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: Identify the distribution from the given condition.
The constant one-step survival property $P(X\geq k+1)=p\,P(X\geq k)$ characterizes the geometric distribution.

Step 2: Derive the pmf from first principles.
$P(X\geq k)=p^k$, so $P(X=k)=p^k(1-p)$, $k=0,1,2,\ldots$

Step 3: Use the known memoryless property.
\[ P(X\geq k+m \mid X\geq k) = \frac{p^{k+m}}{p^k}=p^m=P(X\geq m). \] TRUE, matching (A).

Step 4: Test (B) with the geometric formulas.
$\text{Var}(X)/E(X)=1/(1-p)>1$, so variance always exceeds mean. (B) FALSE.

Step 5: Test (C) and (D) with the exact cdf.
$P(X\leq x)=1-p^{x+1}$, not $1-p^x$; similarly $P(X\leq k+m\mid X\geq k)=1-p^{m+1}$, not $1-p^m$. Both FALSE.

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