In an application, I need to obtain a lower bound on the CDF of the sum of $n$ exponential random variables $(X_i)_{i=1}^n$, with parameters $\lambda_i = {n+2-i \choose 2}$. So let $X = \sum_{i}X_i$, Then I have used the expression given in Prop.3.1 of this paper, to get the expression for cdf

$ F_X(x) = \sum_{j=1}^n \prod_{k\not=j}(\frac{\lambda_k}{\lambda_k-\lambda_j})(1-e^{-\lambda_jx})$

Now I am hoping to get a lower bound on this of the form $C_nx$. Since there are both negative and positive terms in this sum, I did the following:

  1. Lower bound the positive terms by using $e^{-\lambda_jx} \leq 1-x+x^2/2$ which gives me $ (1-e^{-\lambda_jx} \geq \lambda_jx(1-\lambda_jx/2)$. I can make an assumption to restrict $x$ to small enough values to make $1-\lambda_jx \geq 1/2$ for all $j$. So this gives me a lower bound for the positive terms.

  2. For the negative terms, I upper bound $(1-e^{-\lambda_jx})$ with $\lambda_jx$, to make the terms more negative.

From these two, I can say that $F_X(x) \geq C_nx$ for some term $C_n$ depending on $n$, and for $x \leq 1/(2{n \choose 2})$.

However, one issue is that I have not been able to show that $C_n >0$.

Can someone help me to find some $C_n>0$, which satisfies the above.

Alternative, can anyone see a different way of obtaining a linear lower bound on $F_X(x)$.


I think you can use Markov inequality,

$F_X(x_o)=P(X\leq x_o)=1-P(X\geq x_o)\geq1-\frac{E\{X\}}{x_o}=1-\frac{\sum_i\frac{1}{\lambda_i}}{x_o}$

Then you can substitute for the value of $\lambda_i$ .

But take care that this lower bound may give you negative values if ($x_o>\sum_i\frac{1}{\lambda_i}$)


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