Bounds of Q(x)

  • Thread starter iVenky
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Main Question or Discussion Point

I think everyone knows that

Q(x)= P(X>x) where X is a Gaussian Random variable.

Now I was reading about it and it says that Q(x) is bounded as follows

Q(x)≤ (1/2)(e-x2/2) for x≥0

and

Q(x)< [1/(√(2∏)x)](e-x2/2) for x≥0

and the lower bound is

Q(x)> [1/(√(2∏)x)](1-1/x2) e-x2/2 for x≥0

Can you tell me how you get this?


Thanks a lot.
 
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Answers and Replies

  • #2
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One example for the first inequality: It is exact at x=0, as you can check. For [itex]0<x<\frac{1}{\sqrt{2pi}}[/itex], the derivative of the upper estimate is larger (negative with a smaller magnitude) than the derivative of Q(x), which is simply the normal distribution. Therefore, the upper estimate is valid.

In the same way, for all larger x, consider the limit of both for x->inf: It is 0. Now, the upper bound has a smaller derivative (negative with larger magnitude) everywhere, therefore it is valid there, too.

I would expect that you can get the other inequalities with similar methods.
 

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