Help me understand... E(x)

[ingvar]

Here is a problem...
f(x)=2/(x^3), x>1, max 10. E(Y)=?

Y=X, X<10
Y=10, X>10
or Y=min(X, 10)

I know you can solve it this way:
E(Y) = Int(1 to 10) xf(x) dx + 10 Int(10 to inf) f(x) dx = 1.9

I tried to solve using survival function
E(Y)=E(min (X, 10))=Int (1 to 10) S(x) dx = .9
Don't get why I get .9 and not 1.9 as it should be.

Thanks!

[sugarnspice82805]

The def of survival is int (0~inf) Pr X>x. I believe that since you are starting your integral at 1 and not 0, you have to add 1.

[Anu Dhanuka]

another question related to discrete probability:
compute the expected value of the random number of coin tosses until a run of k successive heads occurs when the tosses are independent and each lands on heads with a probability of 1/2
answer given is [2^(k+1)]-2...

[sugarnspice82805]

those questions trip me up everytime!!

[jthias]

Here is a problem...
f(x)=2/(x^3), x>1, max 10. E(Y)=?

Y=X, X<10
Y=10, X>10
or Y=min(X, 10)

I know you can solve it this way:
E(Y) = Int(1 to 10) xf(x) dx + 10 Int(10 to inf) f(x) dx = 1.9

I tried to solve using survival function
E(Y)=E(min (X, 10))=Int (1 to 10) S(x) dx = .9
Don't get why I get .9 and not 1.9 as it should be.

Thanks!

The formula works like this: if X is defined on [a,b] with a>=0 , then

E(X) = a + int[a to b] of Sx(x) dx

So getting back to the problem we have

Y = X on the interval (1,10)

so E(Y) = 1 + int[1 to 10] of Sx(x) dx = 1 + .9 = 1.9

I didn't look at your calculation of the integral above; I'm just showing you how to find the missing value of 1.

X can be defined on open interval (a,b). It doesn't matter, as long as the endpoint 'a' is not a negative number, the formula will work.

[ingvar]

The formula works like this: if X is defined on [a,b] with a>=0 , then

E(X) = a + int[a to b] of Sx(x) dx

So getting back to the problem we have

Y = X on the interval (1,10)

so E(Y) = 1 + int[1 to 10] of Sx(x) dx = 1 + .9 = 1.9

I didn't look at your calculation of the integral above; I'm just showing you how to find the missing value of 1.

X can be defined on open interval (a,b). It doesn't matter, as long as the endpoint 'a' is not a negative number, the formula will work.

If I understand correctly if X is between a and b:
E(min (X, a)) = a + int[a to b] of Sx(x) dx

what about E(max (X, a)) =

[Anu Dhanuka]

another question related to discrete probability:
compute the expected value of the random number of coin tosses until a run of k successive heads occurs when the tosses are independent and each lands on heads with a probability of 1/2
answer given is [2^(k+1)]-2...

Hey guys any answer?

[Anu Dhanuka]

If I understand correctly if X is between a and b:
E(min (X, a)) = a + int[a to b] of Sx(x) dx

what about E(max (X, a)) =

if X = max{T,c} for some value c in (0,infinity),
then E(X) = c + int[c to infinity] of St(t) dt and
If Y = min{T,c} ,then E(Y) = int[0 to c] of St(t) dt

is it right?

[sohpmalvin]

if X = max{T,c} for some value c in (0,infinity),
then E(X) = c + int[c to infinity] of St(t) dt and
If Y = min{T,c} ,then E(Y) = int[0 to c] of St(t) dt

is it right?

You have said it rightly.

[ingvar]

You have said it rightly.

No, it's not right, see my problem above