PDF - June 13

[Amit Goyal]

June13_PDF.png

[Chinni18]

I am getting P as 2d-d^2, but I'm not too sure

The PDF is f(d)=2-2d

[anon_econ]

How r u getting this Chinni? I can't proceed beyond interpreting what v hv to find out!

[Amit Goyal]

This is how you do it Vasudha:
Pr(D ≤ d) = 1 - Pr(D > d)
And Check that
Pr(D > d) = Pr(d/2 < X < 1 - (d/2), d/2 < Y < 1 - (d/2))

[Chinni18]

I only got it because I had done this question a month or so ago and really broke my head over it
The radius of the circle will be either x,y,1-x or 1-y
=>D= 2min{x,y,1-x or 1-y}
So P((D<= d) = 1 - P(D > d) which comes out to be 1-(1-d)^2 ie 2d- d^2
Try drawing the circle in the square as specified in the question, then you'll get it.
For PDF I differentiated cdf wrt d

[anon_econ]

Thanks Amit sir and Chinni. Actually I got this part: "The radius of the circle will be either x,y,1-x or 1-y
=>D= 2min{x,y,1-x or 1-y}
So P((D<= d) = 1 - P(D > d)" but i'm unable to get the final answer..i think i'm not very good at this topic; i'll go through it once more and try it again :)

[Rain Man]

Hi

How do we get the value for P ( D > d ) . Please explain in some detail.

Thanks

[Amit Goyal]

Lets see how to get this:
Pr(D > d) = Pr(d/2 < X < 1 - (d/2), d/2 < Y < 1 - (d/2))

Pr(D > d) is the probability of the event D > d i.e. diameter of the circle inside the unit square is greater than d. Note that for the diameter to be greater than d, the center of the circle must be inside the open square given by (d/2, 1 - (d/2)) x (d/2, 1 - (d/2)) so that when you make the circle of diameter d (and hence radius d/2), the entire circle can lie inside [0, 1] x [0, 1]. That explains the limit.

[anon_econ]

Got it now..thanks :)