Upper bound for the probability of violating a set of conditions












1












$begingroup$


Let $X_1,ldots,X_n$ be independent and normally distributed $mathcal{N}(bar{x},sigma^2)$ random variables. Let $g:{rm R} to [0,bar{g}]$ be a decreasing bounded function. Let $a$, $lambda$ and $l_{i,j}$ be positive constants, with $sum_j l_{i,j}=1$ for all $i$. Let $x^{ast}$ denote the unique solution to



begin{align}
x^{ast}-{rm E}[X_i , | , X_i <x^{ast}]-lambda g(x^{ast})=0
end{align}



Consider the set of conditions:
begin{align}
sum_{j=1}^n textbf{1}_{X_j in (-infty,x^{ast})}l_{i,j}{rm E}[X_j , | , X_j <x^{ast}] + sum_{j=1}^n textbf{1}_{X_j in [x^{ast},+infty)}l_{i,j}X_j -sum_{j=1}^n textbf{1}_{X_j in [x^{ast},+infty)}(l_{i,j-1}+l_{i,j+1})g(X_j) - X_i-textbf{1}_{X_{i-1} in [x^{ast},+infty)}g(X_{i-1})-textbf{1}_{X_{i+1} in [x^{ast},+infty)}g(X_{i+1})geq -a qquad i =1,ldots,n
end{align}



Can someone find an upper bound that does not depend on $x^{ast}$ for the probability that the set of conditions above does not hold (i.e., that at least one of the inequalities does not hold)?



EDIT: If I could show that



begin{align}
sum_{j=1}^n textbf{1}_{X_j in (-infty,x^{ast})}l_{i,j}{rm E}[X_j , | , X_j <x^{ast}] + sum_{j=1}^n textbf{1}_{X_j in [x^{ast},+infty)}l_{i,j}X_j end{align}



is concave in $x^{ast}$, or if I could impose that in terms of the primitives, then I would be able to find that upper bound. For instance, if the above expression is concave, then I can find an upper bound for the probability that the inequality does not hold for $i$ given by



begin{align}
maxleft{1-Phileft(dfrac{a-2bar{g}}{sigmaleft[sum_{jneq i}l_{i,j}^2+left(1-l_{i,i}right)^2right]^{frac{1}{2}}}right),,,1-Phileft(dfrac{a}{sigma}right)right}
end{align}
where $Phi$ is the standard normal CDF.



Can anyone show that the above expression is concave in $x^{ast}$, or can anyone impose concavity in terms of the primitives?






probability normal-distribution conditional-probability upper-lower-bounds







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $X_1,ldots,X_n$ be independent and normally distributed $mathcal{N}(bar{x},sigma^2)$ random variables. Let $g:{rm R} to [0,bar{g}]$ be a decreasing bounded function. Let $a$, $lambda$ and $l_{i,j}$ be positive constants, with $sum_j l_{i,j}=1$ for all $i$. Let $x^{ast}$ denote the unique solution to



    begin{align}
    x^{ast}-{rm E}[X_i , | , X_i <x^{ast}]-lambda g(x^{ast})=0
    end{align}



    Consider the set of conditions:
    begin{align}
    sum_{j=1}^n textbf{1}_{X_j in (-infty,x^{ast})}l_{i,j}{rm E}[X_j , | , X_j <x^{ast}] + sum_{j=1}^n textbf{1}_{X_j in [x^{ast},+infty)}l_{i,j}X_j -sum_{j=1}^n textbf{1}_{X_j in [x^{ast},+infty)}(l_{i,j-1}+l_{i,j+1})g(X_j) - X_i-textbf{1}_{X_{i-1} in [x^{ast},+infty)}g(X_{i-1})-textbf{1}_{X_{i+1} in [x^{ast},+infty)}g(X_{i+1})geq -a qquad i =1,ldots,n
    end{align}



    Can someone find an upper bound that does not depend on $x^{ast}$ for the probability that the set of conditions above does not hold (i.e., that at least one of the inequalities does not hold)?



    EDIT: If I could show that



    begin{align}
    sum_{j=1}^n textbf{1}_{X_j in (-infty,x^{ast})}l_{i,j}{rm E}[X_j , | , X_j <x^{ast}] + sum_{j=1}^n textbf{1}_{X_j in [x^{ast},+infty)}l_{i,j}X_j end{align}



    is concave in $x^{ast}$, or if I could impose that in terms of the primitives, then I would be able to find that upper bound. For instance, if the above expression is concave, then I can find an upper bound for the probability that the inequality does not hold for $i$ given by



    begin{align}
    maxleft{1-Phileft(dfrac{a-2bar{g}}{sigmaleft[sum_{jneq i}l_{i,j}^2+left(1-l_{i,i}right)^2right]^{frac{1}{2}}}right),,,1-Phileft(dfrac{a}{sigma}right)right}
    end{align}
    where $Phi$ is the standard normal CDF.



    Can anyone show that the above expression is concave in $x^{ast}$, or can anyone impose concavity in terms of the primitives?






    probability normal-distribution conditional-probability upper-lower-bounds







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $X_1,ldots,X_n$ be independent and normally distributed $mathcal{N}(bar{x},sigma^2)$ random variables. Let $g:{rm R} to [0,bar{g}]$ be a decreasing bounded function. Let $a$, $lambda$ and $l_{i,j}$ be positive constants, with $sum_j l_{i,j}=1$ for all $i$. Let $x^{ast}$ denote the unique solution to



      begin{align}
      x^{ast}-{rm E}[X_i , | , X_i <x^{ast}]-lambda g(x^{ast})=0
      end{align}



      Consider the set of conditions:
      begin{align}
      sum_{j=1}^n textbf{1}_{X_j in (-infty,x^{ast})}l_{i,j}{rm E}[X_j , | , X_j <x^{ast}] + sum_{j=1}^n textbf{1}_{X_j in [x^{ast},+infty)}l_{i,j}X_j -sum_{j=1}^n textbf{1}_{X_j in [x^{ast},+infty)}(l_{i,j-1}+l_{i,j+1})g(X_j) - X_i-textbf{1}_{X_{i-1} in [x^{ast},+infty)}g(X_{i-1})-textbf{1}_{X_{i+1} in [x^{ast},+infty)}g(X_{i+1})geq -a qquad i =1,ldots,n
      end{align}



      Can someone find an upper bound that does not depend on $x^{ast}$ for the probability that the set of conditions above does not hold (i.e., that at least one of the inequalities does not hold)?



      EDIT: If I could show that



      begin{align}
      sum_{j=1}^n textbf{1}_{X_j in (-infty,x^{ast})}l_{i,j}{rm E}[X_j , | , X_j <x^{ast}] + sum_{j=1}^n textbf{1}_{X_j in [x^{ast},+infty)}l_{i,j}X_j end{align}



      is concave in $x^{ast}$, or if I could impose that in terms of the primitives, then I would be able to find that upper bound. For instance, if the above expression is concave, then I can find an upper bound for the probability that the inequality does not hold for $i$ given by



      begin{align}
      maxleft{1-Phileft(dfrac{a-2bar{g}}{sigmaleft[sum_{jneq i}l_{i,j}^2+left(1-l_{i,i}right)^2right]^{frac{1}{2}}}right),,,1-Phileft(dfrac{a}{sigma}right)right}
      end{align}
      where $Phi$ is the standard normal CDF.



      Can anyone show that the above expression is concave in $x^{ast}$, or can anyone impose concavity in terms of the primitives?






      probability normal-distribution conditional-probability upper-lower-bounds







      share|cite|improve this question











      $endgroup$




      Let $X_1,ldots,X_n$ be independent and normally distributed $mathcal{N}(bar{x},sigma^2)$ random variables. Let $g:{rm R} to [0,bar{g}]$ be a decreasing bounded function. Let $a$, $lambda$ and $l_{i,j}$ be positive constants, with $sum_j l_{i,j}=1$ for all $i$. Let $x^{ast}$ denote the unique solution to



      begin{align}
      x^{ast}-{rm E}[X_i , | , X_i <x^{ast}]-lambda g(x^{ast})=0
      end{align}



      Consider the set of conditions:
      begin{align}
      sum_{j=1}^n textbf{1}_{X_j in (-infty,x^{ast})}l_{i,j}{rm E}[X_j , | , X_j <x^{ast}] + sum_{j=1}^n textbf{1}_{X_j in [x^{ast},+infty)}l_{i,j}X_j -sum_{j=1}^n textbf{1}_{X_j in [x^{ast},+infty)}(l_{i,j-1}+l_{i,j+1})g(X_j) - X_i-textbf{1}_{X_{i-1} in [x^{ast},+infty)}g(X_{i-1})-textbf{1}_{X_{i+1} in [x^{ast},+infty)}g(X_{i+1})geq -a qquad i =1,ldots,n
      end{align}



      Can someone find an upper bound that does not depend on $x^{ast}$ for the probability that the set of conditions above does not hold (i.e., that at least one of the inequalities does not hold)?



      EDIT: If I could show that



      begin{align}
      sum_{j=1}^n textbf{1}_{X_j in (-infty,x^{ast})}l_{i,j}{rm E}[X_j , | , X_j <x^{ast}] + sum_{j=1}^n textbf{1}_{X_j in [x^{ast},+infty)}l_{i,j}X_j end{align}



      is concave in $x^{ast}$, or if I could impose that in terms of the primitives, then I would be able to find that upper bound. For instance, if the above expression is concave, then I can find an upper bound for the probability that the inequality does not hold for $i$ given by



      begin{align}
      maxleft{1-Phileft(dfrac{a-2bar{g}}{sigmaleft[sum_{jneq i}l_{i,j}^2+left(1-l_{i,i}right)^2right]^{frac{1}{2}}}right),,,1-Phileft(dfrac{a}{sigma}right)right}
      end{align}
      where $Phi$ is the standard normal CDF.



      Can anyone show that the above expression is concave in $x^{ast}$, or can anyone impose concavity in terms of the primitives?






      probability normal-distribution conditional-probability upper-lower-bounds




      probability normal-distribution conditional-probability upper-lower-bounds






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Dec 16 '18 at 17:04







      capadocia

















      asked Dec 15 '18 at 21:52









      capadociacapadocia

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