Let $f^B_{j,a}(s)$ be the probability mass function of the binomial distribution, that is $f^B_{j,a}(s) = {j \choose s} a^s (1-a)^s$. And let $F^B_{j+1,b}(s)$ be the cdf of the binomial distribution, that is $F^B_{j+1,b}(s) = \sum_{i=0}^{s} f^B_{j+1,b}(i)$.

In my research, I need a quite precise upper bound on the sum

$$\sum_{s=0}^j \frac{f^B_{j,a}(s)}{F^B_{j+1,b}(s)}.$$

For $a>b$, a useful upper bound would be $1 + O(e^{-(a-b)^2j})$ for $j > 1/(a-b)$. That this bound could be true is supported by experiments (although not very extensive). We can prove the bound $1 + O(e^{-(a-b)^2j})$ on the part of the above sum from $s=\lceil bj \rceil$ to $j$. But for the remaining part of the sum we can only obtain a bound of $O(\frac{e^{-(a-b)^2j}}{a-b})$.

One difficulty in bounding the above sum is that we don't know good lower bounds for $F^B_{j+1,b}(s)$ for small $s$.

Let $f^B_{j,a}(s)$ be the probability mass function of the binomial distribution, that is $f^B_{j,a}(s) = {j \choose s} a^s (1-a)^{j-s}$. And let $F^B_{j+1,b}(s)$ be the cdf of the binomial distribution, that is $F^B_{j+1,b}(s) = \sum_{i=0}^{s} f^B_{j+1,b}(i)$.

In my research, I need a quite precise upper bound on the sum

$$\sum_{s=0}^j \frac{f^B_{j,a}(s)}{F^B_{j+1,b}(s)}.$$

For $a>b$, a useful upper bound would be $1 + O(e^{-(a-b)^2j})$ for $j > 1/(a-b)$. That this bound could be true is supported by experiments (although not very extensive). We can prove the bound $1 + O(e^{-(a-b)^2j})$ on the part of the above sum from $s=\lceil bj \rceil$ to $j$. But for the remaining part of the sum we can only obtain a bound of $O(\frac{e^{-(a-b)^2j}}{a-b})$.

One difficulty in bounding the above sum is that we don't know good lower bounds for $F^B_{j+1,b}(s)$ for small $s$.

Let $f^B_{j,a}(s)$ be the probability mass function of the binomial distribution, that is $f^B_{j,a}(s) = {j \choose s} a^s (1-a)^s$. And let $F^B_{j+1,b}(s)$ be the cdf of the binomial distribution, that is $F^B_{j+1,b}(s) = \sum_{i=0}^{s} f^B_{j+1,b}(s)$.

In my research, I need a quite precise upper bound on the sum

$$\sum_{s=0}^j \frac{f^B_{j,a}(s)}{F^B_{j+1,b}(s)}.$$

For $a>b$, a useful upper bound would be $1 + O(e^{-(a-b)^2j})$ for $j > 1/(a-b)$. That this bound could be true is supported by experiments (although not very extensive). We can prove the bound $1 + O(e^{-(a-b)^2j})$ on the part of the above sum from $s=\lceil bj \rceil$ to $j$. But for the remaining part of the sum we can only obtain a bound of $O(\frac{e^{-(a-b)^2j}}{a-b})$.

One difficulty in bounding the above sum is that we don't know good lower bounds for $F^B_{j+1,b}(s)$ for small $s$.

Let $f^B_{j,a}(s)$ be the probability mass function of the binomial distribution, that is $f^B_{j,a}(s) = {j \choose s} a^s (1-a)^s$. And let $F^B_{j+1,b}(s)$ be the cdf of the binomial distribution, that is $F^B_{j+1,b}(s) = \sum_{i=0}^{s} f^B_{j+1,b}(i)$.

In my research, I need a quite precise upper bound on the sum

$$\sum_{s=0}^j \frac{f^B_{j,a}(s)}{F^B_{j+1,b}(s)}.$$

For $a>b$, a useful upper bound would be $1 + O(e^{-(a-b)^2j})$ for $j > 1/(a-b)$. That this bound could be true is supported by experiments (although not very extensive). We can prove the bound $1 + O(e^{-(a-b)^2j})$ on the part of the above sum from $s=\lceil bj \rceil$ to $j$. But for the remaining part of the sum we can only obtain a bound of $O(\frac{e^{-(a-b)^2j}}{a-b})$.

One difficulty in bounding the above sum is that we don't know good lower bounds for $F^B_{j+1,b}(s)$ for small $s$.