# If $x = y$ what is $p(x\mid y)$?

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...

$$s = w_1 - w_2$$

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
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

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: $p(s|w_1,w_2) = \delta_{w_1-w_2}(s)$.
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