Chebyshev’s inequality states that irrespective of the distribution of the underlying, 1-1/k^2 is used to determine the proportion. Can you use 1/k^2 also? Plz refer the following: With Chebyshev’s inequality, the maximum probability of the exchange rate being at least k standard deviations from the mean is P(| X − μ | ≥ kσ) ≤ (1/k)2. Isn’t the above statement wrong?

Sounds OK to me.

Why do you think that it’s wrong?

Shouldn’t we use 1-1/k^2 in the second statement? Why are we using 1/k^2 here?

Draw a picture. Color in the area to which Chebyshev’s inequality is usually applied. Then color in the area to which the statement in question applies.

Put simply, Chebychev’s inequality says that AT LEAST 1 - 1/*k*^{2} of the distribution’s values are within *k* standard deviations of the mean.

Alternately, it can be said to read that AT MOST, 1/*k*^{2} of the observations are greater than *k* standard deviations from the mean.

To see the equivalence of these two statements:

start with P(| X − μ | ≤ kσ) ≥ 1 - 1/*k*^{2 }(i.e. AT LEAST 1 - 1/*k*^{2} of the distribution’s values are within *k* standard deviations of the mean)

Add 1/*k*^{2} to both sides and subtract P(| X − μ | ≤ kσ) from both sides. you get

1/*k*^{2} ≥ 1 - P(| X − μ | ≤ kσ)

then use the fact that P(| X − μ | > kσ) = 1 - P(| X − μ | ≤ kσ) and with the substitution, you get

1/*k*^{2} ≥ P(| X − μ | > kσ) (i.e. NO MORE THAN 1/*k*^{2} of the distribution’s values are greater than *k* standard deviations from the mean)

Great! Now he doesn’t have to draw a picture.

Thanks both for the explanation.

Actually, I think the “draw a picture” method works better for most people in most circumstances, Just wanted to get my geek on for a bit (been a while since I did much math).

Well . . . your explanation was spot on.

Thankee, sah.