Lottery-type stochastics question

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SUMMARY

The discussion focuses on calculating the joint probability distribution of random variables T1, T2 - T1, ..., Tn - Tn-1, where n balls are drawn from an urn. The solution involves determining the probabilities P[T1 = m1], P[T2 - T1 = m2], and so forth, leading to the conclusion that these random variables are independent. The independence is established by analyzing the distributions of T2 - 1 and T3 - T2, confirming that they follow familiar distributions based on prior draws.

PREREQUISITES
  • Understanding of random variables and probability distributions
  • Familiarity with the concept of independence in probability
  • Knowledge of combinatorial probability related to drawing from an urn
  • Basic experience with stochastic processes
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  • Study the distribution of T2 - 1 in detail
  • Explore the conditional probability P{T3 - T2 = m3 | T2 - 1 = m2}
  • Review the properties of independent random variables in probability theory
  • Investigate applications of stochastic processes in real-world scenarios
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Students and professionals in statistics, mathematics, or data science who are working with stochastic processes and probability theory, particularly those interested in understanding joint distributions and independence of random variables.

sunrah
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Homework Statement



n balls, numbered from 1 to n, are drawn randomly from an urn one after the other (whilst returning each ball before drawing the next). The random variable Tk represents the number of draws until k different balls have been drawn, where k = 1,..., n.

calculate:
a) P[T1 = m1 ^ T2 - T1 = m2 ^ . . . ^ Tn - Tn-1 = mn]

for mi in N.

b) show that random variables T1, T2 - T1, . . . , Tn - Tn-1 are independent

Homework Equations



The Attempt at a Solution



a)
P[T1 = m1 ^ T2 - T1 = m2 ^ . . . ^ Tn - Tn-1 = mn]

= P[m1 ^ m2 ^ . . . ^ mn] = P(m1)P(m2)...P(mn) = \prod^{n}_{i=1} P(m_{i})

ok so i think I have to work out the product from i = 1,..,n of the probabilities P(mi) but how I don't even know what the mi are.

b) I guess this means linearly independent, I don't know?
 
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sunrah said:

Homework Statement



n balls, numbered from 1 to n, are drawn randomly from an urn one after the other (whilst returning each ball before drawing the next). The random variable Tk represents the number of draws until k different balls have been drawn, where k = 1,..., n.

calculate:
a) P[T1 = m1 ^ T2 - T1 = m2 ^ . . . ^ Tn - Tn-1 = mn]

for mi in N.

b) show that random variables T1, T2 - T1, . . . , Tn - Tn-1 are independent

Homework Equations



The Attempt at a Solution



a)
P[T1 = m1 ^ T2 - T1 = m2 ^ . . . ^ Tn - Tn-1 = mn]

= P[m1 ^ m2 ^ . . . ^ mn] = P(m1)P(m2)...P(mn) = \prod^{n}_{i=1} P(m_{i})

ok so i think I have to work out the product from i = 1,..,n of the probabilities P(mi) but how I don't even know what the mi are.

b) I guess this means linearly independent, I don't know?

Presumably, T_1=1, because the definition you gave would be "T_1 = the number of draws until 1 ball is obtained". Is that correct? (Of course, a deterministic quantity is a random variable with a degenerate probability distribution.) So, translating: T_2-1 is the number of draws until we get a ball different from the first one, T_3-T_2 is the number of draws until we get a ball different from the first two, etc. The problem is asking for the joint probability distribution of these numbers, and is then asking you to prove that these numbers are independent random variables. Linear independence has absolutely nothing to do with the problem.

To get started, I suggest you look first at the distribution of T_2-1 (it is a distribution you should have seen already!), then look at the distribution of T_3 - T_2, given a value \{T_1-1 = m_2\},; that is, you need to get
P\{T_3 - T_2 = m_3 | T_2 - 1 = m_2\} This should be a very familiar distribution. Note: here, m_2, \, m_3 = 1, 2, 3, \ldots are not specified; you need to work out what happens for any possible values of m_2, \, m_3.

After you have done these two steps you should see the pattern that emerges.

RGV
 

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