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I've always wondered if this is a sensible way to buy shares. Say I wanted to buy Google shares in the next 30 days. I could monitor the price for the first 36.

by beejiu 12y ago

I've always wondered if this is a sensible way to buy shares. Say I wanted to buy Google shares in the next 30 days. I could monitor the price for the first 36.8%, and then choose the best price in the remainder. Does that make sense as a strategy?

codyb 12y ago
I think the volatility of the stock market adds extra complexity. Presuming you go on dates with 36.8% of your pool, you will have an idea of what you would like in a woman, and from there you can make a proper selection. But with the stock market, you can watch 36.8% of the time, see a rising trend, think it's a good choice, and be totally wrong five days later when their quarterly earnings come out. I think you'd want better heuristics for investing in the stock market than observing a bit of time (unless, apparently, you're a HFT algorithm ;-)).
xutopia 12y ago
Whoa cowboy... this isn't the same at all. We're talking about a whole different ball game here. Stock prices are not like meeting that special someone.
- angersock 12y ago
  Yeah, you can at least get something out of exiting from a relationship with a stock. The positions are more varied. The hours are more consistent. But when they go down it's usually worse. :(
broolstoryco 12y ago
No, it does not. You would essentially just be gambling on the price.
- judk 12y ago
  But how is that different from the original problem? Unknown distributions are unknown. Assuming a uniform shuffle of ranked options isn't actually appropriate in most real world situations. IOW, strategy is bad, but the stick market isn't worse than the mate/employee interviewing situation.
gopi 12y ago
In this case a portfolio accumulation strategy known as Value averaging works better. http://en.wikipedia.org/wiki/Value_averaging http://en.wikipedia.org/wiki/Value_averaging
mseebach 12y ago
No. This method assumes normally distributed potential wives. If a stock was normally distributed (i.e. has volatility) around a stable average, then yes, you could do that - but so can thousands of automated trading algorithms.
- taejo 12y ago
  No, it doesn't assume that at all. It just assumes that they are i.i.d. (so all permutations of ranks are equally likely). Still probably not a good model of stock prices (which presumably are correlated more with the previous day than ten days ago)

by beejiu 12y ago

codyb 12y ago

I think the volatility of the stock market adds extra complexity. Presuming you go on dates with 36.8% of your pool, you will have an idea of what you would like in a woman, and from there you can make a proper selection. But with the stock market, you can watch 36.8% of the time, see a rising trend, think it's a good choice, and be totally wrong five days later when their quarterly earnings come out. I think you'd want better heuristics for investing in the stock market than observing a bit of time (unless, apparently, you're a HFT algorithm ;-)).

xutopia 12y ago

Whoa cowboy... this isn't the same at all. We're talking about a whole different ball game here. Stock prices are not like meeting that special someone.

angersock 12y ago
Yeah, you can at least get something out of exiting from a relationship with a stock. The positions are more varied. The hours are more consistent. But when they go down it's usually worse. :(

broolstoryco 12y ago

No, it does not. You would essentially just be gambling on the price.

judk 12y ago
But how is that different from the original problem? Unknown distributions are unknown. Assuming a uniform shuffle of ranked options isn't actually appropriate in most real world situations. IOW, strategy is bad, but the stick market isn't worse than the mate/employee interviewing situation.

gopi 12y ago

In this case a portfolio accumulation strategy known as Value averaging works better. http://en.wikipedia.org/wiki/Value_averaging http://en.wikipedia.org/wiki/Value_averaging

mseebach 12y ago

No. This method assumes normally distributed potential wives. If a stock was normally distributed (i.e. has volatility) around a stable average, then yes, you could do that - but so can thousands of automated trading algorithms.

taejo 12y ago
No, it doesn't assume that at all. It just assumes that they are i.i.d. (so all permutations of ranks are equally likely). Still probably not a good model of stock prices (which presumably are correlated more with the previous day than ten days ago)