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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. (1/(α + β))
  2. min (α, β)
  3. (αβ/(α + β))
  4. α + β
asked in Probability by Veteran (19k points)   | 827 views
Answer this please

1 Answer

+7 votes
Best answer

$X$ 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 $\frac{\alpha  \beta} {\alpha + \beta}$.

 

Ref: http://ocw.mit.edu/courses/mathematics/18-440-probability-and-random-variables-spring-2011/lecture-notes/MIT18_440S11_Lecture20.pdf 

answered by Veteran (280k points)  
selected 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.
link is broken

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

Please elaborate this.

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
please someone explain..

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?
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