Normalizing a Probability Distribution with Integral Calculations

[Cosmossos]

Homework Statement

The probability is given by:
p(x,y)=6(1-x-y)
x>=0
y>=0
x+y<=1

Show that p is normilized.
calculate <x>

Homework Equations

I found that p is normilized using the integral:

and It's equal to 1.
are the limits of the integral correct?

 

[gabbagabbahey]

Why are you integrating y from x to 1-x? Are you told that y\geq x? If not, you should be integrating from 0 to 1-x instead.

 

[Cosmossos]

I know that y>=x-1
and if I do it for 0 to 1 I don't get in the end that the probabilty =1

 

[vela]

You were given x+y \le 1, so, moving the x over to the other side, you get y \le 1-x, right?

 

[Cosmossos]

right... so that's why I chose the integral limits for dy to be from x to 1-x.
Is it wrong?

 

[gabbagabbahey]

right... so that's why I chose the integral limits for dy to be from x to 1-x.
Is it wrong?

Again, why are you using x as your lower limit? You are told that y\geq0, not y\geq x

 

[vela]

I know that y>=x-1
and if I do it for 0 to 1 I don't get in the end that the probabilty =1

right... so that's why I chose the integral limits for dy to be from x to 1-x.
Is it wrong?

Your response to gabbagabbahey's question was confusing. In your previous post, you answered, it was because you knew y \ge x-1. The "x-1" suggested to me you somehow got that condition from the constraint x+y \le 1, and my point was that the constraint can only lead to y \le 1-x, the upper limit of the integral. There's no way you can get y \ge x-1 from it.

Even if you did have y \ge x-1, your answer didn't make sense. Why would the lower limit of the integral be x? Shouldn't it be x-1 if your inequality were true?