Robin
Posts: 1431
Joined: Sep. 2009
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| Quote (DiEb @ Oct. 22 2012,07:37) | | Quote (Robin @ Oct. 22 2012,02:02) | Got a statistics problem for anyone feeling bored:
(1+mi)*(1+mi+1)*(1+mi+2)*...(1+mn)=(1+X)^n
Suppose each value of M has a standard deviation associated with it. What is the standard deviation of X? Is it a simpler calculation if the standard deviation of each M is the same?
This is from my wife's friend. It's been over 25 years since I've done statistics so I'm rusty and trying to brush up. Any help would be appreciated.
Oh...and the notation is the way it is because I've not figured out how to get series fonts to work. |
Let me clarify:
1) You have a number of random variables M_1 .... M_n and you form the product R= (1+M_1)(1+M_2)...(1+M_n). Those M_i are real valued, independent, and are following an identical distribution.
2) Are asking whether there is a random variable X such that the product R equals (1+X)^n ?
Obviously if n is even and those M are following the Gaussian law then such an X doesn't exist - the right hand side is always positive, while the probability for the left hand to be negative is positive :-) |
Dieb -
1) I confess I'm confused on this point as well, but your summary is my take as well.
2) No, I don't think so. I think the friend is asking what the standard deviation of X would be given a standard deviation of Mi. On top of that, she wants to know whether the calculation is easier if the standard deviation of Mi is the same throughout the series. Seems to me that the calculation has the exact same difficulty either way as "calculation difficulty" seems somewhat subjective once you are dealing with standard deviations, but perhaps she's asking whether the notation of the calculation requires more variables if the standard deviation of Mi is not the same.
Your note on the Gaussian law gave me a chuckle. I don't think that my wife's friend was thinking in terms of physics when she posted the problem (but then, I don't actually know that) and I doubt it applies to any inverse square type calculation, but then that's a bit out of my area of knowledge anyway. I'm tempted to note the X likely can't exist for that reason though just to see her response. :-)
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we IDists rule in design for the flagellum and cilium largely because they do look designed. Bilbo
The only reason you reject Thor is because, like a cushion, you bear the imprint of the biggest arse that sat on you. Louis
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