Let X1 and X2 be two independent exponentially distributed random variables with means 0.5 and 0.25, respectively. Then Y = min (X1, X2) is exponentially distributed with mean x/6 . Find the value of x. (Answer up to the nearest integer)Correct answer is '1'. Can you explain this answer?

Question

Answer

Given:
- X1 and X2 are two independent exponentially distributed random variables with means 0.5 and 0.25, respectively.
- Y = min(X1, X2)

To Find:
The value of x, the mean of Y

Solution:

1. Exponential Distribution:
The exponential distribution is a continuous probability distribution that models the time between events in a Poisson process. It is characterized by a single parameter, λ (lambda), which represents the average number of events in a unit of time.

The probability density function (PDF) of an exponential distribution with parameter λ is given by:
f(x) = λ * exp(-λx) for x ≥ 0
where exp(-λx) is the exponential term and λ > 0.

The mean (μ) of an exponential distribution is given by:
μ = 1 / λ

2. Properties of Minimum:
If X1 and X2 are two independent random variables, then the minimum of X1 and X2, denoted by Y = min(X1, X2), is a new random variable with the following properties:
- The PDF of Y is given by the minimum of the individual PDFs of X1 and X2.
- The CDF (cumulative distribution function) of Y is given by the product of the individual CDFs of X1 and X2.

3. Finding the PDF of Y:
Since X1 and X2 are independent exponential random variables, we can find the PDF of Y by taking the minimum of their individual PDFs.

The PDF of X1 is given by:
f1(x) = λ1 * exp(-λ1x) for x ≥ 0
where λ1 = 1 / 0.5 = 2

The PDF of X2 is given by:
f2(x) = λ2 * exp(-λ2x) for x ≥ 0
where λ2 = 1 / 0.25 = 4

Therefore, the PDF of Y, denoted by fY(x), is given by:
fY(x) = min(f1(x), f2(x))
= min(2 * exp(-2x), 4 * exp(-4x))

4. Finding the Mean of Y:
To find the mean of Y, denoted by μY, we integrate the PDF of Y multiplied by x:
μY = ∫[0,∞] x * fY(x) dx

Substituting the PDF of Y into the above equation, we get:
μY = ∫[0,∞] x * min(2 * exp(-2x), 4 * exp(-4x)) dx

5. Solving the Integral:
To solve the integral, we need to consider two cases based on the minimum function.

Case 1: If 2 * exp(-2x) ≤ 4 * exp(-4x), then min(2 * exp(-2x), 4 * exp(-4x)) = 2 * exp(-2x).
In this case, we have:
μY = ∫[0,∞] x * 2 * exp(-2x) dx

Case 2: If 2

Answer

We know that, if X₁ and X₂ are independent and exponential R. V's with

parameters λ₁and λ₂ then X = min(X₁, X₂) is exponential R.V with parameter

Mean of Y = E(Y) = 1/6

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