What is the Joint Density of C = min(A,B)?

[spitz]

Homework Statement

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?

 

[Ray Vickson]

Homework Statement

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?

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) &gt; c \} = \Pr \{ A &gt; c \mbox{ and } B &gt; c \}<br /> = \Pr \{A &gt; c \} \cdot \Pr \{ B &gt; c \}.

RGV

 

[spitz]

...much easier is to say \Pr \{\min(A,B) &gt; c \} = \Pr \{ A &gt; c \mbox{ and } B &gt; c \}<br /> = \Pr \{A &gt; c \} \cdot \Pr \{ B &gt; c \}.

RGV

=(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?

 

[Ray Vickson]

=(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?

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