Die cast probability

iocal

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Hello, I would appreciate some help in the following:

A fair die is cast at random three independent times. Let the Random Variable Xi be equal to the number of spots on ith trial, i=1,2,3. Let the RV Y be equal to max(Xi). Find the cdf of Y and show that the pmf is equal to \(\displaystyle \displaystyle\frac{y^3-(y-1)^3}{6^3}]\). Hint: Consider the cdf of Y, P(Xi<Y, i=1,2,3).

I think is reasonable to assume that the probability mass function of Xi is uniform and is given by 1/6. But then I get stuck in the derivation of the CDF of Y as P(Xi<Y)= \(\displaystyle \displaystyle \sum_{i=1}^y\frac{1}{6}=y/6.\). And then we can raise the fraction to the power of 3 because each trial has the same probability but that does not seem close to the result I need to show.

Do you think I am missing something here? Thanks in advance.
 
Hello, I would appreciate some help in the following:

A fair die is cast at random three independent times. Let the Random Variable Xi be equal to the number of spots on ith trial, i=1,2,3. Let the RV Y be equal to max(Xi). Find the cdf of Y and show that the pmf is equal to \(\displaystyle \displaystyle\frac{y^3-(y-1)^3}{6^3}]\). Hint: Consider the cdf of Y, P(Xi<Y, i=1,2,3).

I think is reasonable to assume that the probability mass function of Xi is uniform and is given by 1/6. But then I get stuck in the derivation of the CDF of Y as P(Xi<Y)= \(\displaystyle \displaystyle \sum_{i=1}^y\frac{1}{6}=y/6.\). And then we can raise the fraction to the power of 3 because each trial has the same probability but that does not seem close to the result I need to show.

Do you think I am missing something here? Thanks in advance.
ok for P(Xi) = 1/6
If you imagine a tree diagram with 6 branches at each of 3 levels, there are 6^3 paths. Possible outcomes for Y are:
\(\displaystyle Y \in \{1,2,3,4,5,6\}\)

Here is how to construct the cdf for \(\displaystyle Y\):

To get \(\displaystyle Y\le 1\), you must have \(\displaystyle (X_1\le 1) \cup (X_2\le 1) \cup (X_3 \le 1)\)

..........\(\displaystyle Y\le 2\)........................\(\displaystyle (X_1\le 2) \cup (X_2\le 2) \cup (X_3 \le 2)\)

..........\(\displaystyle Y\le n\)........................\(\displaystyle (X_1\le n) \cup (X_2\le n) \cup (X_3 \le n)\)

This is a cdf because each entry includes all preceding cases. Note that the 3 Xs are statistically independent, so the uniion (\(\displaystyle \cup\)) is the product of the three probabilities. That may correspond to your statement about raising fractions to the power 3.

EDIT, to show how close you are to the answer:

\(\displaystyle P(Y\le y) = (y/6)^3 \)
 
Last edited:
ok for P(Xi) = 1/6
If you imagine a tree diagram with 6 branches at each of 3 levels, there are 6^3 paths. Possible outcomes for Y are:
\(\displaystyle Y \in \{1,2,3,4,5,6\}\)

Here is how to construct the cdf for \(\displaystyle Y\):

To get \(\displaystyle Y\le 1\), you must have \(\displaystyle (X_1\le 1) \cup (X_2\le 1) \cup (X_3 \le 1)\)

..........\(\displaystyle Y\le 2\)........................\(\displaystyle (X_1\le 2) \cup (X_2\le 2) \cup (X_3 \le 2)\)

..........\(\displaystyle Y\le n\)........................\(\displaystyle (X_1\le n) \cup (X_2\le n) \cup (X_3 \le n)\)

This is a cdf because each entry includes all preceding cases. Note that the 3 Xs are statistically independent, so the uniion (\(\displaystyle \cup\)) is the product of the three probabilities. That may correspond to your statement about raising fractions to the power 3.

EDIT, to show how close you are to the answer:

\(\displaystyle P(Y\le y) = (y/6)^3 \)


This is most illuminating, thank you very much, my work for the cdf is done. But how do you get the pmf next? Differentiating the cdf does not yield \(\displaystyle \displaystyle \frac {y^3-(y-1)^3}{6^3}\). Could it be that the book has a mistake at this point?
 
"Differentiating" does not make sense at all because this probability distribution is only defined for integers. If the "cdf(y)" (probability the result is less than or equal to y) is f(y) then the "pdf" (probabilty the result is equal to y) is f(y)- f(y-1).
 
"Differentiating" does not make sense at all because this probability distribution is only defined for integers. If the "cdf(y)" (probability the result is less than or equal to y) is f(y) then the "pdf" (probabilty the result is equal to y) is f(y)- f(y-1).

Oh big mistake of my part then. Thank you.
 
Oh big mistake of my part then. Thank you.
I would like to comment on the SIGNIFICANCE of this problem, from the point of view of a practitioner of Monte Carlo simulation, which is used to integrate over many variables for which deterministic computation is impractical. For any uniform process, a sample can be selected by a simple random number. However, in complex problems there are variables that require sampling of non-uniform distributions, and there are many algorithms for common distributions. What you have here is to generate a sample Y from a cubic distribution, by taking the largest of three independent samples from a uniform distribution. Likewise a sample from a parabolic function can be found as the larger of two uniform random numbers.
 
I would like to comment on the SIGNIFICANCE of this problem, from the point of view of a practitioner of Monte Carlo simulation, which is used to integrate over many variables for which deterministic computation is impractical. For any uniform process, a sample can be selected by a simple random number. However, in complex problems there are variables that require sampling of non-uniform distributions, and there are many algorithms for common distributions. What you have here is to generate a sample Y from a cubic distribution, by taking the largest of three independent samples from a uniform distribution. Likewise a sample from a parabolic function can be found as the larger of two uniform random numbers.

Yeah, I understand. I'll have more questions about that when I reach the Monte Carlo chapter, not long from now. Thanks.
 
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