Timeline for exchangeable normal r.v.s
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 20, 2011 at 0:12 | comment | added | David Moews | Substituting $\alpha$ for $y$, this shows that the function is a probability density. The density of the marginals can now be found by multiplying $2\phi(x) f(\alpha x)$ by the density of $\alpha$, $\phi(\alpha)$, and integrating over $\alpha$. Substituting $x$ for $y$ in the equation above shows that $\int 2 f(\alpha x) \phi(\alpha) d\alpha=1$, so the overall integral reduces to $\phi(x)$, giving marginal normality, as desired. | |
| Sep 20, 2011 at 0:12 | comment | added | David Moews | I think, that the example is correct as stated. There is nothing special about the function $\Phi$; you could use any nonnegative function $f$ satisfying $f(x)+f(-x)=1$, and let each $X_i$ have density $2\phi(x) f(\alpha x)$. Since $\phi$ is even, you then get $\int 2\phi(z) f(yz) dz=2 \int_{z\ge 0} (f(yz) + f(-yz))\phi(z) dz=1$ for any $y$. | |
| Sep 19, 2011 at 17:45 | comment | added | Michael Hardy | $E\left(\int_{-\infty}^x 2\alpha \varphi(\alpha x)\Phi(\alpha x)\;dx \right)$ $= E\left(\int_0^{\Phi^{-1}(x)/\alpha} 2u \;du \right)$ $= E\left( ( \Phi^{-1}(x)/\alpha)^2 \right)$ $= \Phi^{-1}(x) E(1/\alpha^2)$. I don′t see this becoming $\Phi(x)$. | |
| Sep 19, 2011 at 17:31 | comment | added | Michael Hardy | @David Moews: I wonder if you meant $2\alpha \varphi(\alpha x)\Phi(\alpha x)$? With that function, I can at least readily verify that it's a probability density. However, I'm having trouble with marginal normality of $X$. We've got $\alpha\sim N(0,1)$ and $X\mid\alpha$ distributed according to this proposed density. We'd like to show that that implies that $X$ is marginally normally distributed, i.e. $\Pr(X\le x)=\Phi(x)$ for all $x$. So $\Pr(X\le x) = E(\Pr(X\le x\mid \alpha)) = E\left(\int_{-\infty}^x 2\alpha \varphi(\alpha x)\Phi(\alpha x)\;dx \right)$. | |
| Sep 12, 2011 at 8:13 | comment | added | David Moews | True, Lauritzen's example is multivariate normal. OK, here's another example: pick $\alpha\sim N(0,1)$, and then let $X_i$ be i.i.d., each with density $2\phi(x) \Phi(\alpha x)$, where $\phi$ and $\Phi$ are the density and cdf of $N(0,1)$. In this example the $X_i$'s have skewed distributions where $\alpha$, a measure of the skew, is itself normally distributed. | |
| Sep 11, 2011 at 20:40 | comment | added | Michael Hardy | .....or maybe I should say "almost always" normal, in the usual probabilistic sense. | |
| Sep 11, 2011 at 20:39 | comment | added | Michael Hardy | Maybe my fourth bullet point isn't clear, and maybe it can't be made clear without a lot more work. Lauritzen's example goes only a tiny distance toward what I had in mind. The distribution given by $\lim\limits_{n\to\infty} \dfrac{|A\cap \lbrace X_1,\ldots,X_n \rbrace|}{n}$ would always be normal. Not the same normal every time, but always normal. And the sequence actually is multivariate normal in the sense that every linear combination of the $X$s is normal. (And the reason why he used the lower-case letter $\rho$ for that parameter seems to be just what you'd guess.) | |
| Sep 9, 2011 at 23:55 | comment | added | David Moews | Another example, in which $\Phi$ is discrete, would be to draw the graph of the density function of $N(0,1)$, divide the region underneath the density function into countably many chunks, and pick $\mu$ by throwing a dart at the picture and looking at the chunk it lands in. To pick the $X_i$'s, we throw more darts at the picture, but condition them to land in the chunk we chose earlier. | |
| Sep 9, 2011 at 23:52 | comment | added | David Moews | Lauritzen gives a simple example in his lecture. He fixes some $\rho$ in $[0,1]$ and then takes $\mu=N(Y,1-\rho)$, where $Y\sim N(0,\rho)$. This interpolates between the i.i.d. case ($\rho=0$) and the case where $X_1$, $X_2$, ... is a constant sequence ($\rho=1$.) | |
| Sep 9, 2011 at 23:06 | comment | added | Michael Hardy | I think you may be right..... I'm wondering if I need to figure out how to get a handle on which distributions $\Phi$ have expectation $N(0,1)$. | |
| Sep 9, 2011 at 23:02 | history | edited | Michael Hardy | CC BY-SA 3.0 |
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| Sep 9, 2011 at 21:50 | history | answered | David Moews | CC BY-SA 3.0 |