# Joint density problem.

1. Feb 7, 2012

### spitz

1. The problem statement, all variables and given/known data

I have:

$$f_A=\lambda e^{-\lambda a}$$
$$f_B=\mu e^{-\mu b}$$

($A$ and $B$ are independent)

I need to find the density of $C=\min(A,B)$

2. The attempt at a solution
$$f_C(c)=f_A(c)+f_B(c)-f_A(c)F_B(c)-F_B(c)f_A(c)$$
$$=\lambda e^{-\lambda c}+\mu e^{-\mu c}-\lambda e^{-\lambda c}(1-e^{-\mu c})-(1-e^{-\lambda c})\mu e^{-\mu c}$$
$$=\lambda e^{-\lambda c}e^{-\mu c}+\mu e^{-\lambda c}e^{-\mu c}$$
$$=2(\lambda+\mu)e^{-c(\lambda+\mu)}$$

Correct or utterly wrong?

2. Feb 7, 2012

### Ray Vickson

Some blunders (probably just typos), but answer is right: your first line should have been
$$f_C(c)=f_A(c)+f_B(c)-f_A(c)F_B(c)-F_A(c)f_B(c),$$
which is what your later lines computed. However, you are doing it the hard way: much easier is to say $$\Pr \{\min(A,B) > c \} = \Pr \{ A > c \mbox{ and } B > c \} = \Pr \{A > c \} \cdot \Pr \{ B > c \}.$$

RGV

3. Feb 8, 2012

### spitz

$$=(1-F_A(c)) \cdot (1-F_B(c))=(1-(1-e^{-\lambda c})) \cdot (1-(1-e^{-\mu c}))=e^{-c(\lambda+\mu)}$$
$$\Rightarrow\frac{d}{dc}e^{-c(\lambda+\mu)}=-(\lambda+\mu)e^{-c(\lambda+\mu)}$$

Although my first answer should have been: $$(\lambda+\mu)e^{-c(\lambda+\mu)}$$

Which is correct?

Last edited: Feb 8, 2012
4. Feb 8, 2012

### Ray Vickson

Your first *answer* was incorrect, but your method was correct up to the second-last line; in my original response, I messed the factor of 2, so should not have said it was correct. Basically, to get your last line you said a+b = 2(a+b), so you made a blunder.

The result $\min(A,B) \leftrightarrow (\lambda+\mu) e^{-(\lambda + \mu)t}$ is correct. It is one of the absolutely standard properties of the exponential. Since the second way of getting it is correct, step-by-step, it cannot fail to be correct; you just need more confidence when making true statements, but you also need to be careful when doing algebraic manipulations.

RGV