Lorenzo Biasi and Michael Aichmüller

Exercise 1.

a.

$\sum_{k=1}^{\infty} P_X(2^{k-1}) = \sum_{k=1}^{\infty} P(k) = \sum_{k=1}^{\infty} (\frac{1}{2})^k = \frac{1}{1 - 1/2} -1 = 1$

b.

$\mu_X = \lim_{N\to\infty} \sum_{k=1}^{N} P_X(2^{k-1}) X(k) = \lim_{N\to\infty} \sum_{k=1}^{N} 2^{-k} 2^{k - 1} = \lim_{N\to\infty} \sum_{k=1}^{N} 2^{-1}$ This sum diverges for N that goes to infinity, for this we can say that the mean of $X$ is not defined.

Exercise 2.

$\mu = \int_{\infty}^{-\infty} x p(x) dx = \int_{-\infty}^{0} x p(x) dx + \int_{0}^{\infty} x p(x) dx = \int_{-\infty}^{0} -(-x) (p(-x)) dx + \int_{0}^{\infty} x p(x) dx$

Sobstituing $x = -u$ we obtain: $\mu = \int_{\infty}^{0} u p(u) dx + \int_{0}^{\infty} x p(x) dx = -\int_{0}^{\infty} u p(u) dx + \int_{0}^{\infty} x p(x) dx$

Being u a dummy variable it can be called however you want, also $x$. By doing this we can see that the two integrals calcel out and $\mu = 0$.

Exercise 3.

a.

$\Omega = \{(0, 0, 1), (1, 0, 0), (0, 1, 0), (1, 1, 1)\}$. Since the dice is fair we know the probability of of each element $\omega_i$ is 1/4. If we