$P[X \leq a] + P\left[Y \leq \frac{1}{a}\right]$ is equal to
F Distribution Probability Transformation Explained
Let $X$ be a random variable following the F distribution with $m$ and $n$ degrees of freedom. We denote this as $X \sim F(m, n)$.
We are given another random variable $Y = \frac{1}{X}$. We need to find the value of the expression $P[X \leq a] + P\left[Y \leq \frac{1}{a}\right]$ for $a > 0$.
Understanding the Variable Transformation
A key property of the F distribution is related to the reciprocal of the random variable. If $X \sim F(m, n)$, then the random variable $\frac{1}{X}$ follows the F distribution with $n$ and $m$ degrees of freedom. That is:
$ \frac{1}{X} \sim F(n, m) $
Since $Y = \frac{1}{X}$, we have $Y \sim F(n, m)$.
Calculating the Probability Expression
The expression we need to evaluate is $P[X \leq a] + P\left[Y \leq \frac{1}{a}\right]$.
Substitute $Y$ with $\frac{1}{X}$: $ P[X \leq a] + P\left[\frac{1}{X} \leq \frac{1}{a}\right] $
Consider the term $P\left[\frac{1}{X} \leq \frac{1}{a}\right]$. Since $X$ represents a value from an F distribution, $X$ is always positive. Also, we are given $a > 0$. Therefore, the inequality $\frac{1}{X} \leq \frac{1}{a}$ is equivalent to $X \geq a$.
So, the expression becomes: $ P[X \leq a] + P[X \geq a] $
For any continuous random variable, the sum of the probabilities of being less than or equal to a value and greater than or equal to that same value covers the entire probability space. Since $X$ is a continuous random variable, $P[X = a] = 0$. Thus, $P[X \geq a] = 1 - P[X < a] = 1 - P[X \leq a]$.
Substituting this back into the expression: $ P[X \leq a] + (1 - P[X \leq a]) $
Simplifying this gives:
$ 1 $Therefore, $P[X \leq a] + P\left[Y \leq \frac{1}{a}\right] = 1$.
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- Who among the following was not a member of the constituent assembly:-
