# Verifying Solution for Exponentially Distributed Random Vars.

• MHB
• user_01
In summary, we have two i.i.d. random variables $X$ and $Y$ with $X \sim \exp(1)$ and $Y \sim \exp(1)$. We are looking for the probability $\Phi$ which is defined as $\mathbb{P}\left[P_v \geq A + \frac{B}{Y}\right]$. However, the analytical solution does not match with the simulation and we are looking for someone to rectify our mistake. The solution is given by $\Phi = \exp\left(-\frac{D+B}{ab}\left(1 + \frac{e^\varphi}{1+c}\right)\right)\exp\left(-\frac{P_a user_01 Given two i.i.d. random variables$X,Y$, such that$X\sim \exp(1), Y \sim \exp(1)$. I am looking for the probability$\Phi$. However, the analytical solution that I have got does not match with my simulation. I am presenting it here with the hope that someone with rectifies my mistake. : $$\Phi =\mathbb{P}\left[P_v \geq A + \frac{B}{Y}\right]$$ $$P_v= \left\{ \begin{array}{ll} a\left(\frac{b}{1+\exp\left(-\bar \mu\frac{P_s X}{r^\alpha}+\varphi\right)}-1\right), & \text{if}\ \frac{P_s X}{r^\alpha}\geq P_a,\\ 0, & \text{otherwise}. \end{array} \right.$$ --- **My solution** \begin{multline} \Phi = \mathbb{P}\left[ a\left(\frac{b}{1+\exp\left(-\bar \mu\frac{P_s X}{r^\alpha}+\varphi\right)}-1\right) \geq A + \frac{B}{Y}\right]\mathbb{P}\left[X \geq \frac{P_ar^\alpha}{P_s}\right] + {\mathbb{P}\left[0 \geq A + \frac{B}{Y}\right] \mathbb{P}\left[X \geq \frac{P_ar^\alpha}{P_s}\right]} \end{multline} Given that$\mathbb{P}[0>A+\frac{B}{Y}] = 0$$$\Phi = \mathbb{P}\left[a\left(\frac{b}{1+\exp\left(-\bar \mu\frac{P_s X}{r^\alpha}+\varphi\right)}-1\right) \geq A + \frac{B}{Y}\right] \exp\left(-\frac{P_ar^\alpha}{P_s}\right)$$ consider$D = A + a$,$c = \frac{\bar{\mu}P_s}{r^{\alpha}}$$$\Phi = \mathbb{P} \left[Y \geq \frac{1}{ab}\mathbb{E}_Y[DY + B]. \mathbb{E}_X [1+e^\varphi e^{-c X}] \right]$$ $$\Phi = \exp\left(-\frac{D+B}{ab}\left(1 + \frac{e^\varphi}{1+c}\right)\right)\exp\left(-\frac{P_a r^\alpha}{P_s}\right)$$ #### Attachments • pic_1.PNG 6.7 KB · Views: 55 I did not include the whole intermediate steps in the above solution which may cause confusion. Hence those steps are now presented below. --- **My solution** \begin{multline} \Phi = \mathbb{P}\left[ a\left(\frac{b}{1+\exp\left(-\bar \mu\frac{P_s X}{r^\alpha}+\varphi\right)}-1\right) \geq A + \frac{B}{Y}\right]\mathbb{P}\left[X \geq \frac{P_ar^\alpha}{P_s}\right] + {\mathbb{P}\left[0 \geq A + \frac{B}{Y}\right] \mathbb{P}\left[X \geq \frac{P_ar^\alpha}{P_s}\right]} \end{multline} Given that$\mathbb{P}[0>A+\frac{B}{Y}] = 0$$$\Phi = \mathbb{P}\left[a\left(\frac{b}{1+\exp\left(-\bar \mu\frac{P_s X}{r^\alpha}+\varphi\right)}-1\right) \geq A + \frac{B}{Y}\right] \exp\left(-\frac{P_ar^\alpha}{P_s}\right)$$ Now consider only the first part of the expression on the right side of the equation, and letting$c = \frac{\bar{\mu}P_s}{r^{\alpha}}$. $$\mathbb{P}\left[a\left(\frac{b}{1+\exp\left(-\bar \mu\frac{P_s X}{r^\alpha}+\varphi\right)}-1\right) \geq A + \frac{B}{Y}\right] = \mathbb{P}\left[\frac{ab}{1+\exp\left(-c X+\varphi\right)}-a \geq A + \frac{B}{Y}\right]$$ consider$D = A + a\$, and with mathematical manipulations, we get:

$$\Phi = \mathbb{P} \left[Y \geq \frac{1}{ab}\mathbb{E}_Y[DY + B]. \mathbb{E}_X [1+e^\varphi e^{-c X}] \right]$$

$$\Phi = \exp\left(-\frac{D+B}{ab}\left(1 + \frac{e^\varphi}{1+c}\right)\right)\exp\left(-\frac{P_a r^\alpha}{P_s}\right)$$

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

An exponentially distributed random variable is a continuous probability distribution that is commonly used to model the time between events in a Poisson process. It is characterized by a single parameter, λ, which represents the rate at which events occur.

## 2. How do you verify a solution for an exponentially distributed random variable?

To verify a solution for an exponentially distributed random variable, you can use the cumulative distribution function (CDF). The CDF for an exponentially distributed random variable is F(x) = 1 - e^(-λx). You can plug in the values for λ and x to calculate the probability of a random variable being less than or equal to x.

## 3. What is the mean of an exponentially distributed random variable?

The mean of an exponentially distributed random variable is equal to 1/λ. This means that on average, the time between events in a Poisson process will be 1/λ units of time.

## 4. How do you interpret the parameter λ in an exponentially distributed random variable?

The parameter λ represents the rate at which events occur in a Poisson process. This means that for every unit of time, on average, λ events will occur. It also affects the shape of the distribution, with larger values of λ resulting in a steeper curve and smaller values resulting in a flatter curve.

## 5. Can an exponentially distributed random variable take on negative values?

No, an exponentially distributed random variable cannot take on negative values. It is a continuous distribution that is only defined for positive values. This is because the time between events in a Poisson process cannot be negative.

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