- Thread starter shanshuiii
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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_distribution#Related_distributions)

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

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.

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.

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

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] \)

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] \)

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

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.