Get inspired by the success stories of our students in IIT JAM 2021. Learn More

For Students who are passionate for Mathematics and want to pursue it for higher studies in India and abroad.

Content

[hide]

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 \(X\) and \(U\) are independent random variables with

\(P(X=k)=\frac{1}{N+1} \) , \( k=0,1,2,......,N\) ,

and \(U\) having a uniform distribution on [0,1]. Let \(Y=X+U\).

(a) For \( y \in \mathbb{R} \), find \( P(Y \le y)\).

(b) Find the correlation coefficient between \(X\) and \(Y\).

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 \(Y\) , where , \(Y=X+U \), and \(X\) and \(Y\) are defined as above.

So, \(P(Y\le y)= P(X+U \le y)=\sum_{k=0}^N P(U \le y-k|X=k)P(X=k) = \frac{1}{N+1} \sum_{i=1}^NP(U \le y-k) \), [ since \(U\) and \(X\) 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 \(y-k <0\) then \(P(U\le y-k)=0\), and if \(y-k>1\) then \(P(U\le y-k)=1\), ( why??)

So, for \( k* \le y \le k*+1 \) the \(P(U \le y-k*)=y-k*\), so when \(k* \le y \le k*+1\), \(P(U\le y-k)=0\) for \(k>k*\), and \(P(U \le y-k)=1\) for \(k<k*\).

So, the required Cdf will depend on the interval , y belongs to, for the above illustrated case, i.e. \(k* \le y\le k*+1\) there will be \(k*\) number of 1's, and \(N-k*-1\) number of 0's in the above summation, derived, so, \(P(Y\le y)\) reduces to,

\(P(Y\le y)= \frac{1}{N+1} ( k*+y-k*)=\frac{y}{N+1}\), \(0<y<N+1\), [ since here \(k*\) 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 \(N+1\) 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 !!

What to do to shape your Career in Mathematics after 12th?

From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:

- What are some of the best colleges for Mathematics that you can aim to apply for after high school?
- How can you strategically opt for less known colleges and prepare yourself for the best universities in India or Abroad for your Masters or Ph.D. Programs?
- What are the best universities for MS, MMath, and Ph.D. Programs in India?
- What topics in Mathematics are really needed to crack some great Masters or Ph.D. level entrances?
- How can you pursue a Ph.D. in Mathematics outside India?
- What are the 5 ways Cheenta can help you to pursue Higher Mathematics in India and abroad?

Cheenta has taken an initiative of helping College and High School Passout Students with its "Open Seminars" and "Open for all Math Camps". These events are extremely useful for students who are really passionate for Mathematic and want to pursue their career in it.

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIAL
Google