This is a simple and elegant sample problem from ISI MStat PSB 2005 Problem 5. It's based the mixture of Discrete and Continuous Uniform Distribution, the simplicity in the problem actually fools us, and we miss subtle happenings. Be careful while thinking !
Suppose and
are independent random variables with
,
,
and having a uniform distribution on [0,1]. Let
.
(a) For , find
.
(b) Find the correlation coefficient between and
.
Uniform Distribution
Law of Total Probability
Conditional Distribution
This ptroblem is quite straight forward enough, and we do what we are told to.
Here, we need to find the Cdf of , where ,
, and
and
are defined as above.
So, , [ since
and
are indepemdent ],
Now, here is where we get fooled often by the simplicity of the problem. The beauty is to observe in the above expression, if then
, and if
then
, ( why??)
So, for the
, so when
,
for
, and
for
.
So, the required Cdf will depend on the interval , y belongs to, for the above illustrated case, i.e. there will be
number of 1's, and
number of 0's in the above summation, derived, so,
reduces to,
,
, [ since here
may vary from 0,1,...., N, hence union of all the nested sub-intervals give the mentioned interval]
Hence the required Cdf. But can you find this Cdf argumentatively, without this algebraic labor ?? What distribution is it ?? Does the Uniform retains its Uniformity ?
I leave the part (b) as an exercise, its quite trivial.
How to deal with some random triangles ??
Suppose, You are given a stick of length units, Now you are to break the stick into 3 pieces, and the breaking points are chosen randomly , What is the chance of constructing a triangle with those 3 broken part as the sides of the constructed triangle ??
Now if you first break the stick into two pieces randomly, and further break the longer piece again into two parts (randomly), How the chance of making a triangle changes ?? Do the chance increases what do you think ??
Lastly, does the length of the stick matters at all , Give it a thought !!
This is a simple and elegant sample problem from ISI MStat PSB 2005 Problem 5. It's based the mixture of Discrete and Continuous Uniform Distribution, the simplicity in the problem actually fools us, and we miss subtle happenings. Be careful while thinking !
Suppose and
are independent random variables with
,
,
and having a uniform distribution on [0,1]. Let
.
(a) For , find
.
(b) Find the correlation coefficient between and
.
Uniform Distribution
Law of Total Probability
Conditional Distribution
This ptroblem is quite straight forward enough, and we do what we are told to.
Here, we need to find the Cdf of , where ,
, and
and
are defined as above.
So, , [ since
and
are indepemdent ],
Now, here is where we get fooled often by the simplicity of the problem. The beauty is to observe in the above expression, if then
, and if
then
, ( why??)
So, for the
, so when
,
for
, and
for
.
So, the required Cdf will depend on the interval , y belongs to, for the above illustrated case, i.e. there will be
number of 1's, and
number of 0's in the above summation, derived, so,
reduces to,
,
, [ since here
may vary from 0,1,...., N, hence union of all the nested sub-intervals give the mentioned interval]
Hence the required Cdf. But can you find this Cdf argumentatively, without this algebraic labor ?? What distribution is it ?? Does the Uniform retains its Uniformity ?
I leave the part (b) as an exercise, its quite trivial.
How to deal with some random triangles ??
Suppose, You are given a stick of length units, Now you are to break the stick into 3 pieces, and the breaking points are chosen randomly , What is the chance of constructing a triangle with those 3 broken part as the sides of the constructed triangle ??
Now if you first break the stick into two pieces randomly, and further break the longer piece again into two parts (randomly), How the chance of making a triangle changes ?? Do the chance increases what do you think ??
Lastly, does the length of the stick matters at all , Give it a thought !!