The page you've uploaded provides an explanation of the probability mass function (PMF) and cumulative distribution function (CDF) for a geometric distribution. Here’s a breakdown of the content and proofs on the page.

1. Geometric Series Sum and PMF Validity

• The geometric distribution PMF is defined as $P(X = k) = p \cdot q^k$ for $k = 0, 1, 2, \ldots$, where $p$ is the probability of the first success and $q = 1 - p$ is the probability of failure on each trial.
• A probability mass function (PMF) must sum to 1 over all possible values of $X$. The sum for the geometric PMF is: $\sum_{k=0}^{\infty} p \cdot q^k = p \sum_{k=0}^{\infty} q^k = p \cdot \frac{1}{1 - q} = 1$
• This demonstrates that the PMF is valid since the series converges to 1, confirming that it represents a proper probability distribution.

2. Theorem 4.3.3: Geometric CDF

• The cumulative distribution function (CDF) for $X \sim \text{Geom}(p)$ is given as: 1 - q^{\lfloor x \rfloor + 1}, & \text{if } x \geq 0; \\ 0, & \text{if } x < 0. \end{cases}$$
• Here, $q = 1 - p$ and $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.

3. Proof Outline for the Geometric CDF

• Step 1: Consider the case where $x \geq 0$. The goal is to find $F(x) = P(X \leq x)$, which is the probability that the number of trials needed to get the first success is less than or equal to $x$.
• Step 2: We rewrite $F(x)$ as the sum of probabilities up to $x$: $F(n) = \sum_{k=0}^{n} P(X = k) = \sum_{k=0}^{n} p \cdot q^k = p \cdot \frac{1 - q^{n+1}}{1 - q} = 1 - q^{n+1}.$
• This uses the formula for the sum of a geometric series, yielding the expression $F(x) = 1 - q^{\lfloor x \rfloor + 1}$ for non-negative $x$.

4. Proof for $x < 0$

• When $x < 0$, $F(x) = P(X \leq x) = 0$ because $X$ (number of trials until the first success) cannot be negative.

5. Verification Using Complementary Probabilities

• An alternative way to think about this is by considering the probability of failing the first $n+1$ trials, which is $q^{n+1}$. Thus: $F(n) = 1 - P(X > n) = 1 - q^{n+1}.$
• Since $X$ only takes integer values, we conclude $F(x) = P(X \leq \lfloor x \rfloor)$, and this confirms the form of $F(x)$ as derived in the theorem.

This section establishes the form of the CDF for a geometric distribution, ensuring it is correctly defined.

Would you like more details on a specific part of this proof, or do you have any other questions?

Here are some related questions to consider:

1. How does the geometric distribution differ from other discrete distributions like the binomial distribution?
2. What is the significance of the geometric series sum formula in probability theory?
3. Can you derive the PMF for the number of failures before the first success in a geometric distribution?
4. How would the CDF change if we considered the distribution of the number of trials including the success?
5. Why is the geometric distribution memoryless, and how does this property relate to its CDF?

Tip: The geometric distribution is often used to model the number of trials until the first success in repeated, independent Bernoulli trials.