Optimized Chernoff Bound I'm trying to find the optimized Chernoff bound of a sum of i.i.d. discrete random variables Xi, 0$\leq$ i < n, with support {-1,0,1}.  I also know that pX0(1) > pX0(-1).  However, I'm not sure if I optimized the bound correctly over t.  
My attempt looks like this:
I computed the moment generating function of X0 as 
MX0(t) = E[$e^{tX0}$] = $\sum_{k=1}^{\infty} e^{tX0(k)}$pX0(k) = $e^{t(-1)}$pX0(-1) + $e^{t(1)}$pX0(1) + pX0(0) 
I then found the moment generating function of the sum of the random variables from 0 to n-1 by multiplying their respective moment generating functions: 
MSn = M[$\sum_{i=0}^{n-1}$ Xi] = MX0 MX1 ... MXn-1 = [$e^{-t}$pX0(-1) + $e^{t}$pX0(1)+pX0(0)]$^{n}$ 
The optimized Chernoff bound for Pr[Sn$\le$a] is given by 
Pr[Sn$\le$a] $\le$ min $e^{ta}$ $\prod _ { i = 0 } ^ n E[e^{-tXi}]$ 
Pr[Sn$\le$0] $\le$ [$e^{-t}$pX0(-1) + $e^{t}$pX0(1) + pX0(0)]$^{n}$
Letting c1 = pX0(-1) and c2 = pX0(1), I then minimize the above expression by taking its derivative and setting it equal to 0 to obtain
-c1$e^{-t}$ + c2$e^{t}$ = 0 
c2$e^{t}$ = c1$e^{-t}$ 
$e^{2t}$ = c1/c2 
2t = ln(c1/c2) 
t = $\frac 1 { 2 }$ln$\frac {c1} {c2}$ 
Substituting the optimized value of t into the generic Chernoff bound I obtain
Pr(Sn$\leq$0) $\leq$ [($\frac {pX0(-1)} {pX0(1)}$) $^{-1/2}$ pX0(-1) + ($\frac {pX0(-1)} {pX0(1)}$) $^{1/2}$ pX0(1) + pX0(0)]$^n$
Is this the correct optimized Chernoff bound for Pr[Sn$\le$0]?  Thanks for your help and insights.

Your calculation seems to be correct but your final value of t  is less than 0 as probability of getting a 1 is greater than getting a  -1 as you stated earlier. So you can't  use that value if you actually try to minimize the expression you have to put t equals to 0 which will  be of no use.
