Mixture Distributions

Hi, I have a bit of problem in calculating DF of mixture distributions

Example -

X|W=w ~ U (0,w)

W ~ U (6,18)

So what is the unconditional distribution of X ?

I calculated it using method given in the book (Ch 3, Pg 28) by integrating the two DF over the range of W.

i.e. Integrate (1/12)*(1/w) over 6 to 18
which gives "0.0915551"
which means uniform dist. over 0 -18

But 0.0915551 is not equal to "1/18"

Can anyone tell me where am I going wrong?

 

I wasted so much time on this . Now I also this think that it'll be some sort of decreasing triangular distribution, touching y-axis at (1/9) so that total area under the curve is 1. It can't be simply uniform over 0-18.

Thanks John for pointing this out.

 

First some notation, which you may or may not know already.\[\mathbf{1}_{(\bullet)}=\left\{\begin{array}{ll}0 & \textrm{if }\bullet\textrm{ is false}\\1&\textrm{if }\bullet\textrm{ is true}\end{array}\right.\]
So,\[\begin{array}{ll}\mathbb{P}(X=x) & =\int_6^{18} {\mathbb{P}(X=x|W=w)\mathbb{}P(W=w)dw}\\& =\int_6^{18} {\frac{\mathbf{1}_{(0\leq x \leq w)}}{w}\times\frac{1}{12}dw}\end{array}\]
If \(x\leq 6\) then,\[\mathbb{P}(X=x)=\int_6^{18} {\frac{1}{w}\times\frac{1}{12}dw}=\frac{1}{12}\ln(3)\]
If \(6\leq x\leq 18\) then,\[\mathbb{P}(X=x)=\int_6^x {0dw}+\int_x^{18} {\frac{1}{w}\times\frac{1}{12}dw}=\frac{1}{12}\ln \left(\frac{18}{x}\right)\]
If \(x>18\) then,\[\mathbb{P}(X=x)=\int_6^{18} {0dw}=0\]
The function this gives for \(\mathbb{P}(X=x)\) is continuous and you should find that it integrates to 1. Let me know if you need any of the above steps clarified or elaborated.

 

My opinion changes every time I see a new soln . Now I think this is correct.
Because the interval (0,6) is common no matter what the value of parameter (w) is, the distribution is simply uniform over (0,6)

After that it starts falling from 6 to 18 which is given by curve "1/12*ln(18/x)"

So the distribution is actually a rectangle of height "0.0915551" in (0,6)
and a falling curve "1/12*ln(18/x)" in (6,18)

Looks like we've to take special care in case of uniform distribution because the conditioning is on the range of distribution unlike in other distributions where we can simply integrate the two DFs over the parameter's range.

Many thanks td290 for the reply. Everything is clear now

 

Only point I would like to make here is to td290.

Although the approach is correct, (being a purist) I would avoid writing P(X=x) or similar given that these are continuous variables.

It is desirable that one expresses this as fx(x).

 

You're right of course. The fact was I'd had enough fun getting MathJax to behave without worrying about that and I was trying to get it done quickly after work before going home. I was also trying to minimise the amount of potentially confusing notation as I had already used indicator functions. I assumed everyone would know what I meant and on that at least I seem to have been proved right.