Skewness

**okthenso** · January 26th 2009, 12:53 AM

have to find the mean, variance and skewness and kurtosis for various distributions.
Take the bernoulli distribution where the mean (u) is p and the variance is p(1-p)

I'm having trouble with the skewness. I know the formula for it...

E[(x - u)^3] / sd^3

and that I cube out the top to get E[x^3 - 3x^2u etc...

My problem where to go from here. I'll end up with a E[x^3], do I have to go back and calculate that then, and the same for every other E[...] I'll end up with?

Also, is the bottom line just the variance to the power of 3/2?

Some similar help on the kurtosis would be useful is possible.
Thanks

**mr fantastic** · January 26th 2009, 01:54 AM

Originally Posted by okthenso

have to find the mean, variance and skewness and kurtosis for various distributions.
Take the bernoulli distribution where the mean (u) is p and the variance is p(1-p)

I'm having trouble with the skewness. I know the formula for it...

E[(x - u)^3] / sd^3

and that I cube out the top to get E[x^3 - 3x^2u etc...

My problem where to go from here. I'll end up with a E[x^3], do I have to go back and calculate that then, Mr F says: Yes.

and the same for every other E[...] I'll end up with?

Also, is the bottom line just the variance to the power of 3/2? Mr F says: Yes.

Some similar help on the kurtosis would be useful is possible.
Thanks

Yes, you have to calculate $E(X^3)$ . Use the definition.

And to get kurtosis you'll need to calculate $E(X^4)$ as well.

But on the bright side, note that:

$E(X) = \mu = p$

$E(X^2) = \sigma^2 + [E(X)]^2 = \sigma^2 + \mu^2 = p(1-p) + p = 2p - p^2$ .

**okthenso** · January 26th 2009, 02:28 AM

Originally Posted by mr fantastic

Yes, you have to calculate $E(X^3)$ . Use the definition.

And to get kurtosis you'll need to calculate $E(X^4)$ as well.

But on the bright side, note that:

$E(X) = \mu = p$

$E(X^2) = \sigma^2 + [E(X)]^2 = \sigma^2 + \mu^2 = p(1-p) + p = 2p - p^2$ .

Thanks... one last question.... when I have the E[x^3 - 3x^2p...]

Will the second part of that be 3pE[x^2], or do I calculate E[3x^2p]?

**mr fantastic** · January 26th 2009, 02:51 AM

Originally Posted by okthenso

Thanks... one last question.... when I have the E[x^3 - 3x^2p...]

Will the second part of that be 3pE[x^2], or do I calculate E[3x^2p]?

E(aX + b) = a E(X) + b.

**mr fantastic** · January 26th 2009, 04:01 AM

Originally Posted by okthenso

have to find the mean, variance and skewness and kurtosis for various distributions.
Take the bernoulli distribution where the mean (u) is p and the variance is p(1-p)

I'm having trouble with the skewness. I know the formula for it...

E[(x - u)^3] / sd^3

and that I cube out the top to get E[x^3 - 3x^2u etc...

My problem where to go from here. I'll end up with a E[x^3], do I have to go back and calculate that then, and the same for every other E[...] I'll end up with?

Also, is the bottom line just the variance to the power of 3/2?

Some similar help on the kurtosis would be useful is possible.
Thanks

Alternatively, you could use the pmf of the Bernoulli distribution to directly get $E((X - p)^3) = (0-p)^3 q + (1 - p)^3 p = p q (q - p)$ ....

Similarly for kurtosis.

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