upper bound for variance when using estimated values

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Let’s say I have a parameterspace $Omega$ with known probability density function $p(omega), omega in Omega$. I want to estimate the expectation $E[x]$ of a variable $x$ depending on $omega in Omega$. I cannot calculate directly $x$ but I can make an estimation $tildex$ such that $tildex- x leq epsilon, epsilon > 0$ and $x leq tildex$.



It’s easy to prove that $|E[tildex] – E[x]| leq epsilon$. Is there a way to calculate a similar upper bound for $|Var[tildex] – Var[x] |$?



Kind regards,
Koen






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    Let’s say I have a parameterspace $Omega$ with known probability density function $p(omega), omega in Omega$. I want to estimate the expectation $E[x]$ of a variable $x$ depending on $omega in Omega$. I cannot calculate directly $x$ but I can make an estimation $tildex$ such that $tildex- x leq epsilon, epsilon > 0$ and $x leq tildex$.



    It’s easy to prove that $|E[tildex] – E[x]| leq epsilon$. Is there a way to calculate a similar upper bound for $|Var[tildex] – Var[x] |$?



    Kind regards,
    Koen






    probability statistics







    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Let’s say I have a parameterspace $Omega$ with known probability density function $p(omega), omega in Omega$. I want to estimate the expectation $E[x]$ of a variable $x$ depending on $omega in Omega$. I cannot calculate directly $x$ but I can make an estimation $tildex$ such that $tildex- x leq epsilon, epsilon > 0$ and $x leq tildex$.



      It’s easy to prove that $|E[tildex] – E[x]| leq epsilon$. Is there a way to calculate a similar upper bound for $|Var[tildex] – Var[x] |$?



      Kind regards,
      Koen






      probability statistics







      share|cite|improve this question













      Let’s say I have a parameterspace $Omega$ with known probability density function $p(omega), omega in Omega$. I want to estimate the expectation $E[x]$ of a variable $x$ depending on $omega in Omega$. I cannot calculate directly $x$ but I can make an estimation $tildex$ such that $tildex- x leq epsilon, epsilon > 0$ and $x leq tildex$.



      It’s easy to prove that $|E[tildex] – E[x]| leq epsilon$. Is there a way to calculate a similar upper bound for $|Var[tildex] – Var[x] |$?



      Kind regards,
      Koen






      probability statistics




      probability statistics






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      asked Sep 9 at 8:18









      Koen

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          begineqnarray*
          mathsfVar[tilde x]-mathsfVar[x]
          &=&
          mathsfVar[x+(tilde x-x)]-mathsfVar[x]
          \
          &=&
          mathsfVar[x]+mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x]-mathsfVar[x]
          \
          &=&
          mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x]
          \
          &le&
          2epsilonsqrtmathsfVar[x]+epsilon^2;.
          endeqnarray*



          A slightly more natural and perhaps more useful result might be



          begineqnarray*
          fracmathsfVar[tilde x]mathsfVar[x]
          &=&
          fracmathsfVar[x+(tilde x-x)]mathsfVar[x]
          \
          &=&
          fracmathsfVar[x]+mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x]mathsfVar[x]
          \
          &le&
          1+2fracepsilonsqrtmathsfVar[x]+left(fracepsilonsqrtmathsfVar[x]right)^2;.
          endeqnarray*



          (I left out all the absolute values signs for readability, but the inequality still holds if you add them.)






          share|cite|improve this answer




















          • Can you explain why $mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x] leq 2epsilonsqrtmathsfVar[x]+epsilon^2$? (the inequality in the first part)
            – Koen
            Sep 10 at 10:00










          • @Koen: $mathsfVar[tilde x-x]leqepsilon^2$ is an immediate consequence of $|tildex-x|leqepsilon$, and $mathsfCov[x,tilde x-x]leqepsilonsqrtmathsfVar[x]$ is a form of the Cauchy–Schwarz inequality; see here in Wikipedia.
            – joriki
            Sep 10 at 10:07











          • I think $mathsfVar[tilde x-x]leqepsilon^2$ is not in general true if $|tildex-x|leqepsilon$. Check the example: $x = 1, 2, epsilon = 0.5$ and $tilde x = x + epsilon, -epsilon = 1.5, 1.5$, then $mathsfVar[tilde x-x] = 0.5$ while $epsilon^2 = 0.25$ (just assume that the probability on both elements in $x$ is 0.5)
            – Koen
            Sep 10 at 12:11











          • @Koen: Perhaps you're confusing the variance with the standard variation? The variance in your example is $epsilon^2=0.25$; the standard variation, the square root of the variance, is $epsilon=0.5$.
            – joriki
            Sep 10 at 14:07










          • Ok, you're right,
            – Koen
            Sep 10 at 14:13











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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

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          up vote
          0
          down vote



          accepted










          begineqnarray*
          mathsfVar[tilde x]-mathsfVar[x]
          &=&
          mathsfVar[x+(tilde x-x)]-mathsfVar[x]
          \
          &=&
          mathsfVar[x]+mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x]-mathsfVar[x]
          \
          &=&
          mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x]
          \
          &le&
          2epsilonsqrtmathsfVar[x]+epsilon^2;.
          endeqnarray*



          A slightly more natural and perhaps more useful result might be



          begineqnarray*
          fracmathsfVar[tilde x]mathsfVar[x]
          &=&
          fracmathsfVar[x+(tilde x-x)]mathsfVar[x]
          \
          &=&
          fracmathsfVar[x]+mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x]mathsfVar[x]
          \
          &le&
          1+2fracepsilonsqrtmathsfVar[x]+left(fracepsilonsqrtmathsfVar[x]right)^2;.
          endeqnarray*



          (I left out all the absolute values signs for readability, but the inequality still holds if you add them.)






          share|cite|improve this answer




















          • Can you explain why $mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x] leq 2epsilonsqrtmathsfVar[x]+epsilon^2$? (the inequality in the first part)
            – Koen
            Sep 10 at 10:00










          • @Koen: $mathsfVar[tilde x-x]leqepsilon^2$ is an immediate consequence of $|tildex-x|leqepsilon$, and $mathsfCov[x,tilde x-x]leqepsilonsqrtmathsfVar[x]$ is a form of the Cauchy–Schwarz inequality; see here in Wikipedia.
            – joriki
            Sep 10 at 10:07











          • I think $mathsfVar[tilde x-x]leqepsilon^2$ is not in general true if $|tildex-x|leqepsilon$. Check the example: $x = 1, 2, epsilon = 0.5$ and $tilde x = x + epsilon, -epsilon = 1.5, 1.5$, then $mathsfVar[tilde x-x] = 0.5$ while $epsilon^2 = 0.25$ (just assume that the probability on both elements in $x$ is 0.5)
            – Koen
            Sep 10 at 12:11











          • @Koen: Perhaps you're confusing the variance with the standard variation? The variance in your example is $epsilon^2=0.25$; the standard variation, the square root of the variance, is $epsilon=0.5$.
            – joriki
            Sep 10 at 14:07










          • Ok, you're right,
            – Koen
            Sep 10 at 14:13















          up vote
          0
          down vote



          accepted










          begineqnarray*
          mathsfVar[tilde x]-mathsfVar[x]
          &=&
          mathsfVar[x+(tilde x-x)]-mathsfVar[x]
          \
          &=&
          mathsfVar[x]+mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x]-mathsfVar[x]
          \
          &=&
          mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x]
          \
          &le&
          2epsilonsqrtmathsfVar[x]+epsilon^2;.
          endeqnarray*



          A slightly more natural and perhaps more useful result might be



          begineqnarray*
          fracmathsfVar[tilde x]mathsfVar[x]
          &=&
          fracmathsfVar[x+(tilde x-x)]mathsfVar[x]
          \
          &=&
          fracmathsfVar[x]+mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x]mathsfVar[x]
          \
          &le&
          1+2fracepsilonsqrtmathsfVar[x]+left(fracepsilonsqrtmathsfVar[x]right)^2;.
          endeqnarray*



          (I left out all the absolute values signs for readability, but the inequality still holds if you add them.)






          share|cite|improve this answer




















          • Can you explain why $mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x] leq 2epsilonsqrtmathsfVar[x]+epsilon^2$? (the inequality in the first part)
            – Koen
            Sep 10 at 10:00










          • @Koen: $mathsfVar[tilde x-x]leqepsilon^2$ is an immediate consequence of $|tildex-x|leqepsilon$, and $mathsfCov[x,tilde x-x]leqepsilonsqrtmathsfVar[x]$ is a form of the Cauchy–Schwarz inequality; see here in Wikipedia.
            – joriki
            Sep 10 at 10:07











          • I think $mathsfVar[tilde x-x]leqepsilon^2$ is not in general true if $|tildex-x|leqepsilon$. Check the example: $x = 1, 2, epsilon = 0.5$ and $tilde x = x + epsilon, -epsilon = 1.5, 1.5$, then $mathsfVar[tilde x-x] = 0.5$ while $epsilon^2 = 0.25$ (just assume that the probability on both elements in $x$ is 0.5)
            – Koen
            Sep 10 at 12:11











          • @Koen: Perhaps you're confusing the variance with the standard variation? The variance in your example is $epsilon^2=0.25$; the standard variation, the square root of the variance, is $epsilon=0.5$.
            – joriki
            Sep 10 at 14:07










          • Ok, you're right,
            – Koen
            Sep 10 at 14:13













          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          begineqnarray*
          mathsfVar[tilde x]-mathsfVar[x]
          &=&
          mathsfVar[x+(tilde x-x)]-mathsfVar[x]
          \
          &=&
          mathsfVar[x]+mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x]-mathsfVar[x]
          \
          &=&
          mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x]
          \
          &le&
          2epsilonsqrtmathsfVar[x]+epsilon^2;.
          endeqnarray*



          A slightly more natural and perhaps more useful result might be



          begineqnarray*
          fracmathsfVar[tilde x]mathsfVar[x]
          &=&
          fracmathsfVar[x+(tilde x-x)]mathsfVar[x]
          \
          &=&
          fracmathsfVar[x]+mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x]mathsfVar[x]
          \
          &le&
          1+2fracepsilonsqrtmathsfVar[x]+left(fracepsilonsqrtmathsfVar[x]right)^2;.
          endeqnarray*



          (I left out all the absolute values signs for readability, but the inequality still holds if you add them.)






          share|cite|improve this answer












          begineqnarray*
          mathsfVar[tilde x]-mathsfVar[x]
          &=&
          mathsfVar[x+(tilde x-x)]-mathsfVar[x]
          \
          &=&
          mathsfVar[x]+mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x]-mathsfVar[x]
          \
          &=&
          mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x]
          \
          &le&
          2epsilonsqrtmathsfVar[x]+epsilon^2;.
          endeqnarray*



          A slightly more natural and perhaps more useful result might be



          begineqnarray*
          fracmathsfVar[tilde x]mathsfVar[x]
          &=&
          fracmathsfVar[x+(tilde x-x)]mathsfVar[x]
          \
          &=&
          fracmathsfVar[x]+mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x]mathsfVar[x]
          \
          &le&
          1+2fracepsilonsqrtmathsfVar[x]+left(fracepsilonsqrtmathsfVar[x]right)^2;.
          endeqnarray*



          (I left out all the absolute values signs for readability, but the inequality still holds if you add them.)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 9 at 9:13









          joriki

          169k10181337




          169k10181337











          • Can you explain why $mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x] leq 2epsilonsqrtmathsfVar[x]+epsilon^2$? (the inequality in the first part)
            – Koen
            Sep 10 at 10:00










          • @Koen: $mathsfVar[tilde x-x]leqepsilon^2$ is an immediate consequence of $|tildex-x|leqepsilon$, and $mathsfCov[x,tilde x-x]leqepsilonsqrtmathsfVar[x]$ is a form of the Cauchy–Schwarz inequality; see here in Wikipedia.
            – joriki
            Sep 10 at 10:07











          • I think $mathsfVar[tilde x-x]leqepsilon^2$ is not in general true if $|tildex-x|leqepsilon$. Check the example: $x = 1, 2, epsilon = 0.5$ and $tilde x = x + epsilon, -epsilon = 1.5, 1.5$, then $mathsfVar[tilde x-x] = 0.5$ while $epsilon^2 = 0.25$ (just assume that the probability on both elements in $x$ is 0.5)
            – Koen
            Sep 10 at 12:11











          • @Koen: Perhaps you're confusing the variance with the standard variation? The variance in your example is $epsilon^2=0.25$; the standard variation, the square root of the variance, is $epsilon=0.5$.
            – joriki
            Sep 10 at 14:07










          • Ok, you're right,
            – Koen
            Sep 10 at 14:13

















          • Can you explain why $mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x] leq 2epsilonsqrtmathsfVar[x]+epsilon^2$? (the inequality in the first part)
            – Koen
            Sep 10 at 10:00










          • @Koen: $mathsfVar[tilde x-x]leqepsilon^2$ is an immediate consequence of $|tildex-x|leqepsilon$, and $mathsfCov[x,tilde x-x]leqepsilonsqrtmathsfVar[x]$ is a form of the Cauchy–Schwarz inequality; see here in Wikipedia.
            – joriki
            Sep 10 at 10:07











          • I think $mathsfVar[tilde x-x]leqepsilon^2$ is not in general true if $|tildex-x|leqepsilon$. Check the example: $x = 1, 2, epsilon = 0.5$ and $tilde x = x + epsilon, -epsilon = 1.5, 1.5$, then $mathsfVar[tilde x-x] = 0.5$ while $epsilon^2 = 0.25$ (just assume that the probability on both elements in $x$ is 0.5)
            – Koen
            Sep 10 at 12:11











          • @Koen: Perhaps you're confusing the variance with the standard variation? The variance in your example is $epsilon^2=0.25$; the standard variation, the square root of the variance, is $epsilon=0.5$.
            – joriki
            Sep 10 at 14:07










          • Ok, you're right,
            – Koen
            Sep 10 at 14:13
















          Can you explain why $mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x] leq 2epsilonsqrtmathsfVar[x]+epsilon^2$? (the inequality in the first part)
          – Koen
          Sep 10 at 10:00




          Can you explain why $mathsfVar[tilde x-x]+2mathsfCov[x,tilde x-x] leq 2epsilonsqrtmathsfVar[x]+epsilon^2$? (the inequality in the first part)
          – Koen
          Sep 10 at 10:00












          @Koen: $mathsfVar[tilde x-x]leqepsilon^2$ is an immediate consequence of $|tildex-x|leqepsilon$, and $mathsfCov[x,tilde x-x]leqepsilonsqrtmathsfVar[x]$ is a form of the Cauchy–Schwarz inequality; see here in Wikipedia.
          – joriki
          Sep 10 at 10:07





          @Koen: $mathsfVar[tilde x-x]leqepsilon^2$ is an immediate consequence of $|tildex-x|leqepsilon$, and $mathsfCov[x,tilde x-x]leqepsilonsqrtmathsfVar[x]$ is a form of the Cauchy–Schwarz inequality; see here in Wikipedia.
          – joriki
          Sep 10 at 10:07













          I think $mathsfVar[tilde x-x]leqepsilon^2$ is not in general true if $|tildex-x|leqepsilon$. Check the example: $x = 1, 2, epsilon = 0.5$ and $tilde x = x + epsilon, -epsilon = 1.5, 1.5$, then $mathsfVar[tilde x-x] = 0.5$ while $epsilon^2 = 0.25$ (just assume that the probability on both elements in $x$ is 0.5)
          – Koen
          Sep 10 at 12:11





          I think $mathsfVar[tilde x-x]leqepsilon^2$ is not in general true if $|tildex-x|leqepsilon$. Check the example: $x = 1, 2, epsilon = 0.5$ and $tilde x = x + epsilon, -epsilon = 1.5, 1.5$, then $mathsfVar[tilde x-x] = 0.5$ while $epsilon^2 = 0.25$ (just assume that the probability on both elements in $x$ is 0.5)
          – Koen
          Sep 10 at 12:11













          @Koen: Perhaps you're confusing the variance with the standard variation? The variance in your example is $epsilon^2=0.25$; the standard variation, the square root of the variance, is $epsilon=0.5$.
          – joriki
          Sep 10 at 14:07




          @Koen: Perhaps you're confusing the variance with the standard variation? The variance in your example is $epsilon^2=0.25$; the standard variation, the square root of the variance, is $epsilon=0.5$.
          – joriki
          Sep 10 at 14:07












          Ok, you're right,
          – Koen
          Sep 10 at 14:13





          Ok, you're right,
          – Koen
          Sep 10 at 14:13


















           

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