1. ## E(X) and Var(X)

1. The random variable X has pdf fX(x) = x^−4for x > 1, and 0, otherwise.
(a) Compute E(X) and Var(X).
(b) Let Y =√X. Compute E(Y ) and Var(Y ).
(c) Let W = 2X + 5. Compute E(W) and Var(W).

By the way, how do I get the Math symbols to work while pasting directly from my Words? Thanks

2. Originally Posted by Ruichan
1. The random variable X has pdf fX(x) = x^−4for x > 1, and 0, otherwise.
This is not a density, as a requirement for a density is that:

$\displaystyle \int_D f(x) dx = 1$

and $\displaystyle f(x)=x^{-4}$ does not meet this requirement if $\displaystyle D=(1, \infty)$.

RonL

3. Originally Posted by CaptainBlack
This is not a density, as a requirement for a density is that:

$\displaystyle \int_D f(x) dx = 1$

and $\displaystyle f(x)=x^{-4}$ does not meet this requirement if $\displaystyle D=(1, \infty)$.

RonL

That's why I couldn't get the answer at all. I emailed my prof this morning. He just replied saying there's a typo. The correct question is
1. The random variable X has pdf fX(x) = 3x^−4for x > 1, and 0, otherwise.
(a) Compute E(X) and Var(X).
(b) Let Y =√X. Compute E(Y ) and Var(Y ).
(c) Let W = 2X + 5. Compute E(W) and Var(W).

Does the 3 make a difference?

4. $\displaystyle \begin{array}{l} \int\limits_1^\infty {\frac{3}{{x^4 }}dx} = 1 \\ E(X) = \int\limits_1^\infty {x\left( {\frac{3}{{x^4 }}} \right)dx} = \int\limits_1^\infty {\left( {\frac{3}{{x^3 }}} \right)dx = ?} \\ E(X^2 ) = \int\limits_1^\infty {x^2 \left( {\frac{3}{{x^4 }}} \right)dx} = \int\limits_1^\infty {\left( {\frac{3}{{x^2 }}} \right)dx = ?} \\ V(X) = E\left( {X^2 } \right) - E^2 (X) \\ \end{array}$

5. Originally Posted by Ruichan
That's why I couldn't get the answer at all. I emailed my prof this morning. He just replied saying there's a typo. The correct question is
1. The random variable X has pdf fX(x) = 3x^−4for x > 1, and 0, otherwise.
(a) Compute E(X) and Var(X).
(b) Let Y =√X. Compute E(Y ) and Var(Y ).
(c) Let W = 2X + 5. Compute E(W) and Var(W).

Does the 3 make a difference?

a) E(X) = 3/2, E(X^2)=3, Var(X)=3/4
b) Can someone help me with this please? All the questions I did before was when X ~Unif(0,1). How do you do it for this when it's not
unif(0,1)? Thanks

c) Using the the values I got from a), I got E(W)=3 and Var(W)=3. ----> Can the expected value and variance be the same?
E(W) = 2E(X) = 2(3/2) = 3. Var(X) = 2^2Var(X) = 4Var(X) = 4(3/4) = 3.
The answers look so wrong though. Did I do something wrongly? I just integrated with the formula my prof gave.

6. Originally Posted by Ruichan
b) Can someone help me with this please? All the questions I did before was when X ~Unif(0,1). How do you do it for this when it's not
unif(0,1)? Thanks
(b) Let Y =√X. Compute E(Y ) and Var(Y ).
E(Y) = integrate (from 0 to 1) x^1/2 dx =2/3
E(Y^2) = Integrate (0 to 1) x dx = 1/2
Var(Y) = 1/2 - 2/3^2 = 1/18

Is it correct if I do it this way?

7. Originally Posted by Ruichan
(b) Let Y =√X. Compute E(Y ) and Var(Y ).
E(Y) = integrate (from 0 to 1) x^1/2 dx =2/3
E(Y^2) = Integrate (0 to 1) x dx = 1/2
Var(Y) = 1/2 - 2/3^2 = 1/18

Is it correct if I do it this way?
$\displaystyle Y=\sqrt{X}$, where the density of $\displaystyle X$ is $\displaystyle f(x)=3x^{-4}$.

Then:

$\displaystyle E(Y)=\int_1^{\infty} \sqrt{x} 3 x^{-4}\ dx=\int_1^{\infty} 3 x^{-3.5}\ dx$

........................$\displaystyle = \left[ -\frac{3}{2.5}x^{-2.5} \right]_1^{\infty}=3/2.5 = 1.2$

Also:

$\displaystyle E(Y^2)= \int_1^{\infty} (x) 3 x^{-4}\ dx=\int_1^{\infty} 3 x^{-3}\ dx$

and you should be able to take things from there

RonL

8. Originally Posted by CaptainBlack
$\displaystyle Y=\sqrt{X}$, where the density of $\displaystyle X$ is $\displaystyle f(x)=3x^{-4}$.

Then:

$\displaystyle E(Y)=\int_1^{\infty} \sqrt{x} 3 x^{-4}\ dx=\int_1^{\infty} 3 x^{-3.5}\ dx$

........................$\displaystyle = \left[ -\frac{3}{2.5}x^{-2.5} \right]_1^{\infty}=3/2.5 = 1.2$

Also:

$\displaystyle E(Y^2)= \int_1^{\infty} (x) 3 x^{-4}\ dx=\int_1^{\infty} 3 x^{-3}\ dx$

and you should be able to take things from there

RonL
Thank you very much.
Arg.....shows how much i understand probability when I don't even know that I'm supposed to use the pdf from part a). Arg.....

Question: Therefore it doesn't matter that if X~unif(0,1)? I just need to use the pdf? Then integrate from whatever values that are given? The unif(0,1) is to integrate values from 0 to 1? If for eg, if not uniform, values given are 2<x<7, just need to integrate from 2 to 7?

9. Originally Posted by Ruichan
Thank you very much.
Arg.....shows how much i understand probability when I don't even know that I'm supposed to use the pdf from part a). Arg.....

Question: Therefore it doesn't matter that if X~unif(0,1)? I just need to use the pdf? Then integrate from whatever values that are given? The unif(0,1) is to integrate values from 0 to 1? If for eg, if not uniform, values given are 2<x<7, just need to integrate from 2 to 7?
Yes, you integrate the function you want the expectation of times the pdf,
over the range that the pdf is defined on (or rather you integrate over the
entire line or whatever, but ignore the regions over which the pdf is zero).

RonL

10. Originally Posted by Plato
$\displaystyle \begin{array}{l} \int\limits_1^\infty {\frac{3}{{x^4 }}dx} = 1 \\ E(X) = \int\limits_1^\infty {x\left( {\frac{3}{{x^4 }}} \right)dx} = \int\limits_1^\infty {\left( {\frac{3}{{x^3 }}} \right)dx = ?} \\ E(X^2 ) = \int\limits_1^\infty {x^2 \left( {\frac{3}{{x^4 }}} \right)dx} = \int\limits_1^\infty {\left( {\frac{3}{{x^2 }}} \right)dx = ?} \\ V(X) = E\left( {X^2 } \right) - E^2 (X) \\ \end{array}$
I do have the formula in front of me when trying to do the questions. I just didn't know how to start.
Thanks for showing me!