find the expected value of

Solution :

*is this process which i am following is it correct or no?*Problem : If the probability of X is given by

find the expected value of

Solution :

*is this process which i am following is it correct or no?*

find the expected value of

Solution :

Hallo, nice to meet you. >I'm new on this forum.

For me this process of calculation is correct, even if there is a way to see the things that I'm not sure about :

We need <g(X)=Y>=<Y>, and then we need the probability density of Y. I don't know if this gives the same result. (We know that Y runs over [-13/4;3] if x runs over [0;3]).

From this we see there could appear a problem if g were not bijective.

See you soon, with friendly greetings.

For me this process of calculation is correct, even if there is a way to see the things that I'm not sure about :

We need <g(X)=Y>=<Y>, and then we need the probability density of Y. I don't know if this gives the same result. (We know that Y runs over [-13/4;3] if x runs over [0;3]).

From this we see there could appear a problem if g were not bijective.

See you soon, with friendly greetings.

Last edited by jk22 on Sat Jun 26, 2010 1:55 pm, edited 1 time in total.

- maggiemagnet
**Posts:**308**Joined:**Mon Dec 08, 2008 12:32 am-
**Contact:**

According to this Wikipedia article, yes, it looks like you have set this up correctly. Do you need any help with the integrations?

Hallo, nice to meet you too.

Thanks for the reference. Me I lost myself in : should I calculate like this :

y=g(x), then :

and hence we find the distribution of Y : f_y(g(x))g'(x)=f(x), or f_y(y)=f(g^-1(y))/g'(g^-1(y)), to get the same as above.

Thanks for the reference. Me I lost myself in : should I calculate like this :

y=g(x), then :

and hence we find the distribution of Y : f_y(g(x))g'(x)=f(x), or f_y(y)=f(g^-1(y))/g'(g^-1(y)), to get the same as above.