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" wrote:
Previously you wrote: TDist(x,df,2) = BetaDist(df/(df+x²), df/2, 0.5) What would be the BetaDist() equivalent for TDist(x,df,1)? Or doesn't the question make sense to ask? The t distribution is symmetric about zero, so TDist(x,df,1) = TDist(x,df,2)/2 €¦ Based on the wikipedia plots and my vague recollection of the t-distribution plots, shouldn't alpha (df/2) be less than beta (0.5) in order to get a curve similar to the t-distribution? Again, the easiest way to verify the relationship between cumulative beta and t distributions is to calculate both formulas for various {x,df} pairs where df is an integer. The first Wikipedia beta distribution formula implies that f(x,a,b) = f(1-x,b,a) Excels BetaDist function has no cumulative=false option, but you can calculate the beta pdf directly from that Wikipedia formula as = EXP(GAMMALN(a+b)-GAMMALN(a)-GAMMALN(b))*x^(a-1)*(1-x)^(b-1) = EXP(GAMMALN(a+b)-GAMMALN(a)-GAMMALN(b)+LN(x)*(a-1)+LN(1-x)*(b-1)) Note that this beta pdf is not the t pdf. By the chain rule http://mathworld.wolfram.com/ChainRule.html you would multiply the Beta pdf corresponding to TDist(x,df,2) by ABS(df)/(df+x^2)^2 To get the t pdf. Jerry |
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