3.1.27. Let $X$ have a geometric distribution. Show that
$$
P(X \geq k+j \mid X \geq k)=P(X \geq j)
$$
where $k$ and $j$ are nonnegative integers. Note that we sometimes say in this situation that $X$ is memoryless.

Q: 3.1.27. Let $X$ have a geometric distribution. Show that \[ P(X \geq k+j \mid X \geq k)=P(X \geq j) \] where $k$ and $j$ are nonnegative integers. Note that we sometimes say in this situation that $X$ is memoryless.

∻Solution by Steps∻ ‖ step 1 ⋮ Let $ X $ be a random variable with a geometric distribution. The probability mass function is given by $ P(X = n) = (1-p)^{n-1}p $ for $ n = 1, 2, 3, \ldots $. We need to show that $ P(X \geq k+j \mid X \geq k) = P(X \geq j) $. ‖ ‖ step 2 ⋮ By the definition of conditional probability, we have: \[ P(X \geq k+j \mid X \geq k) = \frac{P(X \geq k+j \cap X \geq k)}{P(X \geq k)}. \] Since $ X \geq k+j $ implies $ X \geq k $, we can simplify this to: \[ P(X \geq k+j \mid X \geq k) = \frac{P(X \geq k+j)}{P(X \geq k)}. \] ‖ ‖ step 3 ⋮ The survival function for a geometric distribution is given by $ P(X \geq n) = (1-p)^{n-1} $. Therefore, we can express the probabilities as: \[ P(X \geq k+j) = (1-p)^{(k+j)-1} \quad \text{and} \quad P(X \geq k) = (1-p)^{k-1}. \] ‖ ‖ step 4 ⋮ Substituting these into our conditional probability expression gives: \[ P(X \geq k+j \mid X \geq k) = \frac{(1-p)^{(k+j)-1}}{(1-p)^{k-1}} = (1-p)^{j}. \] This is exactly $ P(X \geq j) $, since $ P(X \geq j) = (1-p)^{j-1} $. Thus, we have shown that: \[ P(X \geq k+j \mid X \geq k) = P(X \geq j). \] ‖ ∻Answer∻ ⚹ The statement $ P(X \geq k+j \mid X \geq k) = P(X \geq j) $ is proven for a geometric distribution, demonstrating its memoryless property. ⚹ ∻Key Concept∻ ⚹ The memoryless property of the geometric distribution states that the probability of future events is independent of past events. ⚹ ∻Explanation∻ ⚹ This property is shown by the equality $ P(X \geq k+j \mid X \geq k) = P(X \geq j) $, indicating that the distribution does not "remember" how long it has already been waiting. ⚹◊What is the probability mass function of a geometric distribution?⍭ Generate me a similar question◊

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∻Solution by Steps∻ ‖ step 1 ⋮ Let $ X $ be a random variable with a geometric distribution. The probability mass function is given by $ P(X = n) = (1-p)^{n-1}p $ for $ n = 1, 2, 3, \ldots $. We need to show that $ P(X \geq k+j \mid X \geq k) = P(X \geq j) $. ‖ ‖ step 2 ⋮ By the definition of conditional probability, we have: $$ P(X \geq k+j \mid X \geq k) = \frac{P(X \geq k+j \cap X \geq k)}{P(X \geq k)}. $$ Since $ X \geq k+j $ implies $ X \geq k $, we can simplify this to: $$ P(X \geq k+j \mid X \geq k) = \frac{P(X \geq k+j)}{P(X \geq k)}. $$ ‖ ‖ step 3 ⋮ The survival function for a geometric distribution is given by $ P(X \geq n) = (1-p)^{n-1} $. Therefore, we can express the probabilities as: $$ P(X \geq k+j) = (1-p)^{(k+j)-1} \quad \text{and} \quad P(X \geq k) = (1-p)^{k-1}. $$ ‖ ‖ step 4 ⋮ Substituting these into our conditional probability expression gives: $$ P(X \geq k+j \mid X \geq k) = \frac{(1-p)^{(k+j)-1}}{(1-p)^{k-1}} = (1-p)^{j}. $$ This is exactly $ P(X \geq j) $, since $ P(X \geq j) = (1-p)^{j-1} $. Thus, we have shown that: $$ P(X \geq k+j \mid X \geq k) = P(X \geq j). $$ ‖ ∻Answer∻ ⚹ The statement $ P(X \geq k+j \mid X \geq k) = P(X \geq j) $ is proven for a geometric distribution, demonstrating its memoryless property. ⚹ ∻Key Concept∻ ⚹ The memoryless property of the geometric distribution states that the probability of future events is independent of past events. ⚹ ∻Explanation∻ ⚹ This property is shown by the equality $ P(X \geq k+j \mid X \geq k) = P(X \geq j) $, indicating that the distribution does not "remember" how long it has already been waiting. ⚹◊What is the probability mass function of a geometric distribution?⍭ Generate me a similar question◊

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