Simple lower bound on Gaussian CDF evaluated at sum: $G(s + t)$ in terms of $G(s)$, with $s, t ge 0$ and $s le 1$

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Let $G: s mapsto int_-infty^s g(s)ds$ be the CDF of the standard Gaussian (with $g(s) := (2pi)^-1/2exp(-s^2/2)$ the density) and $s le 0 le t$.



Question




what is a simple lower bound for $G(s + t)$ ?




For example, I'd be satisfied with a small-degree polynomial lower bound like




$$
G(s + t) ge G(s) + tg(s) + sum_k=1^m t^k+1g^(k)(s)/(k+1)! ge 0.
$$




Of course, the larger the upper bound the better.







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    Let $G: s mapsto int_-infty^s g(s)ds$ be the CDF of the standard Gaussian (with $g(s) := (2pi)^-1/2exp(-s^2/2)$ the density) and $s le 0 le t$.



    Question




    what is a simple lower bound for $G(s + t)$ ?




    For example, I'd be satisfied with a small-degree polynomial lower bound like




    $$
    G(s + t) ge G(s) + tg(s) + sum_k=1^m t^k+1g^(k)(s)/(k+1)! ge 0.
    $$




    Of course, the larger the upper bound the better.







    share|cite|improve this question























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      down vote

      favorite









      up vote
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      down vote

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      Let $G: s mapsto int_-infty^s g(s)ds$ be the CDF of the standard Gaussian (with $g(s) := (2pi)^-1/2exp(-s^2/2)$ the density) and $s le 0 le t$.



      Question




      what is a simple lower bound for $G(s + t)$ ?




      For example, I'd be satisfied with a small-degree polynomial lower bound like




      $$
      G(s + t) ge G(s) + tg(s) + sum_k=1^m t^k+1g^(k)(s)/(k+1)! ge 0.
      $$




      Of course, the larger the upper bound the better.







      share|cite|improve this question













      Let $G: s mapsto int_-infty^s g(s)ds$ be the CDF of the standard Gaussian (with $g(s) := (2pi)^-1/2exp(-s^2/2)$ the density) and $s le 0 le t$.



      Question




      what is a simple lower bound for $G(s + t)$ ?




      For example, I'd be satisfied with a small-degree polynomial lower bound like




      $$
      G(s + t) ge G(s) + tg(s) + sum_k=1^m t^k+1g^(k)(s)/(k+1)! ge 0.
      $$




      Of course, the larger the upper bound the better.









      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited yesterday
























      asked yesterday









      dohmatob

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