Expected value of transformation of random variable with specified distribution function

Question

We have $F_Y(y)=y^3, 0\leq y\leq 1$

Find the expected value of $X = 131+56Y$ (increasing function)

So $Y = \frac{X-131}{56}$

We find the CDF of X:

$F_X = F_Y(y) = \left(\frac{x-131}{56}\right)^3 $

And the PDF:

$f_X(x)= \frac{3(x-131)^2 }{56^3}$

I end up with $E[X] = 0.19$ which is not reasonable, given that $E[Y] = 3/4$.

So what am I missing?

Answer 1

$$ \Bbb E(X)= 131+56\,\Bbb E(Y) $$ $$ f_Y(y)=F'(y)=3y^2\quad\text{for }0\le y\le 1 $$ and $$\Bbb E(Y)=\int_0^1 yf_Y(y)\mathrm dy=\int_0^1 3y^3\mathrm dy=\frac34$$ So we find $$ \Bbb E(X)= 131+56\,\Bbb E(Y)=131+56\,\frac34=173 $$

Answer 2

When you calculate the expectation you have to consider that $$0\leq Y=\frac{X-131}{56}\leq 1.$$

This gives you

$$131\leq x \leq 187$$

as integration boundaries. I bet you did $\int_0^1$.

Answer 3

It is not necessary to find the PDF of $X$ in order to answer. (But if you want to do that, please see @zoli's Answer on how to use it correctly.)

Notice that $Y \sim \mathsf{Beta}(\alpha=3,\beta=1),$ which has $E(Y) = \frac{\alpha}{\alpha + \beta} = 3/4.$ See Wikipedia or your text. (Or @Alexjo's Answer.)

Then $E(X) = E(131 + 56Y) = 131 + 56E(Y).$