Nilesh859
Answered

If [tex]\mathrm{F(x)}[/tex] is a continuous distribution function, then show that [tex]\mathrm{F(x)}[/tex] lies between [tex]0[/tex] & [tex]1[/tex].

Answer :

Step-by-step explanation:

To show that \( F(x) \), a cumulative distribution function (CDF), lies between 0 and 1 for all \( x \), we will utilize the properties of CDFs:

1. **Definition of Cumulative Distribution Function (CDF):**

\( F(x) \) is defined as \( F(x) = P(X \leq x) \), where \( X \) is a random variable. By definition, \( F(x) \) gives the probability that \( X \) takes on a value less than or equal to \( x \).

2. **Properties of CDF:**

- \( F(x) \geq 0 \) for all \( x \): This is because probabilities are non-negative.

- \( F(x) \leq 1 \) for all \( x \): This is because the probability of any event cannot exceed 1.

Let's formally prove these properties:

### Proof:

**1. \( F(x) \geq 0 \) for all \( x \):**

Since \( F(x) = P(X \leq x) \), and probabilities are non-negative, we have:

\[ F(x) = P(X \leq x) \geq 0 \]

Therefore, \( F(x) \geq 0 \) for all \( x \).

**2. \( F(x) \leq 1 \) for all \( x \):**

Again, since \( F(x) = P(X \leq x) \), and \( P(X \leq x) \) represents the cumulative probability up to \( x \), it cannot exceed 1:

\[ F(x) = P(X \leq x) \leq 1 \]

Therefore, \( F(x) \leq 1 \) for all \( x \).

### Conclusion:

From the above two properties, we conclude that \( F(x) \) indeed lies between 0 and 1 for any \( x \):

\[ 0 \leq F(x) \leq 1 \]

This completes the proof that \( F(x) \), being a cumulative distribution function (CDF), satisfies the condition of lying between 0 and 1 for all \( x \).

can you mark as brainliest

Other Questions