The problem provided describes a geometric random variable $X$, representing the number of trials before the first success in a Bernoulli experiment, where the probability of success is $p = \frac{1}{2}$.

The question asks to calculate $\mathbb{E}\left(\frac{1}{2^X}\right)$, i.e., the expected value of $\frac{1}{2^X}$.

Solution:

1. Geometric distribution properties: The probability mass function (PMF) of a geometric random variable $X$ is given by: $P(X = k) = (1 - p)^{k-1} p \quad \text{for} \quad k = 1, 2, 3, \ldots$ In this case, since $p = \frac{1}{2}$, the PMF becomes: $P(X = k) = \left( \frac{1}{2} \right)^{k-1} \frac{1}{2} = \frac{1}{2^k} \quad \text{for} \quad k = 1, 2, 3, \ldots$

2. Expected value calculation: We need to compute $\mathbb{E}\left( \frac{1}{2^X} \right)$, which is: $\mathbb{E}\left( \frac{1}{2^X} \right) = \sum_{k=1}^{\infty} \frac{1}{2^k} P(X = k)$ Substituting $P(X = k) = \frac{1}{2^k}$: $\mathbb{E}\left( \frac{1}{2^X} \right) = \sum_{k=1}^{\infty} \frac{1}{2^k} \cdot \frac{1}{2^k} = \sum_{k=1}^{\infty} \frac{1}{4^k}$

3. Sum of a geometric series: The series $\sum_{k=1}^{\infty} \frac{1}{4^k}$ is a geometric series with the first term $a = \frac{1}{4}$ and the common ratio $r = \frac{1}{4}$. The sum of an infinite geometric series is given by: $S = \frac{a}{1 - r}$ Substituting $a = \frac{1}{4}$ and $r = \frac{1}{4}$: $S = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$

Thus, the expected value $\mathbb{E}\left( \frac{1}{2^X} \right) = \frac{1}{3}$.

Would you like further details or have any questions? Here are some related topics you could explore:

1. How is the geometric distribution used in modeling waiting times?
2. How does the sum of an infinite geometric series converge?
3. What is the difference between geometric and binomial distributions?
4. How would the result change if the probability $p$ were different?
5. How is the expectation formula for geometric random variables derived?

Tip: Understanding geometric series is crucial for solving many probability problems involving infinite sums.