Question

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?

• A. \(E(X)=\displaystyle\sum_{n=1}^{\infty}(1-F(n-1))\)
• B. \(F(X)\) has discrete uniform distribution
• C. There exists at least one such random variable \(X\) such that \(X\) and \(\dfrac{1}{X}\) have the same distribution
• D. \(E(X^2)=\displaystyle\sum_{n=1}^{\infty}(2n-1)\big(1-F(n-1)\big)\)

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.

Answer 1

The Correct Option is A, D

Step 1: Recall the tail sum formula for the mean.
For a random variable taking values \(1,2,3,\ldots\), write \(X=\sum_{n=1}^{X}1\), which just counts the integers from 1 up to X. Swapping this into an infinite sum of indicators,
\[ X=\sum_{n=1}^{\infty}\mathbf{1}(X\geq n). \]
Taking expectation term by term,
\[ E(X)=\sum_{n=1}^{\infty}P(X\geq n)=\sum_{n=1}^{\infty}(1-F(n-1)), \]
since \(P(X\geq n)=1-P(X\leq n-1)=1-F(n-1)\). So (A) is TRUE.

Step 2: Check (B), F(X) uniform.
The rule that \(F(X)\sim\text{Uniform}(0,1)\), known as the probability integral transform, needs F to be continuous. For a discrete X, F only jumps at the support points, so F(X) takes just the countably many jump values with unequal probabilities. This is never a uniform distribution in general. So (B) is FALSE.

Step 3: Check (C), X and 1/X same distribution.
Full support on \(\{1,2,3,\ldots\}\) means \(P(X=2)>0\). But \(1/X\) only takes values in \(\{1,\tfrac12,\tfrac13,\ldots\}\), and 2 is never among them, so \(P(1/X=2)=0\) always. Since these two probabilities can never match, X and 1/X can never share a distribution. So (C) is FALSE.

Step 4: Check (D), the formula for E(X^2).
Since \(1+3+5+\cdots+(2X-1)=X^2\), the sum of the first X odd numbers, we can write
\[ X^2=\sum_{n=1}^{X}(2n-1)=\sum_{n=1}^{\infty}(2n-1)\mathbf{1}(X\geq n). \]
Taking expectation,
\[ E(X^2)=\sum_{n=1}^{\infty}(2n-1)P(X\geq n)=\sum_{n=1}^{\infty}(2n-1)(1-F(n-1)). \]
So (D) is TRUE.

Final Answer:
The tail sum identities for E(X) and E(X^2) both hold; F(X) is not uniform and X can never match the distribution of 1/X. \[ \boxed{\text{(A) and (D)}} \]