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Let $X$ and $Y$ be two exponentially distributed and independent random variables with mean $α$ and $β$, respectively. If $Z$ = min $(X, Y)$, then the mean of $Z$ is given by

1. $\left(\dfrac{1}{\alpha + \beta}\right)$
2. $\min (\alpha, \beta)$
3. $\left(\dfrac{\alpha\beta}{\alpha + \beta}\right)$
4. $\alpha + \beta$

This might help:

derivation is not in syllabus

$X$ is an exponential random variable of parameter λ when its probability distribution function is

$$f(x) = \begin{cases}\lambda e^{-\lambda x} & x \geq 0 \\ 0 & x < 0 \end{cases}$$

For a > 0, we have the cumulative distribution function

$$F_x(a) = \int_0^a f(x) dx = \int_0^a \lambda e^{-\lambda x} dx = -e^{-\lambda x} \mid_0^a = 1 - e^ {-\lambda a}$$

So,

$$P\left\{X < a \right \} = 1 - e^ {-\lambda a}$$ and

$$P\left\{X > a \right \} = e^ {-\lambda a}$$

Now, we use $P \left \{X > a \right \}$ for our problem because our concerned variable $Z$ is min of $X$ and $Y$.

For exponential distribution with parameter $\lambda$, mean is given by $\frac{1}{\lambda}$.
We have,

$P \left \{X > a \right \} = e^ {-\frac{1}{\alpha} a}$

$P \left \{Y > a \right \} = e^ {-\frac{1}{\beta} a}$

So, \begin{align*}P\left \{Z > a \right \} &= P \left \{X > a \right \} P \left \{Y > a \right \} \left(\because \text{X and Y are independent events and } \\Z > \min \left(X, Y \right) \right)\\&=e^ {-\frac{1}{\alpha} a} e^ {-\frac{1}{\beta} a} \\&=e^{-\left(\frac{1}{\alpha} + \frac{1}{\beta} \right)a} \\&=e^{-\left(\frac{\alpha + \beta} {\alpha \beta} \right)a}\end{align*}

This shows that $Z$ is also exponentially distributed with parameter $\frac{\alpha + \beta} {\alpha \beta}$ and mean $\frac{\alpha \beta} {\alpha + \beta}$.

by

Mean in exponential disribution is 1/λ.

According to your logic, the answer should be 1/α + 1/β, which leads to option C.

Yes. It is C. I had taken mean for parameter. Changed now :)
@csegate2 >> Reciprocal will be the mean.

Now, we use P{X>a} for our problem because our concerned variable Z is min of X and Y.

because in f(x) we solve for x>=0
I didnt get this line

P{X<a}=1&minus;e&minus;&lambda;aP{X<a}=1&minus;e&minus;&lambda;a

and

P{X>a}=e&minus;&lambda;a

Now, we use P{X>a} for our problem because our concerned variable Z is min of X and Y.

What is the meaning and conclusion of this line.
Sir, what is the role of min(x, y) here ?

If it would have been max instead of min, then what would be the answer?

Arjun sir why are you taking this?

we use P{X>a} for our problem because our concerned variable Z is min of X and Y

Given X and Y are exponential random variables with mean $\alpha$ and $\beta$ respectively.

Since, mean of a random variable corresponds to it's expected value,

Expectation of a exponential random variable with parameter $\lambda$ = $\frac{1}{\lambda }$

So $\lambda$ = $\frac{1}{\alpha }$

and $\lambda$ = $\frac{1}{\beta }$

$\lambda$z = $\lambda$x + $\lambda$y = $\frac{1}{\alpha } + \frac{1}{\beta } = \frac{\alpha +\beta }{\alpha \beta }$

So E[Z] =  $\frac{\alpha \beta }{\alpha +\beta }$

What if Z = MAX(X,Y)..where X,Y are independent and exponentially distributed RVs with parameters $\alpha$ and $\beta$

edited

Tell me if i am wrong here,

When we want to calculate the minimum value of Z. It’s like calculating that the upper bound Big O of Z. that’s why we needed to do {X>a} which implies the the value of “a” could range from a to infinity. and  the minimum probability of the random variable will always occur when we want them to happen in intersection and since independent variable

P(a∩b )=P(a)P(b)

Such a simple a beautiful approach Arjun sir
Thanks for sharing the PDF ayush, I didnt know about the memoryless property.

I had some problem in understanding $Z>a$ iff $X>a$ and $Y>a$

So I tried to draw analogy using calculus: