Visualizing Integration as Summing Over a Variable

[pamparana]

Hello all,

I am sorry for this really basic question but am having trouble with visualizing something in my head...

I read up on interpretation of integration as area under the curve by splitting it in strips with Δx as the length of the strip and we have the integral in the limit as Δx→0.

However, I am having trouble visualizing why an integral is considered as summing over a variable. For example, the marginalization operation is described as:

P(x) = \intP(x, y) dy

I am having even trouble explaining this problem. I am basically having trouble intuitively thinking why this is marginalization. I can picture this in discrete case easily when

P(x) = \sum P(x, y) for all y.

This is basically calculating the probability for all values of x regardless of whatever value y takes. I can see this clearly but having trouble making the same connection with the integral operator.

I would be very grateful if someone can help me get an intuition about this...

Thanks,
Luc

 

[mathman]

I am not sure what you are having trouble with. The interpretation of integration is the definition of the Riemann integral approach. If the function is a function of two variables, you may hold one of them fixed and the integral over the other is no different from the original Riemann integral.
http://en.wikipedia.org/wiki/Riemann_integral
http://mathworld.wolfram.com/RiemannIntegral.html

If you Google "Riemann integral" you can get more.

 

[pamparana]

Thanks for your reply! The problem I am having is that any discussion that I see explains integral as the area under the curve, which is intuitive and I grasp that.

So, I thought about this and if we have

P(x) = ∫P(x, y) dy

Then P(x) is a function (prob. distribution) and each "point" in P(x) is the area under the curve when we hold x as fixed and integrate over all of y.

So, the probability of P(X = x) is the area under the curve when we hold X= x and integrate over y. I think I have it now but I think I always have trouble with this term "area" (despite what I said earlier).