The random variable X has pdf given by f(x)=(1+x),-1<x=0 and f(x)=(1-x),0<x<1

Question

The random variable X has pdf given by 

f(x)=(1+x),-1<x≤0 and f(x)=(1-x),0<x<1

 

a)Find F(x) and sketch its graph?

 

my solution:

F(X)=(x^2/2)+x+1/2 , -1<x≤0 and F(x)=x-(x^2/2) , 0<x<1

Is it correct? and how can I sketch graph since its two functions?

 

b)Find P (0.4<X<0.7)

 

My solution:

F(0.7)-F(0.4) but here two functions for F(x) which one I need to choose?

 

c)Find the mean of the distribution?

 

my solution :

E(X)= the integration of x (1+x) from -1 to o

and E(X)= the integration of x from 0 to 1

 

here I find two solves for E(x) , is it ok?

 

d)Find the 75th percentile of the distribution?

 

here could you explain to me how can I find the percentile because my teacher did not explain it and I did not find anything by google?

Answer 1

The graph of f(x) is a piecewise function; on [-1,0] it's y=x+1;

on (0,1] it's y = 1-x. Draw it.  Obviously the total area under the curve equals 1(one half for each of the two parts).

(This graph looks like a tent, consisting of a segment from x-intercept -1 to y-intercept 1, and another segment fromy-int. 1 down to x-int. 1.)

 

The 50% percentile is at x = 0.

 

To find the 75% percentile:

You have to choose a positive x such that the area in the trapezoid from 0 to x equals ¼.  The trapezoid has base running vertically)  from y=0 up to y=1, i.e. its big base is 1, and other base runs vertically up from (x,0) to its y value, which is 1-x.  And the 'height' of this trapezoid is x (from 0 to x along x-axis).

 

So the trapezoidal area is ½(x)[1 + 1-x] = x - ½x² and we must set this to = ¼ (using formula 'average base' times height.)

The only positive solution in (0,1] is x = 1 - ½√2.

 

So the 75% percentile is at about x=0.2929

 

To evaluate the cdf, it's definite integral from 0 to x of (1+x)dx PROVIDED x ≤0.

If x>0, then it's ½ + def. integral from 0 to x of (1-x)dx PROVIDED x≥0.