Is my following reasoning for conditional probabilities correct?

Question

This may be very trivial but I haven't seen it before and I stumbled across it myself while playing around with conditional probabilities and Bayes's theorem. Please bare with me! Here it goes:

I know that

$$ p(x) = \sum_{y}{p(x,y}), $$

and using conditional probabilities this means

$$ p(x) = \sum_{y} p(x|y)p(y). $$

From Bayes's theorem I know that $$ p(x) = \frac{p(x|y)p(y)}{p(y|x)}. $$

This implies that $$ \sum_{y} p(x|y)p(y) = \frac{p(x|y)p(y)}{p(y|x)} $$

Is this correct reasoning? It just doesn't feel right to me as both sides contain the term $p(x|y)p(y)$ but one has a sum and the other doesn't. Thank you!

Answer 1

The left side is a sum over a number of possibilities, but the right side is divided by a probability, which is $\leq 1$, so this increases the size of the right side compared to the numerator, which is a single term of the sum on the left side. (All of that to say that this shouldn't immediately throw any flags.)

But if you buy the first three equations you wrote, the fourth is just algebra and follows from them.

Answer 2

Yes it is correct. $$p(y|x)={p(x|y)p(y)\over p(x)}={p(x|y)p(y)\over \sum\limits_zp(x|z)p(z)}$$

Now you can see why it is correct, right?

Answer 3

Be careful, it is correct , but the L.H.S of the following equation uses $y$ as a boundvariable(https://en.wikipedia.org/wiki/Free_variables_and_bound_variables), it can be misleading. $$\sum_{y} p(x|y)p(y) = \frac{p(x|y)p(y)}{p(y|x)}$$