This problem aims to familiarize us with the **probability** **density functions.** The concepts required to solve this problem are **continuous random variables** and** probability distributions,** which include **exponential distribution** and** densities** of random variables.

A **probability density function** or **PDF** is used in probability theory to describe the **probability** of a random variable staying within a particular **range** of values. These types of functions describe the **probability** density function of normal distribution and how there exists **mean** and **deviation.**

The **cumulative distribution function** or** CDF** of random $x$ is another way to represent the distribution of **random variable,** defined as:

\[ F_X (x) = P(X \geq x),\forall x\in\mathbb{R}\]

Whereas a **continuous random variable** has an exponential distribution having $\lambda > 0$ if the **density** of the function is:

\[f(x) = \lambda e − \lambda x \space\space\space if \space x \geq 0\]

## Expert Answer

Let’s first compute the **exponential distribution** of $x$:

\[ P(X > 1) = \int e^{-x} dx = e^{-x} \]

\[ F_x = 1 – P(X > 1) = 1 – e^{-x} \]

We are going to use this **approach** to find the **exponential distribution** of our function:

\[ Y = \ln X \]

Since **exponentials** are **memoryless,** we can write:

\[ F_Y (y) = P(Y \leq y) \]

**Plugging** in the value of $Y$:

\[ F_Y (y) = P(\ln X \leq y) \]

As **exponential** is the inverse of the **log,** we can get ride it by:

\[ F_Y (y) = P(X \leq e^y) \]

\[ F_Y (y) = F_X (e^y) \]

Then,

\[ F_x (e^y) = 1 – P(X > e^y) = 1 – e^{-e^y} \]

Now we are going to calculate the **probability distribution function,** which is the derivative of the **cumulative distribution function** $F(x)$:

\[ f(x) = \dfrac{d}{dx} F(x) \]

**Substituting** the values gives us:

\[ f_Y (y) = \dfrac{d}{dy} F_Y (y) \]

\[ f_Y (y) = \dfrac{d}{dy} F_X (e^y) \dfrac{d}{dy} \]

\[ f_Y (y) = \dfrac{d}{dy} \left [1 – e^{-e^y} \right ] \]

\[ f_Y (y) = -(-e^y) (e^{-e^y}) \]

\[ f_Y (y) = e^y e^{-e^y} \]

## Numerical Result

The **probability distribution function** is:

\[ f_Y (y) = e^y e^{-e^y} \]

## Example

Let $X$ be a **discrete random** variable handling **positive** value integers. **Suppose** that $P(X = k) \geq P(X = k + 1) \forall$ **positive** integer $k$. Prove that for any positive integer $k$,

\[ P(X = k) \geq \dfrac{2E [X] }{k^2} \]

Since $P(X = I) \geq 0$, it can be said that for any $k \in \mathbb{N}$,

\[ E [X] = \sum_{i=1}^{\infty} iP(X = i) \geq \sum_{i=1}^{k} iP(X = i) \]

Moreover,

\[ P(X = k) \geq P(X = k + 1) \forall k \in \mathbb{N} \]

We have,

\[ P(X = k) \geq P(X = i) \forall i \geq k \]

**F****inally,**

\[ \sum_{i=1}^k iP(X = i) \geq \sum_{i=1}^k iP(X = k) \]

\[ \dfrac{k(k + 1)}{2} P(X = k) \]

\[ \geq \dfrac{k^2}{2} P(X = k) \]

**Hence,** we can say that,

\[ E [X] \geq k^2 P(X = k)/2 \]

**Proved!**