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Suppose we have two independent exponentially distributed arrival times $X_1$, $X_2$ having rates $\lambda$ and $\mu$. This means their corresponding expected waiting times are $1/\lambda$ and $1/\mu$ accordingly.

Now I'm looking for the expected waiting time for the following case:

Arrival of $X_1$ and then $X_2$ after maximum constant waiting time of $w_1$ ($X_2$ occurs in the interval $[0,w_1]$ after occurring $X_1$) or arrival of $X_2$ and then $X_1$ after maximum constant waiting time of $w_2$. (as shown in the picture)

enter image description here

To be more specific I'm looking for a coincidence of two variables in a given constant interval (interval $[0,w_1]$ if $X_1$ occurs first, $[0,w_2]$ if $X_2$ comes first). I have tried to compute the density function as suggested in the other stackexchange post but that doesn't give me the expected value as I expect. It is rather the density function of the intervals $w_1$ and $w_2$ and delivers the expected interval for $w_1$ or $w_2$ instead of the expected waiting time.

I try and reach no where, can anyone give me a hint on this?

Just_A_User
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Gongotar
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  • Could you be a little more specific ? Do you mean you wait for at most $w_1$ if $X_2>X_1$ (which means you consider variable $Y_1=\min(\min(X_2,X_1+w_1)$, or at most $w_2$ if $X_1>X_2$ ? If this is the case, the computation is quite easy. If not, my knowledge is insufficient to provide help. – Nicolas FRANCOIS Nov 03 '17 at 16:12
  • I have edited the post to be more specific. Also the picture may help you. Please tell me if it's still confusing. Thanks – Gongotar Nov 03 '17 at 16:20
  • So you are waiting for the FIRST time there's a "coincidence" between occurrence of $X_1$ and $X_2$ ? For this, I may need quite a lot of time to answer :-) – Nicolas FRANCOIS Nov 03 '17 at 16:36
  • That is correct. actually a coincidence of two variables in a given constant interval (interval [0, w1] if X1 occurs first, [0, w2] if X2 comes first). I have tried many approaches but reached no where. maybe a hint could work for me – Gongotar Nov 03 '17 at 16:42

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