Using discrete and continuous rvs in a distribution

**FGT12** · May 17th 2011, 08:21 AM

Suppose that W, the amount of moisture in the air on a given day, is a gamma random variable with parameters $(t, \beta)$
Suppose also that given that $W = w$ , the number of accidents during that day - call it N - has a poisson distribution with mean $w$ .
Show that the conditional distribution of W given that N = n is the gamma distribution with parameters $(t+n, \beta +\sum_{i = 1}^n x_i)$

I would like some help to write the formula for the second supposition, as it goes from a continuous rv to a discrete one

Thanks

**theodds** · May 17th 2011, 09:03 AM

Originally Posted by FGT12

Suppose that W, the amount of moisture in the air on a given day, is a gamma random variable with parameters $(t, \beta)$
Suppose also that given that $W = w$ , the number of accidents during that day - call it N - has a poisson distribution with mean $w$ .
Show that the conditional distribution of W given that N = n is the gamma distribution with parameters $(t+n, \beta +\sum_{i = 1}^n x_i)$

I would like some help to write the formula for the second supposition, as it goes from a continuous rv to a discrete one

Thanks

So we know the distribution of W and N|W. To get the distribution of W|N note that $f_{W|N}(w|n) \propto f_{W, N} (w, n) = f_{N|W} (n|w) f_W (w)$ . You should be able to recognize the RHS as the kernal of a gamma with the appropriate parameters.

**FGT12** · May 17th 2011, 09:21 AM

When I evalute the RHS I get
$\frac{e^{-(n+\beta)w} w^{x_1+...+x_n} (\beta)^t }{x_1!x_2!...x_n!\Gamma(t) }$

I do not see how to go further with this question

**theodds** · May 17th 2011, 09:40 AM

That shouldn't be what you get when you evaluate the RHS. Come to think of it, you didn't even define what $x_i$ is...I think what you intended was that $X_i | W$ is distributed Poisson with mean W, $i = 1, ..., n$ and you want the distribution of $W| X_1, ..., X_n$ .

Can you post the question exactly as it is written? Among other things that don't make sense, if you interpret the question as I wrote it above you get $(t + \sum x_i, \beta + n)$ and not $(t + n, \beta + \sum x_i)$ .

**FGT12** · May 18th 2011, 04:51 AM

Question:
Let $W$ be a gamma random variable with parameters $(t, \beta )$ , and suppose that conditional on $W = w, X_1, X_2, ..., X_n$ are independent exponential random variables with rate $w$ .
Show that the conditional distribution of $W$ given that $X_1=x_1, X_2=x_2,..., X_n=x_n$ is gamma with parameters $(t+n, \beta + \sum_{i=1}^{n}x_i )$

**theodds** · May 18th 2011, 06:55 AM

Originally Posted by FGT12

Question:
Let $W$ be a gamma random variable with parameters $(t, \beta )$ , and suppose that conditional on $W = w, X_1, X_2, ..., X_n$ are independent exponential random variables with rate $w$ .
Show that the conditional distribution of $W$ given that $X_1=x_1, X_2=x_2,..., X_n=x_n$ is gamma with parameters $(t+n, \beta + \sum_{i=1}^{n}x_i )$

That is false; it should be that $W|X = x$ is $(t + \sum x_i, \beta + n)$ .

$\displaystyle f_{W|X} (w|x) \propto f_{X|W} (x|w) f_W (w)$

$\displaystyle= \left(\prod_{i = 1} ^ n \frac{w^{x_i} e^{-w}}{x_i !}\right) \frac{\beta^t}{\Gamma(t)} w^{t - 1} e^{-\beta w} = \frac{\beta^t}{\Gamma(t) \prod_{i = 1} ^ n x_i !} w^{t + \sum x_i - 1} e^{-(\beta + n) w}$

$\displaystyle\propto w^{t + \sum x_i - 1} e^{-(\beta + n)w}$

(valid for positive w) which is the kernal of a Gamma $(t + \sum x_i, \beta + n)$ .

**FGT12** · May 18th 2011, 07:29 AM

what exactly do you mean by the kernel in this instance? How can I recognise other kernels for other distributions?

And is there no way of getting to

$f(w)=\frac{(\beta+n)e^{-(\beta+n)w}((\beta+n)w)^{t+\sum x_i -1}}{\Gamma(t+\sum x_i)}$

Could we integrate between infinity and minus infinity so that it equals one

**theodds** · May 18th 2011, 07:45 AM

When I say the "kernal" I mean the part of the density that matters, i.e. everything but a normalizing constant (in this case, the x_i are considered fixed so you can get rid of anything that is only a function of the x_i as well as any other fixed constants). You can retrieve the normalizing constant because the density must integrate to 1. If you can show that a pdf is proportional to the kernal of something you know then you know what the pdf is because you can get the normalizing constant by integrating the kernal.

The technique I used is nice because it saves you from having to calculate the marginal of X which requires integration.

To give another example, here are a couple of useful kernals for the normal distribution: $e^{\frac {-1} {2 \sigma^2} (x - \mu)^2}$ as well as $e^{\frac{-1}{2\sigma^2}(x^2 - 2x\mu)}$ .

**ragnar** · April 11th 2012, 07:18 PM

So I'm working on this same problem and wondering, how is the claim of proportionality that you use here justified? I thought that it would be

$f_{W|X}(w|x) P_{X}(x) = f_{X|W}(x|w)P_{W}(w)$

so that when you divide, you're not dividing by a constant but instead dividing by a function of $x$ .

**theodds** · April 11th 2012, 07:25 PM

Yeah, that's fine. For finding the law of W|X you can think of all the stuff on the right side of the conditioning bar as being constants when you do any proportionality stuff.

**ragnar** · April 11th 2012, 07:29 PM

Oh right, duh, the thing we are to prove is that the resulting distribution has parameters which are themselves functions of $x_{i}$ ! Making sense now, thank you!

**ragnar** · April 12th 2012, 07:07 PM

Okay, I lied, I've been off-and-on staring at this some more and I'm back to not really getting it. I intuitively understand the idea of how, conditional on $X$ , things in terms of $X$ are like a constant, but I'm not sure how to make rigorous use of that idea.

Here's an outline of what I've done, followed by a more detailed description if it's helpful.

By some simple algebraic manipulation and Bayes's Law, I get

$P_{W|X}P(w|x) = \frac{P_{X|W}(x|w)P_{W}(w)}{P_{X}(x)}$

Where the expressions in the numerator are described in the assumptions of the problem. From that, I combine expressions with a base of $w$ and with a base of $e$ . The result is $w^{t+n-1}e^{-w(\beta +\sum x_{i})}$ , which seems to me the (as you call it) "kernel" of a gamma distribution with parameters $t+n, \, \, \beta+\sum x_{i}$ . Now I know that you earlier said this cannot be right, and maybe that's why I'm running into problems--however, I'm not seeing how what I've done is wrong or how anything else could work.

But as a result of so organizing my terms, my "coefficient" is now

$\frac{\beta^{t}}{\Gamma (t)P_{X}(x)}$

For this to truly be a gamma distribution in those parameters, I need my coefficient to be

$\frac{\Big( \beta+\sum x_{i} \Big)^{t+n}}{\Gamma(t+n)}$

So how do I do this? I don't have freedom to choose what any of the terms are, so it doesn't seem like I am able to compensate for this difference by assigning some value to a constant coefficient or anything like that.

A more detailed derivation of the expression that I ultimately obtain:

$P(X_{1}=x_{1}, ..., X_{n}=x_{n}|W=w)P(W=w) \quad = \\\\ P(W=w|X_{1}=x_{1}, ..., X_{n}=x_{n})P(X_{1}=x_{1}, ..., X_{n}=x_{n}) \quad \Longrightarrow \\\\ P(W=w|X_{1}=x_{1}, ..., X_{n}=x_{n}) \quad = \quad \frac{P(X_{1}=x_{1}, ..., X_{n}=x_{n}|W=w)P(W=w)}{P(X_{1}=x_{1}, ..., X_{n}=x_{n})} \quad = \\\\ \frac{w^{n}e^{-w(x_{1}+...+x_{n})}\frac{\beta^{t}}{\Gamma (t)}w^{t-1}e^{-\beta w}}{P(X_{1}=x_{1}, ..., X_{n}=x_{n})} \quad = \quad \frac{\beta^{t}}{\Gamma (t) P(X_{1}=x_{1}, ..., X_{n}=x_{n})}{w^{t+n-1}e^{-w(\beta + \sum_{i=1}^{n}x_{i})}}$

**ragnar** · April 12th 2012, 07:13 PM

By the way, I just noticed that, in your earlier statement you were using the Poisson distribution, which is what the original poster had originally posted.

However, in the original poster's reply (time-stamped May 18th 8:51 AM) when he wrote exactly what the problem was asking, he wrote the exponential distribution. So really, this problem should be about each $X_{i}|W$ being exponential with rate $w$ .

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