Mixture Distributions

[suraj]

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?

 

[John Lee]

A very good question Suraj

It is easy to calculate the mean and variance using E(X) = E[E(X|W)] =6 and var(X) = E[var(X|W)] + var[E(X|W)] = 16.

But obtaining the distribution is a lot harder mainly due to the fact that the range of the PDF depends on the parameter f(x|w) = 1/w for 0 < x< w.

Suffice to say it won't be uniform but probably some sort of decreasing distribution - possibly triangular with max w=0 and heading to zero for w=18.

This would be harder than would be required for the UK, India or US exams.

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?

 

[suraj]

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.

 

[John Lee]

I won't tell you how long I spent on this one....

 

[td290]

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.

 

[suraj]

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

 

[bapan]

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).

 

[td290]

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.