You win $2^n$ dollars if you flip a tail on the $n^{\text{th}}$ flip, aka $P(X=2^n)=.5^n$ for $n$ from $1$ to infinity.
- Let $Y(x)=x^{\frac{1}{k}}$ for $k > 0$. What is $E[Y(X)]$ and for which values of $k$ does this converge? What property does $Y$ have for these values of $k$?
How would I solve this since $Y(x)$ is not a linear function? I know $E(Y)=aE(X)+b$ if $Y=aX+b$. Could this apply to this function as well?
Or would I do $E[Y(X)]=\sum{Y(x)}{p_X(x)}$?
- $Y(X)$ has finite expectation if $k = 2$. Find variance.
This means that the random variable $Y$ is discrete since there's finite expectation. so does this mean
$$\mathrm{var}(Y(X))=E[Y(X)^2]-E[Y(X)]^2=E[0.5^2]-0.5^2 $$
- If $Y(x) = \log_2(x)$ how does expectation and variance differ? Is the value larger or smaller?
I'm trying to learn this material ahead of time but I am not fully grasping these concepts. I appreciate any help
Let us recall the definition of the expectation (for discrete random variables): If $X$ can have only values $x_{1}, x_{2}, \cdots$, then its expectation is defined as
$$ \Bbb{E}X = \sum_{i} x_{i} \, \Bbb{P}(X = x_{i}). $$
So if $f(x)$ is a function (in your case, the notation $Y$ is used instead), then working with this definition a little bit yields
$$ \Bbb{E}f(X) = \sum_{i} f(x_{i}) \Bbb{P}(X = x_{i}) \tag{*}$$
This is actually what you have point out in your own question, with a slightly different notation. Now with $f(x) = x^{1/k}$, we can answer to your first question by making the following calculation:
$$ \Bbb{E}X^{1/k} = \sum_{i=1}^{\infty} (2^{i})^{1/k} 2^{-i} = \sum_{i=1}^{\infty} (2^{(1/k)-1})^{i}.$$
This converges if and only if $2^{(1/k)-1} < 1$, or equivalently, $k > 1$.
Now can you demonstrate how to solve the other problems?
Addendum. Here I show you how $\text{(*)}$ follows solely from the definition. You need not know how it comes, but rather it is sufficient to know that this is indeed possible.
Proof of $\text{(*)}$. Let us enumerate the image of $\{x_{1}, x_{2}, \cdots \}$ under $f$ as $\{ y_{1}, y_{2}, \cdots \}$. Then for each $y_{i}$, there exists a set of indices $J_{i}$ such that $f(x) = y_{i}$ if and only if $x = x_{j}$ for some $j \in J_{i}$. Then \begin{align*} \Bbb{E}f(X) &= \sum_{i} y_{i} \Bbb{P}(f(X) = y_{i}) \\ &= \sum_{i} y_{i} \Bbb{P}(X = x_{j} \text{ for some } j \in J_{i}) = \sum_{i} y_{i} \sum_{j \in J_{i}} \Bbb{P}(X = x_{j}) \\ &= \sum_{i} \sum_{j \in J_{i}} f(x_{j}) \Bbb{P}(X = x_{j}) = \sum_{j} f(x_{j}) \Bbb{P}(X = x_{j}). \end{align*}