Joint problem density function problem

Lewis7879
Messages
6
Reaction score
0
I need help guys I can't understand this
Can anyone explain thoroughly how do I form the range for this question?
f(x,y)= e-x for 0≤x≤y≤∞
0 Otherwise

Find P(x+y≤1)
I attempted this by integrating through the range of
0≤y≤(1-x) and 0≤x≤∞ but that doesn't seem right
 
Physics news on Phys.org
The statement is confusing. You appear to have a density function in x and y, which is a function of x only.
 
mathman said:
The statement is confusing. You appear to have a density function in x and y, which is a function of x only.
Hello mathman there's a slight error with range I made in the question which is 0≤y≤x≤∞
There was no other problems with the question as I was asked this way.
 
No, f(x,y)= e^{-x} for 0\le x\le y\le A is a function of both x and y. To determine "A", use the fact that the "total" probability must be 1:
\int_{y= 0}^A\int_{x= 0}^y e^{-x} dx dy= 1
 
  • Like
Likes Lewis7879
Getting an equation for A is easy enough. A+e^{-A}=2. I am confused as to what is the question.
 
  • Like
Likes Lewis7879
HallsofIvy said:
No, f(x,y)= e^{-x} for 0\le x\le y\le A is a function of both x and y. To determine "A", use the fact that the "total" probability must be 1:
\int_{y= 0}^A\int_{x= 0}^y e^{-x} dx dy= 1
Will I be able to find P(x+y≤1) after determine A?
 
Namaste & G'day Postulate: A strongly-knit team wins on average over a less knit one Fundamentals: - Two teams face off with 4 players each - A polo team consists of players that each have assigned to them a measure of their ability (called a "Handicap" - 10 is highest, -2 lowest) I attempted to measure close-knitness of a team in terms of standard deviation (SD) of handicaps of the players. Failure: It turns out that, more often than, a team with a higher SD wins. In my language, that...
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Back
Top