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

This problem is a beautiful application of the probability theory and cauchy functional equations. This is from ISI MStat 2019 PSB problem 4.

Let \(X\) and \(Y\) be independent and identically distributed random variables with mean \(\mu>0\) and taking values in {\(0,1,2, \ldots\)}. Suppose, for all \(m \geq 0\)

$$

\mathrm{P}(X=k | X+Y=m)=\frac{1}{m+1}, \quad k=0,1, \ldots, m

$$

Find the distribution of \(X\) in terms of \(\mu\).

- Conditional Probability
- Geometric Distribution
- \(a, b, c, d\) are numbers such that \(ac = b^2\) & \(ad = bc\), then \(a, b, c, d\) are in geometric progression.

Let \( P(X =i) = p_i\) where $$\sum_{i=0}^{\infty} p_i = 1$$. Now, let's calculate \(P(X+Y = m)\).

$$P(X+Y = m) = \sum_{i=0}^{m} P(X+Y = m, X = i) = \sum_{i=0}^{m} P(Y = m-i, X = i) = \sum_{i=0}^{m} p_ip_{m-i}$$.

$$P( X = k|X+Y = m) = \frac{P( X = k, X+Y = m)}{P(X+Y = m)} = \frac{P( X = k, Y = m-k)}{\sum_{i=0}^{m} p_ip_{m-i}} = \frac{p_kp_{m-k}}{\sum_{i=0}^{m} p_ip_{m-i}} = \frac{1}{m+1}$$.

Hence,$$ \forall m \geq 0, p_0p_m =p_1p_{m-1} = \dots = p_mp_0$$.

Thus, we get the following set of equations.

$$ p_0p_2 = p_1^2$$ $$ p_0p_3 = p_1p_2$$ Hence, by the thrid prerequisite, \(p_0, p_1, p_2, p_3\) are in geometric progression.

Observe that as a result we get \( p_1p_3 =p_2^2 \). In the line there is waiting:

$$ p_1p_4 = p_2p_3$$. Thus, in the similar way we get \(p_1, p_2, p_3, p_4\) are in geometric progression.

Hence, by induction, we will get that \(p_k; k \geq 0\) form a geometric progression.

This is only possible if \(X, Y\) ~ Geom(\( p\)). We need to find \(p\) now, but here \(X\) counts the number of failures, and \(p\) is the probability of success.

So, \(E(X) = \frac{1-p}{p} = \mu \Rightarrow p = \frac{1}{\mu +1}\).

So, can you guess, what it will be its continous version?

It will be the exponential distribution. Prove it. But, what exactly is governing this exponential structure? What is the intuition behind it?

Observe that the obtained condition

$$ \forall m \geq 0, p_0p_m =p_1p_{m-1} = \dots = p_mp_0.$$ can be written as follows

Find all such functions \(f: \mathbf{N}_0 \to \mathbf{N}_0\) such that \(f(m)f(n) = f(m+n)\) and with the summation = 1 restriction property.

The only solution to this geometric progression structure. This is a variant of the Cauchy Functional Equation. For the continuous case, it will be exponential distribution.

Essentially, this is the functional equation that arises, if you march along to prove that the Geometric Random Variable is the only discrete distribution with the memoryless property.

Stay Tuned!

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.

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