Sums of Independent (but not identically distributed) Random Variables

In summary, the conversation was about finding a Hoeffding-type result for bounding the tail of a sum of independent exponential random variables with distinct parameters. The speaker was looking for tighter bounds than Markov's Inequality and Chebyshev's Inequality and asked if anyone was aware of well-known results in this direction. They also mentioned the possibility of using Hypoexponential distribution and the difficulty in finding a clean tail inequality for the sum. Another person suggested using Bernstein inequalities and mentioned the existence of tail probability inequalities involving an integral of the characteristic function. The speaker thanked them for their input and expressed interest in hearing how it goes.
  • #1
sv79
2
0
I am looking for a Hoeffding-type result that bounds the tail of a sum of independent, but not identically distributed random variables. Let X_1,..,X_n be independent exponential random variables with rates k_1,...,k_n. (Note: X_i's are unbounded unlike the case considered by Hoeffding)

Let S= \sum_{i=1}^{n} X_i. I am interested in bounding P(S>a). I am looking for tighter bounds than Markov's Inequality and Chebyshev's Inequality. Is anyone here aware of well-known results in this direction?
 
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  • #2
sv79 said:
independent exponential random variables

Are the parameters all distinct? (this would simplify the analysis considerably)
 
  • #3
Bpet:
Yes, the parameters are all distinct. Perhaps, you are thinking of a Hypoexponential distribution (also called a Generalized Erlang distribution, I think), which in my case it certainly is. The question is can we get a clean (easily usable like the Hoeffding, Chernoff bounds etc.) tail inequality for the sum?
 
  • #4
I don't know if there are any simple inequalities (simpler than the matrix exponential formula) but maybe some of the Bernstein inequalities (tricky bit is working out the central moment growth rate). Also from memory there are some tail prob inequalities involving an integral of the characteristic function over some small neighbourhood of zero. Sorry I couldn't be of more help but I'm keen to hear how it goes.
 

1. What are independent random variables?

Independent random variables are variables that do not influence each other's outcomes. In other words, the outcome of one variable does not affect the outcome of the other. This means that the variables are not related or dependent on each other in any way.

2. What is the difference between independent and identically distributed random variables?

Independent random variables are variables that are not related or dependent on each other, while identically distributed random variables have the same probability distribution. This means that although the variables are not related, they have the same chances of occurring.

3. How are sums of independent random variables calculated?

To calculate the sum of independent random variables, we simply add up the values of each individual variable. For example, if we have two independent variables X and Y, the sum of these variables would be Z = X + Y.

4. Can independent random variables have different probability distributions?

Yes, independent random variables can have different probability distributions. This means that the variables have different chances of occurring, but they are not related or dependent on each other.

5. Why is it important to consider independence when working with sums of random variables?

Considering independence is important because if the variables are not independent, the calculations for sums of random variables can become more complex. Additionally, if the variables are not independent, the outcome of one variable may affect the outcome of the other, leading to inaccurate results.

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