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I have a sum $Y=X_1u(X_1-t)+\cdots X_Nu(X_N-t)$ where all $X_i's$ are i.i.d with exponential distribution with parameter $1$ and $u(x)$ is the unit step function. As can be seen from the expression of $Y$ that a variable $X_i$ will be added only if it is greater than $t$ otherwise a zero will replaced for that $X_i$. I want to find the CDF of $Y$. I can use following two strategies.

First strategy: For this I can use following $$P(Y<y)=\sum_{n=0}^{N}P(K=n)P(Y_K<y|K=n)$$

where $K$ denotes the number of variables which has value higher than $t$ and $P(Y_K<y|K=n)=F_{Y_k}(y-nt)$. Here $F_{Y_K}(y)$ is the CDF of the sum of $K$ exponentially distributed random variables. It is not difficult to see that there will be $N$ terms in the summation and the second term correspond to the case of $n=1$ that is when only one random variable has a value higher than $t$. Hence, the second term in the above summation can be written as $$N(1-e^{-t})^{N-1}e^{-t}(1-e^{-y+t})=N(1-e^{-t})^{N-1}(e^{-t}-e^{-y})\cdots Eq. (1)$$.

Now we use the second strategy to get the second term in the above summation. Note that second term correspond to the case that only one variable is higher than $t$.

Second Strategy: In this case we use the order statistics. The joint pdf of the highest $(x_{max})$ and second highest variable $(x_{max-1})$ can be written as $$f_{x_{max},x_{max-1}}(x,y)=\frac{N!}{(N-2)!}e^{-x-y}(1-e^{-y})^{N-2}.$$ The probability that $x_{max}$ has value greater than $t$ and $x_{max-1}$ has value less than $t$ can be found by integrating the above joint pdf as follows $$\int_{0}^{t}\int_{t}^{y+t}f_{x_{max},x_{max-1}}(x,y)dxdy.$$ After doing integration the answer will be $$N(1-e^{-t})^{N-1}(e^{-t}-e^{-y-t})\cdots Eq.(2).$$ We can see that the expression in Eq.(1) (obtained through strategy 1) is different from Eq.(2) (obtained through strategy 2).

Is there anything wrong with this reasoning or am I doing it all wrong. Any help in this regard will be highly appreciated.

P.S.: The first strategy is used in CDF of sum of N exponentially distributed random variables with condition by one of the user of stackexchange.

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Frank Moses
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    The integral is $\displaystyle \mathsf P(t\leqslant X_{(N)}\leqslant y, 0\leqslant X_{(N-1)}\lt t) ~=~ \int_0^t \int_t^y f_{X_{(N)},X_{(N-1)}}(u,v)\operatorname d u\operatorname d v$. – Graham Kemp Jun 01 '16 at 07:55
  • but i read it on a research paper that the upper limit is $y+t$. If you say so I can provide you the reference – Frank Moses Jun 01 '16 at 08:00
  • Why would it be $y+t$? You are after the probability that the sum is no more than $y$ and only one variable is above the threshold. – Graham Kemp Jun 01 '16 at 08:16

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