$Y,N$ are a random variables such that $F(y)=y^n, y\in(0,1] , p_N(n)=\frac{5}{6^{(n+1)}}.$ Find $F_Y(\frac{3}{4}),E[Y]$

$Y,N$ are a random variables such that $F(y)=y^n, y\in(0,1] , p_N(n)=\frac{5}{6^{(n+1)}}.$

$n=\{0,1,2...\}$

Find $F_Y(\frac{3}{4}),E[Y]$

My try :

$F_Y(\frac{3}{4})=(\frac{3}{4})^{\sum_{n=0}^\infty \frac{5}{6^{(n+1)}}}$

$\sum_{n=0}^\infty= \frac{5}{6^{(n+1)}}=1$

$F_Y(\frac{3}{4})=(\frac{3}{4})^{1}$

But the answer is $F(\frac{3}{4})=\frac{20}{21}.$

$f_y=F'_y=ny^{(n-1)}$

$E[Y]=\int_0^1y\cdot ny^{(n-1)}dy=\int_0^1ny^ndy=\frac{n}{n+1}$

The answer is $E[Y]=1+5ln(\frac{5}{6}).$

I missed something and I don't know what, help please.

Thank you !

What you are given is not $F_Y(y)$ but it is a conditional CDF

$$F_{Y|N=n}(y|n)=y^n$$

thus what you have to do is to get

$$f_{YN}=p_N\times f_{Y|N}$$

and then summing w.r.t. $N$ (which is a geometric rv)

Same story for the expectation. What you found is $E[Y|N=n]$. To get $E(Y)$ you have to calculate

$$E(Y)=\frac{5}{6}\sum_{n=0}^{\infty}\frac{n}{n+1}\left( \frac{1}{6}\right)^n$$

reading well the exercise, there is something that is not so clear to me: if $F_{Y|N=n}=y^n$, this is the CDF of the maximum of n iid uniform rv's on the unit interval...thus the support of $N$ cannot be $\{0;1;2;\dots;n\}$ but actually $\{1;2;\dots;n\}$

If this is right, $Y$ density becomes

$$f_Y(y)=\frac{5}{6}\cdot\frac{6}{6-y}\underbrace{\sum_{n=1}^{\infty}n\left(\frac{y}{6} \right)^{n-1}\cdot\frac{6-y}{6}}_{=\frac{6}{6-y}}=\frac{30}{(6-y)^2}\cdot\mathbb{1}_{(0;1]}(y)$$

with expectation

$$E(Y)=\int_0^{1}yf(y)dy=6\left(1+5\ln\frac{5}{6}\right)$$

and

$$F_Y(0.75)=\int_0^{3/4}f(y)dy=\frac{5}{7}$$

...but these results do not match...