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I have two types of variables $X_{l,m}$ and $Y_m$. All $X_{l,m}$ are i.i.d distributed according to some probability distribution. Similarly all $Y_m$ are i.i.d distributed according to some other probability distribution. I want to find the following probability $$P\left(\max_{l=1 \cdots L}\left\{\min_{m=1\cdots M}\left(X_{l,m}+Y_m\right)\right\}

My Strategy:

1- First find the PDF and CDF of a generic sum $Z=X+Y$.

2- Find the PDF and CDF of $$T=\min_{m=1\cdots M}Z_m$$ 3- Take the $L$th power of the CDF of $T$ to find the overall probability.

Is there something fundamentally wrong with this strategy? Please clarify if there is something wrong. On the other hand if my strategy is right then how can I verify it. Many many thanks in advance.

Second Strategy:

In this strategy I will fix values of all $Y_m$ and find the conditional probability and then average it over all $Y_m$'s. Obviously I will have to take the limits of at least one of the $Y_m$'s from $0-->C$. But this strategy seems to be much more difficult then the first one. Any comments on my strategies will be highly appreciated. Thanks in advance.

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    Inside the probability expression, do you mean for each $l$, you find a corresponding $m^*(l)$ (depends on $l$) which minimize the sum $X_{l, m^*(l)} + Y_{m^*(l)}$, then find the $l^*$ which maximize this sum. If it is like this then the strategy you list seems not quite matching what required from the question. – 2017-01-14
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    @BGM thank you so much. Your explanation is very clear. – 2017-01-14
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    @BGM it seems that if I am able to write the CDF and PDF of $\min_{1\cdots M}(X_{l,m}+Y_M)$ in the terms of $l$ then I can find the whole probability. Can you please suggest me a way to find the inner probability in terms of $l$. Thanks again for your help. – 2017-01-14
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    @BGM I am extremely sorry for interrupting you again and again but I just want to clear one thing. Suppose if all $Y_M$'s are zero (that is all $Y_M$'s are constants and not random variables and has a value zero). In this case again the inner variable will again be a function of $l$ (as you mentioned in your comment) and hence according to simple use of my strategy (in which the inner variables $X_{l,m}$ will all follow same PDF and CDF irrespective of value of $l$) will result in some expression $[1-e^{-MC\lambda}]^L$ if the variables $X_{l,m}$ are i.i.d. with $exp(\lambda)$... – 2017-01-14
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    .... You can verify this answer by applying my first strategy. But this answer is wrong. I have seen at least two papers where this kind of strategy is applied. So, is this a special case (the case when all $Y_m$'s are zero) when my strategy becomes valid or is it just too difficult to find exact answer and therefore those papers have not considered the dependence on $l$ (to get tractable answer)? – 2017-01-14
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    For your special example, when those $Y$ vanish, it makes the variables minimized in the first stage mutually independent to each other which greatly simplify the matter. In such case the CDF is $(1 - (1 - F_X(x))^M)^L$ (so is this the answer you proposed?) When those $Y$ are there, the sum $X_{l,m^*(l)}+Y_{m^*(l)}$ are dependent.. – 2017-01-15
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    @BGM yes this is the answer I proposed and now I see how inclusion of $Y$ make the problem more difficult. Thanks alot for your explanation. – 2017-01-16

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