# probability integral transformation and distribution of P= P[ |T| <= |t|] .

The task is to find the distribution of P. where , P=P[ |T| <= |t|]. (T is a continuous random variable with PDF f(t)).

now , I tried to make the following two arguments :

1.P= P[ |T| <= |t|] ( the range of P is (0,1))

= P[ U <= |t|] (where U=|T|, a continuous random variable with range (0,1) )

= F(|t|) (F(.) being the CDF of U)

so, by probability integral transformation F(|t|)follows uniform (0,1) distribution. (as in probability integral transformation the distribution of F(x) doesn't depend on the value of x.so we need not to worry about the '|t|' thing .)

2.P= P[ |T| <= |t|] ( the range of P is (0,1))

= P[-|t| <= T <= |t|] ( the range of P is (0,1))

= P[-t<=T<= t]

= G(t)-G(-t) = X-Y (say)

but , X and Y (again by probability integral transformation ) follows uniform (0,1) distribution each . and then, the range of X-Y is (-1,1).whereas P can't be negative. moreover X-Y doesn't follow uniform distribution.(in fact, no distribution over (0,1)). we can calculate that the PDF of V =X-Y is

f(v) = 1- |v| (when -1< V < 1 ) and o (otherwise) .

conclusion : i think in the 2nd argument what i wrote that "(X-Y) has a range of (-1,1)" ,was wrong. Because here X and Y are ,though identical, NOT INDEPENDENT. and, that's the source of the fallacy . Am i correct ?

have I done any other logical mistake ?

• In your story concepts as distribution, CDF and value of CDF at a some $x$ or $|t|$ are quite mixed up. "distribution of $P=P[|T|\leq |t|]$" What exactly is this $P$? A probability measure maybe? A value? A function? "the distribution of $F(x)$" Do you mean the distribution characterized by CDF $F$ here? And there is more of this. This makes it difficult to read. At too much places the reader of the question must guess what you mean to say (and it is likely that he gives up). Make a good distinction between these concepts. – drhab Jan 17 '15 at 10:26
• that 'P' is a random variable. if you are not aware of probability integral transformation I'm politely requesting you to Google or leave with this problem. thank you. – saudade Jan 19 '15 at 7:28