Expected value from a density function

mind0nmath
Messages
19
Reaction score
0
Hey,
I know how to find the expected value from the density function when it is in the form:

(example)

| y^2 -1<y<1
|
fy =|
| 0 elsewhere

Ey = integral(upper limit 1, lower limit -1)[y*y^2 dy)

but, what if the density function looks like this:

| y^2 -1<y<0
|
fy =| y^2 - y 0<y<1
|
| 0 elsewhere

how do you approach here?
 
Physics news on Phys.org
The expectation value of Y is given by

E(Y) = \int_{-\infty}^{+\infty}yf(y)dy[/itex]<br /> <br /> If I understood your question correctly, you just have to split the integral into disjoint intervals and apply the different definitions of f(y) in each such interval. This is immediate from the linearity of the Riemann integral and the continuity of the integrand.
 
E(Y) = \int_{-\infty}^{+\infty}yf(y)dy
 
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