Timeline for Calculating a specific joint probability involving sums of binomial distributions
Current License: CC BY-SA 3.0
4 events
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| Oct 5, 2012 at 20:16 | comment | added | val11 | As far as I can see, when applying your strategy to find an upper bound for $f(x,k)$, I get something like: $\leq \Pr[Z(X) \leq k-\frac{X}{2}]^2\Pr[Y(X) \geq \frac{X}{2}] + \Pr[Z(X) \leq k]^2 \Pr[Y(X) \leq \frac{X}{2}]$ which is not very tight (at a first glance - $\Pr[Z(X) \leq k]$ seems to get quite big).. I'll calculate things more thoroughly tomorrow - maybe this is actually good enough... | |
| Oct 5, 2012 at 20:16 | comment | added | val11 | What I am searching for is an upper bound on $\Pr[Y(X) + Z_1(X) \leq k \wedge Y(X) + Z_2(X) \leq k]$. I originally had the problem of $\Pr[Y(X) + Z_1(X) \leq k \wedge Y(X) + Z_2(X) \geq k+1]$, which I showed was equal to the following difference: $\Pr[Y(X) + Z(X) \leq k] - \Pr[Y(X) + Z_1(X) \leq k \wedge Y(X) + Z_2(X) \leq k]$. The first term is trivial to compute, so I only need an upper bound on the second term. | |
| Oct 5, 2012 at 20:16 | comment | added | val11 | Hi Magnus. I am searching for the average value of $f(x,k)$ over all $x$. I am just looking at it for fixed $x$, if that could make things easier. | |
| Oct 5, 2012 at 18:22 | history | answered | Magnus Find | CC BY-SA 3.0 |