# Thread: solve distribution function

1. ## solve distribution function

Question :

expected value-->
E(g(x))= $\int_0^{1 }g(x^2) ~dx$ for any given g(x);

find P( $1/4$ <X < $9/16$)

my answer ( am i right...?):

put g(x)= x,suppose there is a p.d.f h(x)= x;

then E(g(x))= $\int_0^{1 }g(x)h(x) ~dx$
= $\int_0^{1 }(x^2) ~dx$

so, P( $1/4$ <X < $9/16$)
= $\int_{1/2}^{3/4} (x^2) ~dx$

2. what is the density?

$E(g(X))=\int g(x)f(x)dx$

3. Originally Posted by matheagle
what is the density?

$E(g(X))=\int g(x)f(x)dx$
definition of epected value...

4. Originally Posted by matheagle
what is the density?

$E(g(X))=\int g(x)f(x)dx$

perhaps ignore my answer,
what will u do...?

5. Originally Posted by kin
Question :

expected value-->
E(g(x))= $\int_0^{1 }g(x^2) ~dx$ for any given g(x);

find P( $1/4$ <X < $9/16$)

my answer ( am i right...?):

put g(x)= x,suppose there is a p.d.f h(x)= x;

then E(g(x))= $\int_0^{1 }g(x)h(x) ~dx$
= $\int_0^{1 }(x^2) ~dx$

so, P( $1/4$ <X < $9/16$)
= $\int_{1/2}^{3/4} (x^2) ~dx$
Please type the question exactly as it appears from where you got it from.

6. Originally Posted by mr fantastic
Please type the question exactly as it appears from where you got it from.
it's exactly the question....

for any g(x) satisfies the given condition

7. Originally Posted by kin
Question :

expected value-->
E(g(x))= $\int_0^{1 }g(x^2) ~dx$ for any given g(x);

find P( $1/4$ <X < $9/16$)

my answer ( am i right...?):

put g(x)= x,suppose there is a p.d.f h(x)= x; Mr F says: This is not a pdf - it fails one of the necessary criteria.

then E(g(x))= $\int_0^{1 }g(x)h(x) ~dx$
= $\int_0^{1 }(x^2) ~dx$

so, P( $1/4$ <X < $9/16$)
= $\int_{1/2}^{3/4} (x^2) ~dx$
Well, the best I can come up with right now is that the pfd is given by f(x) = g(x^2)/g(x) for 0 < x < 1 and zero otherwise ....

8. Originally Posted by mr fantastic
Well, the best I can come up with right now is that the pfd is given by f(x) = g(x^2)/g(x) for 0 < x < 1 and zero otherwise ....
f(x) = g(x^2)/g(x) for 0 < x < 1

the above is given by E(g(x))= $\int_0^{1 }g(x^2)~dx$= $\int_0^{1 }g(x)f(x) ~dx$

to satisfy f(x) = g(x^2)/g(x) for 0 < x < 1,
take random variables 1/4 < X < 9/16 ,
then, g(x)=x, f(x)=x ,
so P( 1/4 < X < 9/16)= $\int_{1/4}^{9/16 }x~dx$ ????