Let \(X\) be a discrete random variable with support \(S=\{1,2,3,\ldots\}\) such that \(E(X^2)<\infty\). Let \(F\) be the distribution function of \(X\). Then which of the following statements is/are correct?
Show Hint
Write X, and separately X squared, as a sum of indicators 1(X greater than or equal to n), then swap the order of summation with the expectation to reach the tail sum formulas.
Step 1: Expand E(X) directly and swap the order of summation.
By definition $E(X)=\sum_{k=1}^{\infty}kP(X=k)$. Write $k=\sum_{n=1}^{k}1$ and swap the order of the double sum:
\[ E(X)=\sum_{k=1}^{\infty}\sum_{n=1}^{k}P(X=k)=\sum_{n=1}^{\infty}\sum_{k=n}^{\infty}P(X=k)=\sum_{n=1}^{\infty}P(X\geq n). \]
Since $P(X\geq n)=1-F(n-1)$, this is exactly option (A), so (A) is TRUE.
Step 2: Test (B) with a concrete example.
Take X geometric with $P(X=k)=p(1-p)^{k-1}$. Then $F(k)=1-(1-p)^k$, so $F(X)=1-(1-p)^X$. As X ranges over the integers, $F(X)$ takes unequally spaced values with unequal probabilities, never uniform on $(0,1)$. So (B) fails in general.
Step 3: Test (C) using the support condition.
Since X must place positive probability on every integer $2,3,4,\ldots$, we need $P(1/X=2)=P(X=2)>0$. But $1/X=2$ needs $X=1/2$, which is impossible for an integer-valued X. So no valid X can share a distribution with its reciprocal, and (C) is FALSE.
Step 4: Expand E(X^2) the same way.
Write $k^2=\sum_{n=1}^{k}(2n-1)$, the sum of the first k odd numbers. Then
\[ E(X^2)=\sum_{k=1}^{\infty}\sum_{n=1}^{k}(2n-1)P(X=k)=\sum_{n=1}^{\infty}(2n-1)\sum_{k=n}^{\infty}P(X=k)=\sum_{n=1}^{\infty}(2n-1)(1-F(n-1)). \]
This matches (D) exactly, so (D) is TRUE.
Final Answer:
Swapping the order of summation proves both tail sum formulas; the uniform claim and the reciprocal claim both break down.
\[ \boxed{\text{(A) and (D)}} \]
