doubts

[maahi]

 1) P(a1 < X <= a2; b1 < Y<=  b2) =P(X <= a2; Y <= b2)  - P(X<=a2; Y <= b1)
-P(X <= a1; Y<= b2) + P(X<= a1; Y <= b1)
=FXY (a2; b2) - FXY (a1; b2) - FXY (a2; b1) + FXY (a1; b1)
 what is d reason for including d last term ?



2 ) The percentages of copper and iron in a certain kind of ore are, respectively,
X and Y: If the joint density of these two random variables is given by
fXY (x; y) =
 3/11(5x + y) x; y > 0; x + 2y < 2
0 elsewhere
use the distribution function technique to find the probability density of
Z = X + Y.



3 ) Suppose you and me are tossing two fair coins independently, and we will
stop as soon as each one of us gets a head.
(a) Find the chance that we stop simultaneously.
(b) Find the conditional distribution of the number of coin tosses given that
we stop simultaneously.

[Sinistral]

1)
 we are trying to find the probability of the region bounded by x=a1 & x=a2 and y=b1 & y=b2
 note that when we subtract [P(X<=a2; Y <= b1)  + P(X <= a1; Y<= b2)] ,which is nothing but 2nd and 3rd term of the expression; from P(X <= a2; Y <= b2) , which is nothing but 1st term; we are actually subtracting the region  P(X<= a1; Y <= b1) which is the last term, twice. hence we are adding it again in the end.


2)
f(x,y) = 3/11(5x + y)   x+2y<2 & x,y>0
Z=X+Y

F(z)=P(Z<z)=P(X+Y<z)
now we need to integrate the region overlapped by x+2y<2 and x+y<z (as z varies from 0 to 2) . why 0 to 2? it is evident from the graph.

F(z) = 0 if z<0
when 0<z<1 it is simple integration
F(z)= Integral (x=0 to z) Integral (y=0 to z-x) f(x,y)dy dx   0<z<1

but when 1<z<2 we need to be careful. First we will find P(Z>z) when 1<z<2 and then we will find P(Z<z)for the same region by applying the formula P(Z<z)=1-P(Z>z). by seeing the graph we can split it into 2 double integrals.

when 1<z<2:
P(Z>z)= Integral (x=z-1 to z) Integral (y= z-x to (2-x)/2] f(x,y) dy dx  + Integral [x= z to 2] Integral [y=0 to (2-x)/2] f(x,y) dy dx
so now we can find F(z)=P(Z<z)=1-P(Z>z) for 1<z<2

F(z)=1 , z>2

differentiate to get f(z)

for the limit 0<z<1 I integrated it to get F(z)= (3/11)*z^3
for 1<z<2 I didnt integrate. u can integrate and check the answer.
this question was more of integration than of statistics.
they bounded the examinees by telling that one can only use distribution function technique.

today i realised how useful is the transformation technique for finding pdfs :D

[Sinistral]

3)
a) p(x)= (1/2)^x ,  x=1,2,3,4...
    p(y)= (1/2)^y , y=1,2,3,4...  

where x is the number of tosses in which you stop and y is the number of tosses in which I stop.

If we both stop simultaneouly we will either stop after 1st toss or 2nd toss or 3rd toss and so on.
so required Pr= .5^2  +  [.5^2]^2   +  [.5^3]^2 ....
which is a GP whose sum = 1/3

b) I didnt quite understand the question. means, finding the conditional  probability of a number (and not event)??

[maahi]

hi
 thanks a ton for replying

2 ) one very bad ques :: why do we need to integrate it seperately ? once for [0,1) and then 1 to 2 ??


 3. b ) no of coin tosses will be x or y as its given X=Y .

[Sinistral]

because it wont be possible to integrate it from 1 to 2 in one go.
draw both the graphs to see for yourself.

m still clueless about 3 (b). if you got it, plz solve it here.

[maahi]

1)Suppose a box contains 10 green, 10 red and 10 black balls. We draw 10
balls from the box by sampling with replacement. Let X be the number of
green balls, and Y be the number of black balls in the sample.
(a) Find E(XY ):(357)

2)Let Y be a random variable with a density fn given by
fY (y) = a-1 /y^a   ,y >1  otherwise 0
where a > 1: Given Y = y; let X be a random variable which is Uniformly
distributed on (0; y),
how do we find joint pdf ??are x and y independent here ?

3)summation  n to infinity (lambda e ^t )^n/ n !

4)Let X and Y be two random variables with joint distribution function
fXY (x; y) =

e-x-y x > 0; y > 0
0 otherwise
Find E(XY );E(X);E(Y ) and Cov(X; Y )


5)On a single tank of gasoline, you expect to be able to drive 240 miles before
running out of gas.
(a) Let p be the probability that you will NOT be able to make a 300 mile
journey on a single tank of gas. What can you say about p?