Problem Restatement:

Given that the cumulative distribution function (CDF) of a random variable $X$ is:

$F(x) = \frac{e^x}{1 + e^x} \quad \text{for} \quad -\infty < x < \infty,$

determine the probability density function (PDF) $f(x)$, and then specifically find:

1. $f(0)$
2. $f(1)$

Solution:

To find the PDF $f(x)$, we use the fact that the PDF is the derivative of the CDF. That is:

$f(x) = \frac{d}{dx} F(x)$

The given CDF is:

$F(x) = \frac{e^x}{1 + e^x}$

We need to differentiate this with respect to $x$. Applying the quotient rule:

$f(x) = \frac{d}{dx} \left( \frac{e^x}{1 + e^x} \right)$

The quotient rule is given by:

$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

For this case:

• $u = e^x$ and $v = 1 + e^x$
• $\frac{du}{dx} = e^x$
• $\frac{dv}{dx} = e^x$

Using the quotient rule:

$f(x) = \frac{(1 + e^x)(e^x) - e^x(e^x)}{(1 + e^x)^2}$

Simplifying the numerator:

$f(x) = \frac{e^x + e^{2x} - e^{2x}}{(1 + e^x)^2}$ $f(x) = \frac{e^x}{(1 + e^x)^2}$

Thus, the PDF is:

$f(x) = \frac{e^x}{(1 + e^x)^2}$

Part 1: Find $f(0)$

Substitute $x = 0$ into the PDF:

$f(0) = \frac{e^0}{(1 + e^0)^2} = \frac{1}{(1 + 1)^2} = \frac{1}{4}$

So, $f(0) = \frac{1}{4}$.

Part 2: Find $f(1)$

Substitute $x = 1$ into the PDF:

$f(1) = \frac{e^1}{(1 + e^1)^2} = \frac{e}{(1 + e)^2}$

So, $f(1) = \frac{e}{(1 + e)^2}$.

Let me know if you need further clarification or detailed steps! Here are some related questions to explore:

1. How do we interpret the PDF in terms of probabilities?
2. What is the CDF's behavior as $x \to -\infty$ and $x \to \infty$?
3. How would you find the mode of the distribution given this PDF?
4. Can the PDF be normalized, and why is this important?
5. What would the PDF look like if we transformed $X$ using a function, e.g., $Y = 2X + 1$?

Tip: Understanding the relationship between the CDF and PDF is crucial. The PDF describes the rate of change of the CDF, which gives insights into how probabilities are distributed.