Help on to find probability density function

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To find the probability density function (pdf) of the variable i defined as i = x + (x^2 - y)^(1/2), where x and y are independent uniform distributions, the joint distribution f_{X,Y}(x,y) can be established using their independence. The cumulative distribution function (cdf) must be calculated first by evaluating the integral P{X + √(X^2 - Y) ≤ a}, which involves determining the region where this inequality holds. The equation x + √(x^2 - y) = a can be rearranged to express y in terms of x, yielding y = 2ax - x^2. Understanding this relationship is crucial for setting up the integral needed to derive the pdf.
musademirtas
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hey guys, i am really confused on something.here is the thing:
i have;

i=x+(x^2-y)^(1/2)

and here x is uniform distribution on (a,b)
y is uniform distribution on (c,d)
x and y independent
i need to find the probability density function of i but how?
actually i don't know how to start!
 
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Hi musademirtas! :smile:

First you will need to know the joint distribution of X and Y. This is easy because of independence:

f_{X,Y}(x,y)=f_X(x)f_Y(y)

Now, to find the pdf, you will need to find the cdf first. That is, for each a, you will want to calculate

P\{X+\sqrt{X^2-Y}\leq a\}=\iint_{\{(x,y)~\vert~x+\sqrt{x^2-y}\leq a\}}{f_{X,Y}(x,y)dxdy}

To evaluate this integral, you'll need to know the region \{(x,y)~\vert~x+\sqrt{x^2-y}\leq a\} somewhat better.

So I suggest you first find out for which tuples the equality holds. That is, for which x and y does it hold that

x+\sqrt{x^2-y}=a

(hint: the answer will be a straight line!)
 
hi micromass
thanks for the help but can you solve it? because i used your help to solve it but i couldn't do it.
 
Well, first you're going to need to write

x+\sqrt{x^2-y}=a

in function of y. What do you get for that?
 
ok.i got y=2ax-x^2 then what? how am i going to use this?
 
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