Find $E[X]$, pdf of $Y=\sqrt X$ and calculate $E[Y]$

Question

$X$ has pdf $f(x)=2x^{-2}, x>2$ and $0$ otherwise. Find $E[X]$, pdf of $Y=\sqrt X$ and calculate $E[Y]$ using two ways.

a) Find $E[X]$
I got $$E[X]=\int_2^\infty x2x^{-2}=2[\ln(\infty)-\ln(2)]$$

b) Find pdf of $Y=\sqrt X$

$$F_Y(y)=P(Y\le y)=P(\sqrt{X}\le y)=^{X>2}P(X\le y^2)=F_X(y^2)$$

Now take the derivative to find $f_Y(y)$. It is $$f_Y(y)=\frac{d}{dy}F_Y(y)=\frac{d}{dy}F_X(y^2)=f_X(y^2)\cdot(y^2)'=2(y^2)^{-2}2y=4y$$

c)Find $E[Y]$ from pdf and $E[g(x)]=\int g(x)f(x)$
I got $$E[Y]=\int_2^\infty 4y^2dy= (4/3)\infty^3-(4/3)2^3$$

Both my $E[Y]$ and $E[X]$ seem off.

Answer 1

Your mistake is in thinking that $(y^{2})^{-2}=1$. Actually $(y^{2})^{-2}=y^{-4}$. Now try to redo your calculations.

Answer 2

We have $$ \mathbb E[X] = \int_2^\infty 2x^{-1}\ \mathsf dx = \lim_{b\to\infty} \int_2^b 2x^{-1} \ \mathsf dx = \lim_{b\to\infty}2\log x|_2^b = +\infty. $$ By a similar computation, $$ \mathbb E[Y] = \mathbb E[\sqrt X] = \int_2^\infty 2x^{-3/2}\ \mathsf dx = 2\sqrt 2. $$ To find the distribution of $Y$, we note that $$ F_Y(y)=\mathbb P(Y\leqslant y) = \mathbb P(\sqrt X\leqslant y) = \mathbb P(X\leqslant y^2) = F_X(y^2), $$ where we use the positive square root since $\mathbb P(X>0)=1$. Hence for $y^2>2$ or $y>\sqrt 2$ we have $$ f_Y(y) = F_X'(y^2) = f_X(y^2)\cdot 2y = 2y^{-4}\cdot 2y = 4y^{-3} $$ and so $$ \mathbb E[Y] = \int_{\sqrt 2}^\infty 4y^{-2}\ \mathsf dy = 2\sqrt 2, $$ as expected.

Answer 3

No, all looks good, for $X$.

It is permissable to have expectations that do not converge, and indeed $X$ does not have a convergent expectation.

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HOWEVER. You can find a real expectation for $Y$ .

The support for $Y$'s pdf should be $[\surd 2..\infty)$, and $(y^{-2})^2=y^{-4}$.

There is also another method you can try. (LoTE)