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Thread: Standard deviation of the Ratio of two means

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    Standard deviation of the Ratio of two means




    Hi, guys

    I have two sets of data: A and B.

    I want to calculate a ratio of average(A)/average(B).

    I wonder if there is a way to get the standard deviation of this ratio?

    thanks.

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    Re: Standard deviation of the Ratio of two means

    Suppose avg(A) and avg(B) are normally distributed. This will be exactly true if A and B are normally distributed, because a sum of normally distributed deviates is itself normally distributed. It will be approximately true for large samples however A and B are distributed, by the central limit theorem.

    Then what you are asking is how the ratio of two independent normal deviates is distributed. The answer is the Cauchy distribution. (http://en.wikipedia.org/wiki/Normal_..._distributions)

    The Cauchy distribution has infinite variance, so your ratio's standard deviation is infinite.

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    Re: Standard deviation of the Ratio of two means

    http://en.wikipedia.org/wiki/Taylor_...ndom_variables

    You may try to replace the theoretical moments by the sample moments, provided that
    you have collected the sample means, sample variances and the sample covariance of the
    two data sets.

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    Re: Standard deviation of the Ratio of two means

    Quote Originally Posted by BGM View Post
    http://en.wikipedia.org/wiki/Taylor_...ndom_variables

    You may try to replace the theoretical moments by the sample moments, provided that
    you have collected the sample means, sample variances and the sample covariance of the
    two data sets.
    Thanks, this may work.

    Just one question, in the equation E[Y]^2, does it mean E([Y]^2) or (E[Y])^2?

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    Re: Standard deviation of the Ratio of two means

    Using Taylor Expansion, you can easily see the result:

    Let f(x, y) = \frac {x} {y}

    Then \frac {\partial f} {\partial x} = \frac {1} {y},
\frac {\partial f} {\partial y} = - \frac {x} {y^2}

    Taylor Expand at (\mu_X, \mu_Y), we have

    Var\left[\frac {X} {Y}\right]
\approx Var\left[\frac {\mu_X} {\mu_Y} + \frac {1} {\mu_Y}(X - \mu_X)
- \frac {\mu_X} {\mu_Y^2}(Y - \mu_Y) \right]

    = Var\left[\frac {1} {\mu_Y} X - \frac {\mu_X} {\mu_Y^2}Y\right]

    = \frac {1} {\mu_Y^2}Var[X] + \frac {\mu_X^2} {\mu_Y^4}Var[Y]
- 2\frac {\mu_X} {\mu_Y^3}Cov[X, Y]

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    Re: Standard deviation of the Ratio of two means

    Quote Originally Posted by BGM View Post
    Using Taylor Expansion, you can easily see the result:

    Let f(x, y) = \frac {x} {y}

    Then \frac {\partial f} {\partial x} = \frac {1} {y},
\frac {\partial f} {\partial y} = - \frac {x} {y^2}

    Taylor Expand at (\mu_X, \mu_Y), we have

    Var\left[\frac {X} {Y}\right]
\approx Var\left[\frac {\mu_X} {\mu_Y} + \frac {1} {\mu_Y}(X - \mu_X)
- \frac {\mu_X} {\mu_Y^2}(Y - \mu_Y) \right]

    = Var\left[\frac {1} {\mu_Y} X - \frac {\mu_X} {\mu_Y^2}Y\right]

    = \frac {1} {\mu_Y^2}Var[X] + \frac {\mu_X^2} {\mu_Y^4}Var[Y]
- 2\frac {\mu_X} {\mu_Y^3}Cov[X, Y]
    Thanks, I seem to get it.

    For my data, A and B are independent, does it mean that Cov(A,B)=0?

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    Re: Standard deviation of the Ratio of two means

    Yes if you accept that the independent assumption.

    Reminder: As what Ichbin mentioned above, the variance of the ratio is usually
    large if the random variable in the denominator have a high probability to
    fall in the neighborhood of 0. So make sure, if you have make any distributional
    assumptions, check whether the moment exist or not.

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    Re: Standard deviation of the Ratio of two means


    that ratio should follow the ratio normal dist. See this paper: D. V. Hinkley (December 1969). "On the Ratio of Two Correlated Normal Random Variables". Biometrika 56 (3): 635–639 and substitute the corresponding parameters. If the means are zero you should Cauchy dist even the variances of the A and B are not one. Also Cauchy dist is a special case of the ratio normal dist.

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