I'm trying to solve a question about the coxian distribution. There are two phases, phase 1 and phase 2. Because coxian is not so famous, I will give the definition as in our book:

Let $B(x)~\sim~C_2(\alpha,\mu_1,\mu_2)$. This means that the random variable $X$ can be represented as \begin{array}{ll} X = X_1+X_2 & \mbox{with probability $\alpha$}\\ X = X_1 & \mbox{with probability $1-\alpha$ }\end{array} where $X_1$ and $X_2$ are independent random variables having exponential distributions with means $1/\mu_1$ and $1/\mu_2$

Now I want to solve the question: what is the probability that the first phase is below T, so $P(X_1<T)$

My first thought was that $P(X_1<T)=1-e^{-\mu T}$, because $X_1$ is exponentially distributed. But maybe this is too simple. Hope one of you can help me...

  • $\begingroup$ The definition of $X$ is clear, what tt's not clear is what's "the first phase" $\endgroup$ – leonbloy Apr 9 at 16:56
  • $\begingroup$ Hello Leonbloy, the first phase is $X_1$, the first exponential random variable. :) $\endgroup$ – MathMe Apr 10 at 8:32

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