Propability Density (I'm stuck need assistance)

[Mathman23]

Hi
I have been given this following problem:
X is a stochastic variable with the distribution Function F_X which is given by:
<br /> P(X \leq x) = F_{X}(x) = \left\{ \begin{array}{ll}<br /> 0 &amp; \textrm{if} \ x&gt;0 \\<br /> \frac{{1- e^{-x}}}{{1 - e^{-1}}}&amp; \textrm{if} \ x \in [0,1]\\<br /> 1 &amp; \textrm{if} \ x \geq 1\\<br /> \end{array} \right.<br />
Now I'm supposed to show that X is absolutely continuous and then next calculate the propability density f_x.
I now then dealing with the propability density is found by
F&#039;_{(X)}(x) = \frac{e^{1-x}}{e-1}
But what is the next step from here which will allow me to find the propability density?
Secondly how do I go about showing that X is absolutly continuous ??
Sincerely and Best Regards
Fred

 

[Mathman23]

Hello again,

My own solution:

Since F_{X} (x) is differentiable, thereby according to the differention its also continious (but how is it absolutely continious??))

The density is therefore

p(x) = \frac{e^{1-|x|}}{e-1}

Could anyone please inform me if I'm on the right course?

Best Regards,

Fred

Hi
I have been given this following problem:
X is a stochastic variable with the distribution Function F_X which is given by:
<br /> P(X \leq x) = F_{X}(x) = \left\{ \begin{array}{ll}<br /> 0 &amp; \textrm{if} \ x&gt;0 \\<br /> \frac{{1- e^{-x}}}{{1 - e^{-1}}}&amp; \textrm{if} \ x \in [0,1]\\<br /> 1 &amp; \textrm{if} \ x \geq 1\\<br /> \end{array} \right.<br />
Now I'm supposed to show that X is absolutely continuous and then next calculate the propability density f_x.
I now then dealing with the propability density is found by
F&#039;_{(X)}(x) = \frac{e^{1-x}}{e-1}
But what is the next step from here which will allow me to find the propability density?
Secondly how do I go about showing that X is absolutly continuous ??
Sincerely and Best Regards
Fred