The problem presents a scenario of an infinite series of independent coin tosses where the probability of obtaining "heads" (pile) is $p \in ]0,1[$. Let’s go through the solution step-by-step:

Problem Analysis

1. We denote:

   • $p$: probability of obtaining "heads" (pile).
   • $q = 1 - p$: probability of obtaining "tails" (face).
   • $X$: the random variable representing the number of "tails" obtained before the first "heads".

2. The tasks are:

   • 1. Recognize the law of $Y = X + 1$.
   • 2. Deduce the law of $X$, its expected value, and its variance.

Solution

1. Recognizing the Law of $Y = X + 1$

• $X$ is defined as the number of "tails" (face) obtained before the first "heads" (pile).

• This setup is a typical scenario of a geometric distribution with parameter $p$, where $X$ counts the number of failures (tails) before the first success (heads).

• Therefore, $X \sim \text{Geometric}(p)$.

• For a geometric random variable $X \sim \text{Geometric}(p)$, the probability mass function is: $P(X = k) = (1 - p)^k \cdot p$ where $k$ represents the number of "tails" before the first "heads".

• Since $Y = X + 1$, $Y$ represents the position (or index) of the first "heads" in the sequence of tosses. This follows a shifted geometric distribution. Thus, $Y \sim \text{Shifted Geometric}(p)$.

2. Deduce the Law of $X$, Its Expectation, and Its Variance

• Since $X \sim \text{Geometric}(p)$, we know:

  • The expected value (mean) of $X$ is: $\mathbb{E}[X] = \frac{1 - p}{p}$
  • The variance of $X$ is: $\text{Var}(X) = \frac{1 - p}{p^2}$

• These results are based on standard properties of the geometric distribution.

Summary of the Results

• Law of $Y$: $Y \sim \text{Shifted Geometric}(p)$.
• Law of $X$: $X \sim \text{Geometric}(p)$.
• Expectation of $X$: $\mathbb{E}[X] = \frac{1 - p}{p}$.
• Variance of $X$: $\text{Var}(X) = \frac{1 - p}{p^2}$.

Would you like further details on any specific part of the solution, or do you have any questions?

Related Questions:

1. How does the geometric distribution apply in real-life scenarios?
2. What is the difference between a geometric and shifted geometric distribution?
3. How would the solution change if we were interested in the second "heads" instead?
4. Can you explain the derivation of the expectation and variance for the geometric distribution?
5. What would happen if $p$ was very close to 1? How does it affect the results?

Tip:

For probability distributions, remember that the geometric distribution is memoryless, meaning each trial's probability remains constant regardless of previous outcomes.