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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!

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!

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