Finding an upper bound for a probability

AI Thread Summary
The discussion revolves around finding a tight upper bound for the probability mass function (pmf) of a random variable k_1, given the constraint of knowing only the sum p_1 + p_2. The user seeks assistance in deriving a more effective upper bound than the one provided by the inequality involving factorials and the combined probability. The current approach is deemed insufficiently tight for their needs. Participants are encouraged to contribute methods or insights that could lead to a better solution. The focus is on enhancing the accuracy of the upper bound for the pmf under the specified conditions.
sabbagh80
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Hi,

I will thank If somebody help me solving this problem.

Consider a random variable k_1 with the given pmf as:

Pr[k_1=l]=\sum_{l_1+2l_2=l} \frac{N!}{(N-l_1-l_2)!l_1!l_2!}p_1^{l_1} p_2^{l_2} (1-(p_1+p_2))^{N-l_1-l_2}where l_1,l_2 \in [0,1,...,l].

but we don't have p_1 and p_2 separately and I know just the value of p_1+p_2.

I want to find at least a good and tight upper bound for the above pmf.

For example; we can use the inequality of p_1^{l_1} p_2^{l_2} \leq \frac{l_1!l_2!}{(l_1+l_2)!}(p_1+p_2)^{l_1+l_2}, but it is not that much tight.

Can everybody help me?
 
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Hi,

I will thank If somebody help me solving this problem.

Consider a random variable k_1 with the given pmf as:

Pr[k_1=l]=\sum_{l_1+2l_2=l} \frac{N!}{(N-l_1-l_2)!l_1!l_2!}p_1^{l_1} p_2^{l_2} (1-(p_1+p_2))^{N-l_1-l_2}


where l_1,l_2 \in [0,1,...,l].

but we don't have p_1 and p_2 separately and I know just the value of p_1+p_2.

I want to find at least a good and tight upper bound for the above pmf.

For example; we can use the inequality of p_1^{l_1} p_2^{l_2} \leq \frac{l_1!l_2!}{(l_1+l_2)!}(p_1+p_2)^{l_1+l_2}, but it is not that much tight.

Can everybody help me?
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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