Why aren't those two probabilities equal (exponential dist)

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Hamad
I don't know how to type latex in the forums so I took a picture and uploaded it
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I'm a little concerned by the wording "failure time of A is ##\lambda_1##. I trust you are not saying that is the expected time until failure, since that is in fact the inverse of this, i.e. ##\frac{1}{\lambda}##.

I believe you're saying that exponentially distributed A is parameterized by ##\lambda_1##.

Note that the exponential function is strictly convex over reals. (How do we know this?)

This means that if there is any variation at all in some random variable ##Y## (note in this case you could say that ##Y## takes on a value of ##\lambda_1## with probability p and ##\lambda_2## with probability 1-p, and assert that ##\lambda_1 \neq \lambda_2## and that ##0 \lt p \lt 1## but the idea more generally is that ##Y## is a random variable that is non-degenerate (which in this case I mean it is non deterministic and has a first moment).

Thus we have ##exp\big(E[Y]\big) \lt E\big[exp(Y)\big]## for strictly convex functions like the exponential function.

Do you see how this means your ##C ## cannot be equal to your convex combination of ##A## and ##B##? (Again setting aside cases where ##C## is equal to just ##A## or just ##B##, i.e. when p =0 or p =1 or ##\lambda_1 = \lambda_2##)

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n.b. your instincts are pretty good here though. There is something pretty similar that you can do in terms of splitting and combining Poisson processes. (on the real line:) Poisson processes can be interpreted as a counting process with inter-arrival times that are exponentially distributed. Put differently, the exponential distribution is a building block for the Poisson and you can do splitting and combining there. You may want to read up on Poisson processes.

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For Latex in the forums, see:

https://www.physicsforums.com/help/latexhelp/

It's basically like what you'd think, but you need to encapsulate things in double hashtags instead of single dollars. Plus some BB code or markdown collides with a couple LaTeX commands, but again basically what you'd think.
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and welcome to PF!
 
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Oh I know about poisson processes, and yes I meant ##\lambda## is the rate of failture, but anyway yeah we did study about that inequality in my old course, but I am still curious does that exist no ##\lambda_3=f(\lambda_1,\lambda_2)##to satisfy what I am trying to do such that F(##\lambda_3##) is the cdf of x
 
Hamad said:
Oh I know about poisson processes, and yes I meant ##\lambda## is the rate of failture, but anyway yeah we did study about that inequality in my old course, but I am still curious does that exist no ##\lambda_3=f(\lambda_1,\lambda_2)##to satisfy what I am trying to do such that F(##\lambda_3##) is the cdf of x

You may want to re-read parts of my post. By strict convexity the ##\lambda_3## you are seeking exists iff ##\lambda_1 = \lambda_2## for any ##0\lt p \lt 1##.

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edit: I have lingering concerns that I'm answering a slightly different problem than what you really want, and I keep thinking that Poisson splitting is the solution.
 
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StoneTemplePython said:
You may want to re-read parts of my post. By strict convexity the ##\lambda_3## you are seeking exists iff ##\lambda_1 = \lambda_2## for any ##0\lt p \lt 1##.

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edit: I have lingering concerns that I'm answering a slightly different problem than what you really want, and I keep thinking that Poisson splitting is the solution.
I wouldn't say you are answering a different problem than what I want but I also don't think poisson splitting is the answer. My objective is to know whether I can replace objects A,B with an equivalent object C to model the effects. As for example in circuit elements you can have an equivalent circuit describing your total system, and you know my background is electrical, so I was curious whether I could have an equivalent experiment of object C to model A and B, but I guess again non-linearity is probably why it's not possible, but I will still look for ways out of interest to even approximate/reduce problems with these composite probabilities into a simplified equivalent system if possible.
 
Hamad said:
whether I can replace objects A,B with an equivalent object C
As you found, C would have the distribution function ##1-pe^{-\lambda_1x}-(1-p)e^{-\lambda_2x}##. That cannot be written as ##1-e^{-\lambda_3x}##, no matter what you choose for ##\lambda_3##. So C cannot be Poisson.