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Expected value of a random variable for a continuous random variable

I’ve been having a lot of problems with this exercise that I need to solve to fulfill my practicum.
I have to calculate the expected value of a random variable X given a random variable Y.
$$E(X) = E(X|Y=y)p(Y=y) + \int_{y_{\rm{min}}}^{y_{\rm{max}}} p(X=x|Y=y) dx = y \cdot 10 + \int_{y_{\rm{min}}}^{y_{\rm{max}}} 0.1\cdot (x)dx = y\cdot 11.5$$
The problem is that my prof doesn’t like the second part at all and has told me that this is wrong because the difference is made between the two values and should be considered as a lower/upper bound.
Should I ignore this part of the problem then?

A:

This is an exercise about induction.
In the first part of your calculation, you have $E(X|Y=y) = y \cdot 10$ for all $y \in [y_{\rm min}, y_{\rm max}]$, hence $E(X | Y \in [y_{\rm min}, y_{\rm max}]) = \int_{y_{\rm min}}^{y_{\rm max}} y \cdot 10 \, \mathrm dy = y_{\rm max} \cdot 10 – y_{\rm min} \cdot 10 = 10y_{\rm max} – 10y_{\rm min}$.
But the expected value of the random variable

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How to find the asymptotic growth rate of the following sequence?

I am attempting to determine how to find the asymptotic growth rate of the sequence and I am not sure how to do so given the information I have been given.
Using a text in a lecture notes:
$x_0 = 0.5$, $x_1 = 0.7$, $x_2 = 0.667$, $x_3 = 0.632$, $x_4 = 0.611$ and so on.
I have noticed that $x_n = 1 – \frac{1}{2} \cdot \left( 1 – x_{n-1}\right)$ for some $n>0$.
I understand this can represent a binary search with $n$ and $x_n$ as the search term, but $x_n – x_{n-1}$ is not constant so it’s difficult to see how to take the asymptotic growth rate of the above sequence.
The same text gives the following for the sequence $x_0=1$, $x_n = 2^{n-1}$ that has the asymptotic growth rate $log_2(x_n) = \theta(log_2 n)$.
Could someone help with finding the asymptotic growth rate of the sequence above?

A:

This is not the answer you may be looking for, but it is what you have been asked to calculate.
It is true that $x_n – x_{n-1} = 1/2^{n-1}$ for each $n$. But that is not even relevant. If $x$ is

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