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Rainshadow

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Incredible.

Many forecasts back in November didn't see this coming.

2 nice winters in a row.

Should be a 3rd next season ;).

Even 2008-09 wasn't a complete disaster; it was fairly cold with 22.8" of snow at the airport...a fraction above normal.

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What's the average for back to back winters? At first I was thinking just double the yearly average, but I for some reason doubt that this is correct.

E(x + y) = E(x) + E(y), assuming x and y are identically, independently distributed - which I think is a safe assumption. Therefore,

E(Year1 + Year2) = E(Year1) + E(Year2) = xbar + xbar = 2*xbar

Can you tell I've been doing statistical research all day?

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lol that would of been my guess to lol

i actually think it's lower than the double of the yearly average....where annual is calculated as the yearly snowfall per year over a 60 year period (or something like that)

back to back years would have to be calculated differently and there aren't as many back to back winters that hit the yearly average or more....

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i actually think it's lower than the double of the yearly average....where annual is calculated as the yearly snowfall per year over a 60 year period (or something like that)

back to back years would have to be calculated differently and there aren't as many back to back winters that hit the yearly average or more....

I've never looked at the data, is there an anti-autocorrelation for annual snowfall?

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i actually think it's lower than the double of the yearly average....where annual is calculated as the yearly snowfall per year over a 60 year period (or something like that)

back to back years would have to be calculated differently and there aren't as many back to back winters that hit the yearly average or more....

my guess it just take each back to back winter add the 2 up come up with an average. Then take the next two winters and do the samething. Then ater you got all your averages, take those averages and average that out...

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It's actually not going to work for a 30 year sample set.

E(x) = (1/n)*sum(x)

E(xi + xi+1) = (1/(n-1))*sum(xi + xi+1)

If n=30, E(x) = 1/30*sum(x)

but E(xi + xi+1) = 1/29*((x1+x2) + (x2+x3) + (x3+x4) ... + (x28 + x29) + (x29 + x30))

=1/29*(30*2*xbar - x1- x30) *[note: sum(x) = n*xbar]*

So there is a bias because of the way we are sampling the data. But as n -> infinity, E(xi + xi+1) -> 2*E(x)

Wheeeeeeeee.

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Which better equal 2*average annual snowfall :)

Yeah I see what you're saying...didn't need to do the calculations. I was wondering though since we are dealing with back to back years in a 40 year span wouldn't you divide everything by 20 since you have 20 sets of back to back years. For example: (1970+1971) +(1971 + 1972) would all be considered in 1 set of back to back years...do that over a 40 year period and you have 20 sets...no?

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Yeah I see what you're saying...didn't need to do the calculations. I was wondering though since we are dealing with back to back years in a 40 year span wouldn't you divide everything by 20 since you have 20 sets of back to back years. For example: (1970+1971) +(1971 + 1972) would all be considered in 1 set of back to back years...do that over a 40 year period and you have 20 sets...no?

39, actually, see above.

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