Question

Let $G_x(\cdot)$ be the distribution function of an arbitrary random variable symmetric about $0$ (zero) and $G_x^{\leftarrow}$ is the inverse function of $G_x$ then for $p \in (0, 1)$ value of $G_x^{\leftarrow}(p) + G_x^{\leftarrow}(1-p)$ is

• A. $1$
• B. $0$
• C. $2p$
• D. $p$

Hint:

For any distribution symmetric about a point $\mu$, the quantile function $Q(p)$ satisfies $Q(p) + Q(1-p) = 2\mu$. In this specific case, $\mu = 0$, so the sum is $0$.

Answer 1

The Correct Option is B

This problem explores the relationship between the symmetry of a distribution and its quantile function (the inverse of the CDF).

Step 1: \color{redDefining the Symmetry Relation
Since the random variable is symmetric about $0$, the distribution function $G_x$ satisfies:
$G_x(x) = 1 - G_x(-x)$
Let $G_x(x) = p$, where $p$ is a probability in the range $(0,1)$.

Step 2: \color{redExpressing x in terms of p
By definition of the inverse function (quantile function):
$x = G_x^{\leftarrow}(p)$

Step 3: \color{redAnalyzing the Complementary Probability
Substitute $G_x(x) = p$ into the symmetry equation:
$p = 1 - G_x(-x)$
Rearranging this gives:
$G_x(-x) = 1 - p$

Step 4: \color{redFinding the Inverse for (1-p)
Now, take the inverse $G_x^{\leftarrow}$ of the probability $(1-p)$:
$-x = G_x^{\leftarrow}(1 - p)$

Step 5: \color{redSumming the Quantiles
We now have expressions for both terms in the question:
$G_x^{\leftarrow}(p) = x$
$G_x^{\leftarrow}(1 - p) = -x$
Adding them together:
$G_x^{\leftarrow}(p) + G_x^{\leftarrow}(1 - p) = x + (-x) = 0$
Therefore, the sum is $0$.