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I don't speak maths too well (engineer) so simple language preferred or could you describe it as a graph please?

This has probably been asked but I have no idea what to search...

Follow up question:

$$s = w_1 - w_2$$

What is $p(s \mid w_1,w_2)$?

Edit: my top guess if that both are just uniform distributions, as you have no other priors.

edit2: the wider question:

$p(y \mid w_1,w_2) = \int\int p(y \mid t) p(t \mid s) p(s \mid w_1,w_2) dsdt$

y is the outcome of a 'game'

$t \sim N(t \mid s, 1)$

$s = w_1 - w_2$

So the explanation i have jotted down is that $p(y \mid t)$ is a step function. I reckon $p(t \mid s)$ is a gaussian centred wherever s (the skill difference) is. this all makes sense. im just trying to get a handle on the last term...

Furthermore, the notes go on to day the above integral simplifies:

$p(y \mid w_1,w_2) = \int\int p(y \mid t) p(t \mid s) p(s \mid w_1,w_2) dsdt = \int p(y \mid t) p(t \mid w_1,w_2) dt$

Using the delta function explanation that was the accepted answer, this sort of makes sense in my head as "integrating over s sifts out a gaussian probability distribution of t that is dependent on w1 and w2". the whole variable vs known stuff that is eluded to in the answer comments goes over my head a bit but sort of explains the form above.

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What is s, w1, w2 ? –  Inquest Feb 25 '13 at 21:19
    
variables. like x, y z. (continuous variables) –  Sam Feb 25 '13 at 21:20
    
By the letter $P$ I guess these are random variables. –  Berci Feb 25 '13 at 21:22
    
i think some kind person is formatting my question with latex =D –  Sam Feb 25 '13 at 21:22
1  
Careful with the notations.. $t\sim \mathcal{N}(t| s,1)$ doesn't mean anything to me. Use $t\sim\mathcal{N}(\mu,\sigma^2)$ for a univariate normal distribution. –  Sh3ljohn Feb 25 '13 at 21:45

1 Answer 1

up vote 1 down vote accepted

Whether x and y are random or deterministic variables, if they are equal, then $p(x|y) = \delta_y(x)$ where $\delta$ is the Kronecker delta function. Same goes for s: $p(s|w_1,w_2) = \delta_{w_1-w_2}(s)$.

See http://en.wikipedia.org/wiki/Kronecker_delta for more details :)

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great! i was just thinking of a delta function. so i could describe it as a delta function that sits wherever y is –  Sam Feb 25 '13 at 21:39
    
this tallies with what i was thinking for the middle term of the integral - it is a gaussian that moves around to wherever s is –  Sam Feb 25 '13 at 21:41
    
Exactly, but honestly I didn't understand the "wider question". I hope this answer is sufficient. –  Sh3ljohn Feb 25 '13 at 21:41
    
so, just to be clear on this answer: $\delta_{w_1-w_2}(s) = \delta_{s}(s)$? –  Sam Feb 25 '13 at 21:48
    
Formally, no. But I think you understand. s is a variable, therefore $\delta_s(s)$ doesn't mean anything. In the above, $w_1,w_2$ are known, and so the way I wrote it is the correct way. –  Sh3ljohn Feb 25 '13 at 21:49

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