Question

Let \(X\) be a random variable with support \(S=\{0,1,2,\ldots\}\) and
\[ P(X\geq k+1\,|\,X\geq k)=p, \quad k\in S,\ \ 0<p<1. \]
Then which of the following statements is correct?

• A. \(P(X\geq k+m\,|\,X\geq k)=P(X\geq m)\), for all \(m,k\in S\)
• B. \(E(X)>\text{Var}(X)\)
• C. \(P(X\leq x)=1-p^x\), for all \(x\in S\)
• D. \(P(X\leq k+m\,|\,X\geq k)=1-p^m\), for all \(k,m\in S\)

Hint:

The condition \(P(X\geq k+1|X\geq k)=p\) for all \(k\) forces \(P(X\geq k)=p^k\); this is the discrete memoryless property.

Answer 1

The Correct Option is A

Step 1: Set up the survival function.
Let \(S(k)=P(X\geq k)\). We are told \(S(k+1)/S(k)=p\) for every \(k\), giving \(S(k+1)=p\,S(k)\).

Step 2: Solve the recurrence.
Since \(S(0)=1\), \(S(k)=p^k\) for every \(k=0,1,2,\ldots\).

Step 3: Check option (A).
\[ P(X\geq k+m|X\geq k)=\frac{p^{k+m}}{p^k}=p^m=P(X\geq m). \] TRUE, this is the discrete memoryless property.

Step 4: Check option (B).
This is a geometric pmf with \(q=1-p\): \(E(X)=p/(1-p)\), \(\text{Var}(X)=p/(1-p)^2\). \(\text{Var}(X)/E(X)=1/(1-p)>1\), so \(\text{Var}(X)>E(X)\), opposite of (B). FALSE.

Step 5: Check option (C).
\[ P(X\leq x)=1-P(X\geq x+1)=1-p^{x+1}, \] not \(1-p^x\). FALSE.

Step 6: Check option (D).
\[ P(X\leq k+m|X\geq k)=1-p^{m+1}, \] not \(1-p^m\). FALSE.

Final Answer:
\[ \boxed{P(X\geq k+m|X\geq k)=P(X\geq m),\ \text{for all } m,k\in S} \]