I ran into some problems with this:
Suppose that two electronic components in the guidance system for a missile operate independently and that each has a length of life governed by the exponential distribution with mean 1 (measured in hundreds of hours). Find the pdf for the average length of life of the two components.
I've tried to create a function of these two variables => U = Y1+Y2
which gives E(Y1+Y2) = 2
and then tried to use that f(y) for an exponential distribution is 1/beta *e^-y/beta. I then try to put in beta = 2 and put up an integral:
but it doesn't seem like I'm heading in the right direction. Cam someone please help me out? Thanks a lot!
Just trying to bring some new attention to this thread. Would be grateful if anyone could help me out with this.
The notion of "length of life of the two components" is not completely clear in your text. In this kind of problem, there are typically two possibilities:
Originally Posted by approx
- either the components are "in parallel", meaning that if one of them fails, then the other one compensates for it. And the lifetime of the system is the maximum of the lifetimes of the components;
- or the components are "in series", meaning that as soon as one fails, then the whole system crashes. And the lifetime of the system is then the minimum of the lifetimes of the components. [This is the most plausible interpretation]
There's even another possibility: if one component dies, the other starts working to replace it (like a brand new one). And the lifetime of the system is the sum of the lifetimes of the components (That's what you did). This latter possibility seems however less likely to me since the components are said to work "independently". But I'll say a word about it anyway.
And the question should be "find the pdf and the average of the length of life (...)" since the average is a constant (so that its pdf makes poor sense, if any).
Let be the lifetimes, , , .
- In order to find the pdf of or , you should consider and (for ) and differentiate this. You can then deduce the average. In fact, you can notice that this computation shows that is exponentially distributed with parameter 2. (but not )
- In order to find the pdf of , you must look for the convolution of the pdf's of and : it is when and else. This is called a gamma distribution (of parameters 1 and 2 in some order I don't remember now).