Question

If $X$ has the $F$ distribution with $m, n$ degree of freedoms and let $Y=\frac{1}{X}$ then for $a>0$ $P[X\le a]+P[Y\le\frac{1}{a}]$ is equal to

• A. $0$
• B. $a$
• C. $1$
• D. $2a$

Hint:

The reciprocal property $1/F_{m,n} = F_{n,m}$ is a fundamental characteristic of the F-distribution. This question simplifies significantly once you realize $P(1/X \le 1/a)$ is just the complement of $P(X \le a)$.

Answer 1

The Correct Option is C

This problem tests the reciprocal property of the $F$-distribution and basic probability axioms.

Step 1: \color{redAnalyze the Reciprocal Property
If a random variable $X$ follows an $F$-distribution with $(m, n)$ degrees of freedom ($X \sim F_{m,n}$), then its reciprocal $Y = 1/X$ follows an $F$-distribution with swapped degrees of freedom ($Y \sim F_{n,m}$).

Step 2: \color{redEvaluate the Second Probability Term
We are given the term $P[Y \le \frac{1}{a}]$.
Substituting $Y = 1/X$:
$P[\frac{1}{X} \le \frac{1}{a}]$.
Since $X$ and $a$ are positive ($a > 0$), we can invert the inequality:
$P[X \ge a]$.

Step 3: \color{redCombine the Probabilities
Now, substitute this result back into the original expression:
$P[X \le a] + P[Y \le \frac{1}{a}] = P[X \le a] + P[X \ge a]$.

Step 4: \color{redApply Probability Axioms
For any continuous random variable $X$, $P(X \le a) + P(X > a) = 1$.
Since $P(X=a) = 0$ for continuous distributions, $P[X \le a] + P[X \ge a]$ is effectively the probability of the entire sample space.
$P[X \le a] + P[X \ge a] = 1$.
Therefore, the sum is equal to $1$.