GATE CS 2024 Set 2 Q34 — Probability: Random Variables and Distributions

We are given that $x$ and $y$ are random variables taking values in the interval $[0, 1]$. Thus, we have:

Let us analyze the relationship between $z = xy$ and $x$:

Since $y \le 1$ and $x \ge 0$, multiplying both sides of $y \le 1$ by $x$ preserves the inequality:

Using the monotonicity property of expectation, taking the expectation on both sides gives:

This inequality always holds. Thus, Option D is always true.

Why other options are incorrect:

Let $x$ be a random variable uniformly distributed on $[0, 1]$, so $\bar{x} = E[x] = \frac{1}{2}$.

• If $x$ and $y$ are positively correlated (e.g., $y = x$):
  Here, $\bar{y} = \frac{1}{2}$ and $\bar{x}\bar{y} = \frac{1}{4}$.
  $\bar{z} = E[x^2] = \int_0^1 x^2 dx = \frac{1}{3}$.
  Since $\frac{1}{3} > \frac{1}{4}$, we have $\bar{z} > \bar{x}\bar{y}$. This disproves Option B.


• If $x$ and $y$ are negatively correlated (e.g., $y = 1 - x$):
  Here, $\bar{y} = 1 - \frac{1}{2} = \frac{1}{2}$ and $\bar{x}\bar{y} = \frac{1}{4}$.
  $\bar{z} = E[x(1-x)] = E[x] - E[x^2] = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$.
  Since $\frac{1}{6} < \frac{1}{4}$, we have $\bar{z} < \bar{x}\bar{y}$. This disproves Option C.

Since $\bar{z}$ can be either greater than, less than, or equal to $\bar{x}\bar{y}$ depending on the joint distribution of $x$ and $y$, Option A is also false in general.