# Principle of single big jump

I'm new here, i know that from the principle of single big jumps it follows,

$$\Pr[X_1+ \cdots +X_n>x] \sim \Pr[\max(X_1, \ldots,X_n)>x] \quad \text{as } x \to \infty.$$

So let $F$ be a subexponential destribution with $F(0)=0$ and $X_i$ independent random variables for $i=1,\ldots,n$ with this distribution.

Now i want to determine for $n,k$ with $0\leq k\leq n$

$$\Pr[X_1+\ldots+X_k>x\mid X_1+X_2+\cdots+X_n>x],$$

so the difference is that its not the same parameter, can O still use the lemma above? I dont really know where to start. Thanks in advance

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The probability approaches $1$ that exactly one variable is large given that at least one is large. If they are i.i.d., then the large one is equally likely to be any of the $n$. So the probability you want approaches $k/n$. – mjqxxxx Jan 23 '13 at 18:44
why is the probability 1? – nordmann Jan 23 '13 at 20:17
If they're independent, then $P[\max(X_1,...,X_n) > x]=1-P[X_1\le x]\cdots P[X_n\le x]=1-P[X\le x]^n=1-(1-P[X>x])^n$, which is $\sim nP[X>x]$. Write the conditional probability as a ratio, apply the single-big-jump lemma to the top and bottom, and then use this to get the numerator $\sim kP[X>x]$ and the denominator $\sim nP[X>x]$. – mjqxxxx Jan 23 '13 at 20:55
thanks i will try – nordmann Jan 23 '13 at 21:00
as ratio $P(A | B)=\frac{P(A\cap B )}{P(B)}$ with $A=(X1+...Xk>x)$ and $B=(X1+...Xn>x)$ can i say that $P(A\cap B)=P(A)$ in this case? – nordmann Jan 23 '13 at 23:11