Question

Suppose that X and Y are random variables the following joint PDF: fxy(x,y) = otherwise Determine fx, the marginal PDF of X. a. etermine Fx, the marginal CDF of X.

Answer 1

Marginal PDF of X can be calculated by integrating the joint PDF with respect to all variate except X

fx(x)- fx.y(x, y)dy

Now our y only vary from 0 to 1 so we need to integrate only from 0 to 1

  

fx(x)-| fxy(x,y)dy

  

fx(x)4.rydy

  

y2 AT fx (r) = 4x-

  

fx(x) = 4x × 2

Here one thing you must notice that x belongs from [1,1] , i.e is not an interval that is clearly one value x=1 so when ever we integrate it from 1 to 1 we will get 0 and that violates the rule that say integrating PDF over its range must be zero , and just for a second lets take X is a discrete random variable so if we add 2x on all the range it must be zero which it is clearly not, so through my experience i have done a lot of question with the same joint PDf as given in the question so i think there is a typo the range of x must be from 0 to 1 because it is the value if we integrate the joint PDF it will come 1 so further i am solving by assuming x belongs to [0,1]

So now we want joint CDF to get this we need to following

Fx(x)= | fx(x)da

  

Fx(x)2rdr

  

F, (z) = 2

  $F_X(x)= 2 \times \frac {x^2}{2}$

So our marginal CDF for X is

  

Fx(x) = x