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Let, $X,Y$ are independent rv with pdf.s

$f_X(x\mid \lambda)=\dfrac{1}{\lambda}e^{-x/\lambda}$, $x>0$ and $f_Y(y\mid \mu)=\dfrac{1}{\mu}e^{-x/\mu}$, $y>0$.

Let $Z=\min\{X,Y\}$ and

$W=1$ if $Z=X$ and $0$ if $Z=Y$

Find joint distribution of $(Z,W)$?

I found pmf of $W$ which is $f_W(w)=\Big(\dfrac{\mu}{\lambda+\mu}\Big)^w\Big(\dfrac{\lambda}{\lambda+\mu}\Big)^{1-w}$

But have problem while calculating $P(Z\leq z, W=1)$.

$P(Z\leq z, W=1)=P(Z\leq z\mid W=1)P(W=1)=P(X\leq z)P(W=1)$

The last line does not look right to me. What should I do? Thanks for any help!

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Stat_prob_001
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1 Answers1

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First, develop joint distribution of $W$ and $Z$ for both $W$ cases:

$$\begin{align}f(Z=z,W=0)&=f(Z=z,Z=Y)\\&=f(Y=z,Y<X)\\&=f(Y=z,X>z)\\&=f(Y=z)~P(X>z)\\[2ex]f(Z=z,W=1)&=f(Z=z,Z=X)\\&=f(X=z,X<Y)\\&=f(X=z,Y>z)\\&=f(X=z)~P(Y>z)\end{align}$$

Last two terms can be written as follow:

$$\begin{align}f(Z=z,W=0) &= f(Y=z)~P(X>z) \\&= \frac{1}{\mu}e^{-z/\mu} e^{-z/\lambda}\\[2ex]f(Z=z,W=1) &= f(X=z)~P(Y>z) \\&= \frac{1}{\lambda}e^{-z/\lambda} e^{-z/\mu}\end{align}$$

Finally, one can write: $$\begin{align}f(Z=z,W=w)&=e^{\raise{1ex}{-\left(\frac{1}{\lambda} + \frac{1}{\mu} \right)z}}~\left(\frac{1}{\lambda}\right)^w\left(\frac{1}{\mu}\right)^{1-w}\end{align}$$

Graham Kemp
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Carlos Campos
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    Is that it? Actually I did almost same thing. But problem I am getting is that if that's the answer, can we get $f_Z(z)=P(Z=z, W=0)+P(Z=z, W=1)$? I am totally confused. – Stat_prob_001 Jan 26 '18 at 02:35
  • here, $f_Z(z)=\Big(\dfrac{1}{\lambda}+\dfrac{1}{\mu}\Big)e^{-(z/\lambda+z/\mu)}$ is pdf of $Z=\min{X,Y}$ – Stat_prob_001 Jan 26 '18 at 02:41
  • Yes, you are completely right. I think I found the error. Let me correct the answer – Carlos Campos Jan 26 '18 at 09:50
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    Every event involving $X=z$ or $Y=z$ has probability zero, thus none of the computations in this post is legit, I am afraid. – Did Jan 26 '18 at 10:13
  • I will change P by f to make it clear that it is a pdf – Carlos Campos Jan 26 '18 at 10:20
  • Thanks for the answer. BTW this might not be big deal, but if you write $f$, it makes a big problem (since in your solution $f$ might stand for pdf, but it complicates the part $W$ is a discrete rv.) To avoid such confusion, (learnt from my experience), you might try to find $$\lim_{h\to 0} \frac{P\Big(x-\dfrac{h}{2}< Z\leq x+\dfrac{h}{2}, W=1\Big)}{h}$$ – Stat_prob_001 Jan 26 '18 at 14:56
  • Yes, I see your point. So, how would you write the joint distribution of a continuous RV and a discrete RV? I don't know what notation would be right – Carlos Campos Jan 27 '18 at 10:27
  • "I will change P by f to make it clear that it is a pdf" But "it" is NOT a PDF. – Did Jan 27 '18 at 17:05
  • @Did, is for you $f(Z=z,W=w) = e^{-z \left( \frac{1}{\lambda} + \frac{1}{\mu} \right)} \left( \frac{1}{\lambda} \delta(w-1) + \frac{1}{\mu} \delta(w) \right)$ a pdf? For me, it is. Not differentiable, but it is a pdf – Carlos Campos Jan 27 '18 at 18:41
  • @CarlosCampos Of course not, please review the definition of the PDF of a random variable. The trouble with your suggestion is not even that the RHS is not differentiable, but that it is not a function. – Did Jan 27 '18 at 18:49
  • Wow @Did, Is it not a function $f: \mathbb{R} \times {0,1} \rightarrow [0,1]$ ? I am getting lost... – Carlos Campos Jan 27 '18 at 18:59
  • Hmmm... how to break this mental blockage? Maybe simply recall what D and F mean in PDF? – Did Jan 27 '18 at 19:03
  • Other people share my point of view: https://math.stackexchange.com/questions/54197/can-a-dirac-delta-function-be-a-probability-density-function-of-a-random-variabl – Carlos Campos Jan 27 '18 at 19:11
  • That you think the link you provide shows what you say is stunning. Did you read the page you link to? (Unrelated: Please use @, that is, unless you want the users you address your comments to, to be made aware of your comments.) – Did Jan 30 '18 at 07:32
  • @Did, I was referring to this particular link which appears in that page: https://en.wikipedia.org/wiki/Probability_density_function#Link_between_discrete_and_continuous_distributions It is a different way to consider probability distributions of discrete variables. Link between continuous and discrete variables. It makes sense for me.sense for me – Carlos Campos Jan 30 '18 at 10:16
  • Right, I had missed this paragraph. Mathematically, this is horrible. I know that some physicists like to consider discrete measures as functions, but for which practical advantages? And if we turn to pedagogy, the effects are disastrous... – Did Jan 30 '18 at 10:23
  • @Did, For me, using this notation eases the way to express joint probability distributions of discrete and contineous variables (i.e. $f(X=x,W=w)$. Maybe, the only other alternative would be to express $P(X<x,W=w)$, which avoids the use of Dirac artifact, but taking the cdf over X – Carlos Campos Jan 30 '18 at 11:15
  • That, but rather as $P(X\leqslant x,W=w)$. Or, for a completely rigorous way of writing down the joint distribution of $(X,W)$ when the distribution of $X$ is purely absolutely continuous and the distribution of $W$ is purely discrete, $$P_{(X,W)}(dxdw)=\sum_kf_k(x)\delta_{w_k}(dw)dx$$ – Did Jan 30 '18 at 12:39
  • Yes @Did, I strongly agree. Coming back to the original question, could $f(Z=z,W=w) = e^{-z \left( \frac{1}{\lambda} + \frac{1}{\mu} \right)} \left( \frac{1}{\lambda} \delta(w-1) + \frac{1}{\mu} \delta(w) \right)$ be the right answer? – Carlos Campos Jan 30 '18 at 14:49
  • Sorry but, as already explained, one cannot reasonably declare that a formula involving the horror $$f(Z=z,W=w)$$ is "correct"... But if your question is whether the formula $$P_{(Z,W)}(dzdw)=e^{-\lambda^{-1}z-\mu^{-1}z}(\lambda^{-1}\delta_1(dw)+\mu^{-1}\delta_0(dw))dz$$ is correct, then the answer is yes. – Did Jan 30 '18 at 17:48