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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

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

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Mean in exponential disribution is 1/λ.

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