Re: discontinuity confusion

patrick wrote:

Hello

I am a bit confused about the relation between the
density and distribution functions at the point of discontinuity. when I have a discontinuity in the
distribution, that requires a delta function in the
density, is the reverse true? it seam "yes" for me.

the reason for my question, I want to derive the
distribution F(x) for the density;
f(x) = 4 e^(-4x)  if x>0
    = 0          if x=<0
so it look we have a discontinuity at 0 for this f(x)

F(x) = integral f(x) dx = -4 e^(-4x) if x>0
    = 0 if x=<0



it looks like you're using x for two different things here.  i think what you want is
F(x) = integral_{0}^{x}f(t) dt if x>0
which gives you something different from what you have.


so, given that I understand, to calculate Fx(u) given f(x) = 4 e^(-4x)

Fx(u) = integral_(0)^(u) 4e^(-4x) du, that does not look it would lead to the book answer "after correcting it from my OP"
F(x) = 1 - e^(-4x) for x=>0




so F(x) = 0 - 4 e^(-4x) for x=>0 


this answer is clearly wrong, as F(infinity) must be 1 and F(0) must be 0.


the book answer is
F(x) = 1 - 4 e^(-4x) for x=>0



the same goes for this "book answer" which is also clearly wrong.


am I missing things up at discontinuity?


the discontinuity really doesn't matter since the density is 0 to the left, so the distribution is also 0 to the left. the distribution is not necessarily even discontinuous.


thanks


you're welcome! 
.

Relevant Pages

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