the density for an exponentially distributed RV X with parameter λ>0
is given as:
f(x)=(1/λ )exp(−x/λ)
for non-negative x and 0 otherwise. The expected value of the maximum of two independent exponentially distributed random variables X and Y with parameters λ1=1 and λ2=2 is...?
Whenever you need to find the Expected Value of the Maximum of a random variables it is always wise and easy to solve the problem in the following 3 steps:
Lets do this together!
$\textbf{Step 1:}$
The cumulative density function $F_M(x)$ is the probability that the random variable $M$ (defined to be the maximum of these 2 R.Vs) is less than $x$. I.E $F_M(x):= \mathbb{P}(M\leq x)$ In order for the maximum of the two random variables to be less than $x$ we need $\textbf{both}$ of them to be less than $x$. This should be obvious but let us just stop to think about it for a second . If we both have a dice and I roll a 4 and you roll a 6 is the maximum of our rolls less than or equal to 5? No. Why not? Because they are not $\textbf{both}$ less than 5. Hence $ \mathbb{P}(M\leq x) = \mathbb{P}(X\leq x \cap Y\leq x) $ And as they are independent: $ \mathbb{P}(X\leq x \cap Y\leq x)= \mathbb{P}(X \leq x) \cdot \mathbb{P}(Y \leq x)$ If you are familiar with the exponential distribution you should know that $\mathbb{P}(X \leq x) = 1 - e^{-\lambda_1 x}$ and equally $\mathbb{P}(Y \leq x) = 1 - e^{-\lambda_2 x}$
Putting this all together completes Step $1$ and yields: $F_M(x):= \mathbb{P}(M\leq x) = \mathbb{P}(X\leq x \cap Y\leq x) = \mathbb{P}(X \leq x) \cdot \mathbb{P}(Y \leq x) = (1 - e^{-\lambda_1 x}) \cdot (1 - e^{-\lambda_2 x}) = 1 -e^{-\lambda_1x} - e^{-\lambda_2x} + e^{-(\lambda_1+\lambda_2)x}$ And using $\lambda_1 = 1$ and $\lambda_2 = 2$ yields the final CDF : $F_M(x) = (1-e^{-x}-e^{-2x}+e^{-3x})$
$\textbf{Step 2:}$
It is lightwork from here. We differentiate the CDF with to yield the PDF:
$f_M(x) = \frac{d}{dx}F_m(x) = \frac{d}{dx}(1-e^{-x}-e^{-2x}+e^{-3x}) = e^{-x} + 2e^{-2x}-3e^{-3x} $
$\textbf{Step 3:}$
We are after the expected value of $M$ (the maximum) so all we would need to do is integrate:
$\mathbb{E}[M] = \int_{0}^{+\infty}x\cdot f_M(x)dx = \int_{0}^{+\infty}x\cdot (e^{-x} + 2e^{-2x}-3e^{-3x})dx $
This is fairly easy to do via parts however I am very lazy so lets be naughty and recognise that this is the "sum" of exponential R.V's:
$ \int_{0}^{+\infty}x\cdot (e^{-x} + 2e^{-2x}-3e^{-3x})dx = \int_{0}^{\infty}x\cdot e^{-x} + \int_{0}^{\infty}x\cdot2e^{-2x} - \int_{0}^{\infty}x\cdot3e^{-3x} $
Which we can see the simply be the calculations required to calculate the expected value of $\lambda = 1 , \lambda = 2, \lambda = 3$ exponential R.Vs respectively. And as the Expected value of an exponential R.V is $\frac{1}{\lambda}$ we have:
$ \int_{0}^{\infty}x\cdot e^{-x} + \int_{0}^{\infty}x\cdot2e^{-2x} - \int_{0}^{\infty}x\cdot3e^{-3x} = \frac{1}{1} + \frac{1}{2} - \frac{1}{3} = \frac{6+3-2}{6} = \frac{7}{6}$
And hence the expected value is $\frac{7}{6} = 1 + \frac{1}{6} \approx 1.167 $
