Expectation problem with geometric random variable?
If I have a stream of integers from 0 to 9(each with equal probability and repetition is allowed) that are spewed out until I get the first instance of 4-5 next to each other in this order, how do I find the expected value of the number of integers that will be printed?
I set up a random variable, X, that is the number of integers printed, and an indicator random variable, $X_i$, that is 1 if the $I^{th}$ number does not complete the first instance of 4-5, and 0 otherwise. I think that the probability that $X_i$ = 1 is calculated as 1 - (1/10)^2, so would the expected value of X just be 1/p?
probability random-variables expected-value geometric-probability
asked Dec 2 '18 at 19:45
Justin Dee
615
add a comment |
If I have a stream of integers from 0 to 9(each with equal probability and repetition is allowed) that are spewed out until I get the first instance of 4-5 next to each other in this order, how do I find the expected value of the number of integers that will be printed?
I set up a random variable, X, that is the number of integers printed, and an indicator random variable, $X_i$, that is 1 if the $I^{th}$ number does not complete the first instance of 4-5, and 0 otherwise. I think that the probability that $X_i$ = 1 is calculated as 1 - (1/10)^2, so would the expected value of X just be 1/p?
probability random-variables expected-value geometric-probability
asked Dec 2 '18 at 19:45
Justin Dee
615
add a comment |
If I have a stream of integers from 0 to 9(each with equal probability and repetition is allowed) that are spewed out until I get the first instance of 4-5 next to each other in this order, how do I find the expected value of the number of integers that will be printed?
I set up a random variable, X, that is the number of integers printed, and an indicator random variable, $X_i$, that is 1 if the $I^{th}$ number does not complete the first instance of 4-5, and 0 otherwise. I think that the probability that $X_i$ = 1 is calculated as 1 - (1/10)^2, so would the expected value of X just be 1/p?
probability random-variables expected-value geometric-probability
asked Dec 2 '18 at 19:45
Justin Dee
615
If I have a stream of integers from 0 to 9(each with equal probability and repetition is allowed) that are spewed out until I get the first instance of 4-5 next to each other in this order, how do I find the expected value of the number of integers that will be printed?
I set up a random variable, X, that is the number of integers printed, and an indicator random variable, $X_i$, that is 1 if the $I^{th}$ number does not complete the first instance of 4-5, and 0 otherwise. I think that the probability that $X_i$ = 1 is calculated as 1 - (1/10)^2, so would the expected value of X just be 1/p?
probability random-variables expected-value geometric-probability
probability random-variables expected-value geometric-probability
asked Dec 2 '18 at 19:45
Justin Dee
615
asked Dec 2 '18 at 19:45
Justin Dee
615
asked Dec 2 '18 at 19:45
Justin Dee
615
asked Dec 2 '18 at 19:45
Justin Dee
615
asked Dec 2 '18 at 19:45
Justin Dee
615
615
add a comment |
add a comment |
1 Answer
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I don't understand your computation. It would appear that the expected value of the $i^{th}$ term would be a function of $i$, since you would need to ensure that the game wasn't over before the $i^{th}$ move. Anyway, here is another approach:
There are three states here...$emptyset,4,{4,5}$ according to how much of the $4,5$ block occurs at the end of the running string. Letting $E_S$ deonte the expected number it will take to finish assuming you are starting in state $S$, we note that the answer we want is $E=E_{emptyset}$. Of course $E_{{4,5}}=0$.
Noting that $emptyset$ can only transition to $emptyset$ (probability .9) or $4$ (probability .1) we compute $$E_{emptyset}=frac 9{10}times (E_{emptyset}+1)+frac 1{10}times (E_4+1)implies E_{emptyset}=E_4+10$$
Similarly $$E_4=frac 1{10}times 1+frac 1{10}(E_4+1)+frac 8{10}times (E_{emptyset}+1)implies 9E_4=8E_{emptyset}+10$$
This system is easy to solve and we get $$boxed {E_{emptyset}=100}quad &quad boxed {E_4=90}$$
answered Dec 2 '18 at 19:59
lulu
39.2k24677
add a comment |
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I don't understand your computation. It would appear that the expected value of the $i^{th}$ term would be a function of $i$, since you would need to ensure that the game wasn't over before the $i^{th}$ move. Anyway, here is another approach:
There are three states here...$emptyset,4,{4,5}$ according to how much of the $4,5$ block occurs at the end of the running string. Letting $E_S$ deonte the expected number it will take to finish assuming you are starting in state $S$, we note that the answer we want is $E=E_{emptyset}$. Of course $E_{{4,5}}=0$.
Noting that $emptyset$ can only transition to $emptyset$ (probability .9) or $4$ (probability .1) we compute $$E_{emptyset}=frac 9{10}times (E_{emptyset}+1)+frac 1{10}times (E_4+1)implies E_{emptyset}=E_4+10$$
Similarly $$E_4=frac 1{10}times 1+frac 1{10}(E_4+1)+frac 8{10}times (E_{emptyset}+1)implies 9E_4=8E_{emptyset}+10$$
This system is easy to solve and we get $$boxed {E_{emptyset}=100}quad &quad boxed {E_4=90}$$
answered Dec 2 '18 at 19:59
lulu
39.2k24677
add a comment |
I don't understand your computation. It would appear that the expected value of the $i^{th}$ term would be a function of $i$, since you would need to ensure that the game wasn't over before the $i^{th}$ move. Anyway, here is another approach:
There are three states here...$emptyset,4,{4,5}$ according to how much of the $4,5$ block occurs at the end of the running string. Letting $E_S$ deonte the expected number it will take to finish assuming you are starting in state $S$, we note that the answer we want is $E=E_{emptyset}$. Of course $E_{{4,5}}=0$.
Noting that $emptyset$ can only transition to $emptyset$ (probability .9) or $4$ (probability .1) we compute $$E_{emptyset}=frac 9{10}times (E_{emptyset}+1)+frac 1{10}times (E_4+1)implies E_{emptyset}=E_4+10$$
Similarly $$E_4=frac 1{10}times 1+frac 1{10}(E_4+1)+frac 8{10}times (E_{emptyset}+1)implies 9E_4=8E_{emptyset}+10$$
This system is easy to solve and we get $$boxed {E_{emptyset}=100}quad &quad boxed {E_4=90}$$
answered Dec 2 '18 at 19:59
lulu
39.2k24677
add a comment |
I don't understand your computation. It would appear that the expected value of the $i^{th}$ term would be a function of $i$, since you would need to ensure that the game wasn't over before the $i^{th}$ move. Anyway, here is another approach:
There are three states here...$emptyset,4,{4,5}$ according to how much of the $4,5$ block occurs at the end of the running string. Letting $E_S$ deonte the expected number it will take to finish assuming you are starting in state $S$, we note that the answer we want is $E=E_{emptyset}$. Of course $E_{{4,5}}=0$.
Noting that $emptyset$ can only transition to $emptyset$ (probability .9) or $4$ (probability .1) we compute $$E_{emptyset}=frac 9{10}times (E_{emptyset}+1)+frac 1{10}times (E_4+1)implies E_{emptyset}=E_4+10$$
Similarly $$E_4=frac 1{10}times 1+frac 1{10}(E_4+1)+frac 8{10}times (E_{emptyset}+1)implies 9E_4=8E_{emptyset}+10$$
This system is easy to solve and we get $$boxed {E_{emptyset}=100}quad &quad boxed {E_4=90}$$
answered Dec 2 '18 at 19:59
lulu
39.2k24677
I don't understand your computation. It would appear that the expected value of the $i^{th}$ term would be a function of $i$, since you would need to ensure that the game wasn't over before the $i^{th}$ move. Anyway, here is another approach:
There are three states here...$emptyset,4,{4,5}$ according to how much of the $4,5$ block occurs at the end of the running string. Letting $E_S$ deonte the expected number it will take to finish assuming you are starting in state $S$, we note that the answer we want is $E=E_{emptyset}$. Of course $E_{{4,5}}=0$.
Noting that $emptyset$ can only transition to $emptyset$ (probability .9) or $4$ (probability .1) we compute $$E_{emptyset}=frac 9{10}times (E_{emptyset}+1)+frac 1{10}times (E_4+1)implies E_{emptyset}=E_4+10$$
Similarly $$E_4=frac 1{10}times 1+frac 1{10}(E_4+1)+frac 8{10}times (E_{emptyset}+1)implies 9E_4=8E_{emptyset}+10$$
This system is easy to solve and we get $$boxed {E_{emptyset}=100}quad &quad boxed {E_4=90}$$
answered Dec 2 '18 at 19:59
lulu
39.2k24677
answered Dec 2 '18 at 19:59
lulu
39.2k24677
answered Dec 2 '18 at 19:59
lulu
39.2k24677
answered Dec 2 '18 at 19:59
lulu
39.2k24677
39.2k24677
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