# Expectation of Conditional Expression for Exponentially Distributed RV

• MHB
• user_01
In summary: This approach is valid for any exponentially distributed random variable with parameter $\lambda = 1$, where $g(x)$ is the function defined in the conversation.
user_01
Given an Exponentially Distributed Random Variable $X\sim \exp(1)$, I need to find $\mathbb{E}[P_v]$, where $P_v$ is given as:$$P_v= \left\{ \begin{array}{ll} a\left(\frac{b}{1+\exp\left(-\bar \mu\frac{P_s X}{r^\alpha}+\varphi\right)}-1\right), & \text{if}\ \frac{P_s X}{r^\alpha}\geq P_a,\\ 0, & \text{otherwise}. \end{array} \right.$$

**My Take:**

First, let's solve the equation for $P_v$. For that, let's assume g(x) to be:

$$g(x) = \frac{P^0}{\exp(\overline{\mu}P_{th} + \varphi)}\left( \frac{1+\exp(-\overline{\mu}P_{th} + \varphi)}{1 + \exp(-\overline{\mu}P_s x r^{-\alpha} + \varphi)} - 1\right),$$Then,

$$P_v = \begin{cases} g(x) & x \geq \frac{P_{th}}{P_s}r^\alpha\\ 0 & x < \frac{P_{th}}{P_s}r^\alpha \end{cases}$$Then, with the knowledge that the PDF for Exponentially distributed RV is $f(x) = e^{-\lambda x}$ (with $\lambda = 1$ for our case), we can find $\mathbb{E}[P_v]$.

$$\mathbb{E}[P_v]= \int_Q^\infty g(x)f(x)dx \ \ \ \ \ \ \ \ \ \ \ (1)$$
where $Q = \frac{P_{th} r^{\alpha}}{P_s}$.

Is this method correct or am I making any mistake?

**Answer:**Yes, this method is correct. To calculate $\mathbb{E}[P_v]$, you need to integrate $g(x)f(x)$ as you have done in (1).

## 1. What is an exponentially distributed random variable?

An exponentially distributed random variable is a continuous probability distribution that models the time between events that occur at a constant rate. It is often used in situations where the probability of an event occurring decreases as time goes on, such as radioactive decay or the waiting time between arrivals of customers at a store.

## 2. How is the expectation of a conditional expression calculated for an exponentially distributed random variable?

The expectation of a conditional expression for an exponentially distributed random variable is calculated by taking the integral of the conditional expression multiplied by the probability density function (PDF) of the random variable. The PDF for an exponential distribution is given by λe-λx, where λ is the rate parameter and x is the value of the random variable. The integral is then taken over the range of possible values for the random variable.

## 3. Can the expectation of a conditional expression for an exponentially distributed random variable be negative?

Yes, the expectation of a conditional expression for an exponentially distributed random variable can be negative. This is because the conditional expression can take on negative values, and when multiplied by the negative values of the PDF, it can result in a negative expectation.

## 4. What is the significance of the expectation of a conditional expression for an exponentially distributed random variable?

The expectation of a conditional expression for an exponentially distributed random variable is a measure of the average value of the conditional expression given the random variable. It can be used to make predictions about the behavior of the random variable and to compare different distributions.

## 5. How is the expectation of a conditional expression for an exponentially distributed random variable related to the mean of the distribution?

The expectation of a conditional expression for an exponentially distributed random variable is equal to the mean of the distribution when the conditional expression is equal to the random variable itself. In other words, if the conditional expression is simply the value of the random variable, then the expectation will be equal to the mean. However, in general, the expectation and mean may not be equal for an exponentially distributed random variable.

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