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I'm wondering why the delta method,

delta method,

"is only used for asymptotic distributions", as @mpiktas write in this post: Variance of a function of one random variable,

or as is written here: "The delta method is a general method for deriving the variance of a function of asymptotically normal random variables with known variance"

I'm wondering what it means that a random variable is asymptotic. For me, a random variable is a random variable.

HeyJane
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    Well, asymptotically it becomes exact while for finite samples it is only an approximation. – JohnK Oct 04 '17 at 11:09
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    is added something: I'm wondering what it means that a random variable is asymptotic. For me, a random variable is a random variable. – HeyJane Oct 04 '17 at 11:15
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    To say "asymptotically normal random variables" is a kind of sloppy language which is an abbreviation for a much longer technically correct expression. – kjetil b halvorsen Oct 04 '17 at 11:18
  • Ok, I see :) I was under the impression that when I had a random variable *X* and knew its variance, I could for example find the variance of f(*X*) through the delta method, and I never thought about asymptotic or not...? – HeyJane Oct 04 '17 at 11:21
  • It really should be referring to a sequence of random variables rather than just one. – Michael R. Chernick Oct 04 '17 at 12:06
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    You can always *estimate* the variance with the Delta method. Whether that estimate is any good depends on the random variable. When the variable is "close" to a given one--such as a standard Normal--then one can often find out how accurate the estimate might be. (And it could still be horrible.) That's all that's going on. Sequences, normality, and asymptotics are irrelevant. When you *are* working in that situation, though, you can often make stronger statements, such as the one you quoted. – whuber Oct 04 '17 at 12:53
  • I'm really, really sorry, but I just don't get it. if I have a variable **X**, being standard normal, and I want to find an estimate for exp(**2X**), then the delta method gives me variance 1*[2*exp(0)]^2=4. This is just _one random variable_, not a sequence. @whuber: you say the quality of the estimate depends on the random variable, whether it is "close" to a given one or not. Could this be expanded on? Sorry, I just do not get it :/ – HeyJane Oct 05 '17 at 06:44

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