Let Xi be an absolutely continuous random variable having the probability density function$f_i(x)=\left\{\begin{array}{cl} i e^{-i x}, & \text { if } x \geq 0 \ 0, & \text { if } x

Question

Let Xi be an absolutely continuous random variable having the probability density function$f_i(x)=\left\{\begin{array}{cl} i e^{-i x}, & \text { if } x \geq 0 \ 0, & \text { if } x<0 \end{array}, \right.$i = 1, 2 .Consider a series system comprising of independent components having random lifetimes described by random variables X1 and X2. Let X denote the lifetime of the series system. Then which of the following statements are true?

Prepp Team · Accepted Answer

Series System Lifetime and Probability Distribution

The question describes a series system composed of two independent components, $X_1$ and $X_2$, with given probability density functions (PDFs). For a series system, the system fails if any one of its components fails. Therefore, the lifetime of the series system, denoted by $X$, is the minimum of the lifetimes of the individual components.

The PDFs for $X_1$ and $X_2$ are given as:

$f_1(x) = 1 e^{-1x}$ for $x \geq 0$, and $0$ otherwise. This is the PDF of an exponential distribution with rate parameter $\lambda_1 = 1$.
$f_2(x) = 2 e^{-2x}$ for $x \geq 0$, and $0$ otherwise. This is the PDF of an exponential distribution with rate parameter $\lambda_2 = 2$.

For an exponential distribution with rate $\lambda$, the probability $P(X > x)$, also known as the survival function, is given by $e^{-\lambda x}$ for $x \geq 0$.

For $X_1$, $P(X_1 > x) = e^{-1x} = e^{-x}$.
For $X_2$, $P(X_2 > x) = e^{-2x}$.

Lifetime Distribution of the Series System

The lifetime of the series system is $X = \min(X_1, X_2)$. Since $X_1$ and $X_2$ are independent, the survival function of $X$ is:

$P(X > x) = P(\min(X_1, X_2) > x)$

This means both $X_1$ and $X_2$ must be greater than $x$:

$P(X > x) = P(X_1 > x \text{ and } X_2 > x)$

Due to independence:

$P(X > x) = P(X_1 > x) P(X_2 > x)$

Substituting the survival functions of $X_1$ and $X_2$:

$P(X > x) = e^{-x} \times e^{-2x} = e^{-(x+2x)} = e^{-3x}$, for $x \geq 0$.

The survival function $P(X > x) = e^{-3x}$ for $x \geq 0$ corresponds to an exponential distribution with rate parameter $\lambda = 3$. Thus, $X \sim \text{Exponential}(3)$.

Evaluating the Statements

Now let's evaluate each given statement based on the fact that $X$ follows an exponential distribution with rate 3.

Statement 1: $P(X > 4) = P(X > 1) P(X > 2)$

Using $P(X > x) = e^{-3x}$:

Left side: $P(X > 4) = e^{-3 \times 4} = e^{-12}$.
Right side: $P(X > 1) = e^{-3 \times 1} = e^{-3}$.
$P(X > 2) = e^{-3 \times 2} = e^{-6}$.

The right side is $P(X > 1) P(X > 2) = e^{-3} \times e^{-6} = e^{-3-6} = e^{-9}$.

Comparing both sides: $e^{-12} \neq e^{-9}$. Therefore, Statement 1 is false.

Statement 2: $P(X > 4| X > 2) = P(X > 2)$

This statement involves conditional probability. We can use the property of exponential distributions that they are memoryless. The memoryless property states that for an exponentially distributed random variable $X$, $P(X > t+s | X > s) = P(X > t)$ for any $t, s \geq 0$.

In this statement, we have $s = 2$ and $t = 4 - 2 = 2$. So the statement becomes $P(X > 2+2 | X > 2) = P(X > 2)$, which is exactly the memoryless property $P(X > t+s | X > s) = P(X > t)$ with $t=2$.

Alternatively, using the conditional probability formula $P(A|B) = P(A \cap B) / P(B)$:

$P(X > 4| X > 2) = \frac{P(X > 4 \text{ and } X > 2)}{P(X > 2)}$

Since $X > 4$ implies $X > 2$, the event $X > 4 \text{ and } X > 2$ is simply $X > 4$.

$P(X > 4| X > 2) = \frac{P(X > 4)}{P(X > 2)}$

Using $P(X > x) = e^{-3x}$:

$P(X > 4| X > 2) = \frac{e^{-3 \times 4}}{e^{-3 \times 2}} = \frac{e^{-12}}{e^{-6}} = e^{-12 - (-6)} = e^{-6}$.

Now, let's calculate $P(X > 2)$:

$P(X > 2) = e^{-3 \times 2} = e^{-6}$.

Comparing both sides: $e^{-6} = e^{-6}$. Therefore, Statement 2 is true.

Statement 3: $E(X)=\frac{1}{3}$

For an exponentially distributed random variable with rate $\lambda$, the expected value (mean) is given by $E(X) = 1/\lambda$.

Since $X \sim \text{Exponential}(3)$, the rate parameter is $\lambda = 3$.

$E(X) = \frac{1}{3}$.

Therefore, Statement 3 is true.

Statement 4: $6 X \sim \chi_3^2$

We know that if a random variable $Y$ follows an exponential distribution with rate $\lambda$, i.e., $Y \sim \text{Exponential}(\lambda)$, then $2\lambda Y$ follows a chi-squared distribution with 2 degrees of freedom, i.e., $2\lambda Y \sim \chi^2(2)$.

In our case, $X \sim \text{Exponential}(3)$. So, $\lambda = 3$.

According to the property, $2 \times 3 X = 6X$ follows a chi-squared distribution with 2 degrees of freedom, i.e., $6X \sim \chi^2(2)$.

The statement claims that $6X \sim \chi_3^2$, which is a chi-squared distribution with 3 degrees of freedom.

Since $6X \sim \chi^2(2)$ and not $\chi^2(3)$, Statement 4 is false.

Based on the analysis, Statement 2 and Statement 3 are true.