XN(0,1) \sim N(\alpha \mu, \beta \mu^2 + \alpha^2) + \sqrt{\beta}N(0,1)N(0,1)$$ You thus have a sum between a normal distribution and a normal product distribution.

Improve this answer
$\endgroup$
0
$\begingroup$

Let $\xi_1,\xi_2 \sim \mathcal{N}(0,1)$. Then $X = \alpha + \xi_1 \sqrt{\beta}$ and $Y = \mu X + \xi_2 X$ $ = \mu \alpha + \mu \xi_1 \sqrt{\beta} + \xi_2 \alpha + \xi_1 \xi_2 \sqrt{\beta}$. Hence $E(Y) = \mu \alpha$, $Var(Y) = \mu^2 \beta + \alpha^2 + \beta$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you for your response. Deriving the moments of $Y$ is, as you show, not so difficult. I am, however, interested in the distribution of $Y$. Or may be on a related note I want to know how reasonable it is to assume $Y$ normally distributed. $\endgroup$
    – Vincent Lous
    Sep 19, 2012 at 14:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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