It does not matter if the coefficients a and b sum pp to one or not. What counts is that the modified Weibull rv aX is again of Weibull distribution. Thus the question should be posted as follows: What is the pd of X+Y where X and Y possess Weibull distribution? However, in this formulation there is a wide spectrum of possibilities depending on the correlation between X and Y.

The answer is simple if X and Y are related increasingly and deterministically and if e.g if RX(x) = exp( - a xp) and RY(x) = exp( - b xq), then

Pr{ a Xp > t} = Pr{ b Yq > t} = exp{-t}, i.e a Xp = b Yq and therefore

X+Y = X + ( a Xp / b )1/q := f(X).

Knowing the inverse function f-1(z), which is increasing, we can write:

Pr{ X+Y > z} = Pr{ f(X) >z} = Pr{ X > f-1(z)} = exp( - a [f-1(z)]p ).

If the random variables X and Y are sochastically independent, then the convolution is to be calculated, which is not elementary (except some particular cases):

Pr{X+Y>z} = a/p \int0z exp( - b (z-x)q) exp( - a xp) xp-1 dx

Dear Professor Domsta,

Thank you very much for you comprehensive answer to my question. In my case, the variables are independent. I should have mentioned that in my question. To be more specific, the two variables represent the lifetime of a perishable product under two different conditions from which the decision maker should choose. According to your answer I think these two are independent variables and therefore, I should try to calculate the convolution. Please let me know your opinion. Thank you again.

Yours Truly,

Javid Jouzdani

@ Javid

Thanks for thanks. Note however that the calculation leads to non-elemetary functions. Probably you should switch to partial info, e.g. to moments of the resulting pd. Due to independence, e.g.

E{(X+Y)4} = E{X4} + 4 E{X3} E{Y} + . . .

Regards, Joachim

PS This advice is given not personally to you but rather to those P.T.Followers who are less familiar with the probability calculus. JoD

Sharing link;

https://www.statlect.com/probability-distributions/normal-distribution-linear-combinations

Regards!!

Let me complete the info from the link by @Adriana , that the sum of normal rv-s X+Y is normal under the assumption that the joint pd of the pair (X,Y) is normal, according to formulas given e.g. in section "Density function" at

https://en.wikipedia.org/wiki/Multivariate_normal_distribution

It is important to stress, that it is NOT sufficient that the marginals are normal. A simple example provides the following density function

f(x,y) = exp{ - (x2+y2)/2 } / \pi for x y > 0, and =0 for x y 0.

Obviously, this is not the normal distribution.

Regards, Joachim

... and it should be also clear that the marginals are given by the integrals:

fX(x) = exp{ - x2/2 }/ \pi \inty>0 exp{ - y2/2 } dy = exp{ - x2/2 }/ \sqrt{2 \pi},

if x > 0, and for x< 0 it is given by

fX(x) = exp{ - x2/2 }/ \pi \inty

Dear Professor,

Follow this link https://arxiv.org/pdf/1312.2799 .

Dear Professor Domsta,

I greatly appreciate the time and effort you spent on answering my question. My final goal is to perform the following calculations:

p(t = 1-alpha -> t

The problem of geting the critical set with poor info about the exact pd is based e.g on the Chebyshev tipe inequalities. For your needs, the simplest example reads as follows, if Z is a say positive rv, for elementary outline refere to https://en.wikipedia.org/wiki/Chebyshev%27s_inequality ) :

Pr{ Z > t } 0,

which makes sense (is useful) for p >0 for which the expectation is finite, Best, JoD

Dear professor Domsta,

Thank you for your answer. As far as I know, the inequalities such as Chebyshev, is that they provide a crude bound (obviously due to lack of information). Is there any distribution-specific technique to obtain a more specific inequality?

Yours Truly,

Javid

Dear Dr Jouzdani,

Good question! You are right, some better localization can be obtained, but I don't remember where I have seen them. Let me suggest to seek within math- and/or stat-encyclopedias under concentration function, heavy tailes (when low moments are finite only), etc. Definitely the more moments or other characteristics are in use, the more exact are the estimations. As soon as I find in my deeply hidden books some better and practical formulas, I'll inform you.

Best regards, Joachim

Dear professor Domsta,

I greatly appreciate your help. I will also inform you if I have something to worthy of sharing with you and other guys. Thank you.

Yours Truly,

Javid