Finding the probability of two random variables being equal to 1
Question:
A die is thrown $n+2$ times. After each throw a '$+$' is recorded for $4$, $5$, or $6$ and '$-$' for $1$,$2$, or $3$, the signs forming an ordered sequence. To each, except the first and the last sign, is attached a characteristic random variable which takes the value $1$ if both the neighboring signs differ from the one between them and $0$ otherwise. If $X_1, X_2, ldots , X_n$ are the characteristic random variables, find the mean and variance of $X = sum_{i=1}^n X_i$
Problematic part:
$$ V(X) = V(X_1+X_2+cdots+X_n) = sum_{i=1}^n V(X_i) +2sum_{i=1}^nsum_{j=1, j>i}^n Cov(X_i,X_j) $$
Calculating the variance of $X$ requires calculating the covariance of two arbitrarily chosen random variables $X_i$ and $X_j$.
The formula then used is $Cov(X_i,X_j) = E(X_iX_j) - E(X_i)E(X_j)$
Which brings us to the essence of my problem -- How to find $E(X_iX_j)$?
It is certain that $X_iX_j$ can take only two values, namely, $0$ and $1$. Therefore $E(X_iX_j) = 1.P(X_iX_j=1) + 0.P(X_iX_j=0)= P(X_iX_j=1)$
But at this point I'm not sure how to compute the probability. The book I'm using says the probability is $1/8$, but I can't seem to wrap my head around the reasoning. An intuitive explanation would be highly appreciated!
probability statistics random-variables covariance expected-value
asked Nov 26 at 8:28
s0ulr3aper07
625
add a comment |
Question:
A die is thrown $n+2$ times. After each throw a '$+$' is recorded for $4$, $5$, or $6$ and '$-$' for $1$,$2$, or $3$, the signs forming an ordered sequence. To each, except the first and the last sign, is attached a characteristic random variable which takes the value $1$ if both the neighboring signs differ from the one between them and $0$ otherwise. If $X_1, X_2, ldots , X_n$ are the characteristic random variables, find the mean and variance of $X = sum_{i=1}^n X_i$
Problematic part:
$$ V(X) = V(X_1+X_2+cdots+X_n) = sum_{i=1}^n V(X_i) +2sum_{i=1}^nsum_{j=1, j>i}^n Cov(X_i,X_j) $$
Calculating the variance of $X$ requires calculating the covariance of two arbitrarily chosen random variables $X_i$ and $X_j$.
The formula then used is $Cov(X_i,X_j) = E(X_iX_j) - E(X_i)E(X_j)$
Which brings us to the essence of my problem -- How to find $E(X_iX_j)$?
It is certain that $X_iX_j$ can take only two values, namely, $0$ and $1$. Therefore $E(X_iX_j) = 1.P(X_iX_j=1) + 0.P(X_iX_j=0)= P(X_iX_j=1)$
But at this point I'm not sure how to compute the probability. The book I'm using says the probability is $1/8$, but I can't seem to wrap my head around the reasoning. An intuitive explanation would be highly appreciated!
probability statistics random-variables covariance expected-value
asked Nov 26 at 8:28
s0ulr3aper07
625
add a comment |
Question:
A die is thrown $n+2$ times. After each throw a '$+$' is recorded for $4$, $5$, or $6$ and '$-$' for $1$,$2$, or $3$, the signs forming an ordered sequence. To each, except the first and the last sign, is attached a characteristic random variable which takes the value $1$ if both the neighboring signs differ from the one between them and $0$ otherwise. If $X_1, X_2, ldots , X_n$ are the characteristic random variables, find the mean and variance of $X = sum_{i=1}^n X_i$
Problematic part:
$$ V(X) = V(X_1+X_2+cdots+X_n) = sum_{i=1}^n V(X_i) +2sum_{i=1}^nsum_{j=1, j>i}^n Cov(X_i,X_j) $$
Calculating the variance of $X$ requires calculating the covariance of two arbitrarily chosen random variables $X_i$ and $X_j$.
The formula then used is $Cov(X_i,X_j) = E(X_iX_j) - E(X_i)E(X_j)$
Which brings us to the essence of my problem -- How to find $E(X_iX_j)$?
It is certain that $X_iX_j$ can take only two values, namely, $0$ and $1$. Therefore $E(X_iX_j) = 1.P(X_iX_j=1) + 0.P(X_iX_j=0)= P(X_iX_j=1)$
But at this point I'm not sure how to compute the probability. The book I'm using says the probability is $1/8$, but I can't seem to wrap my head around the reasoning. An intuitive explanation would be highly appreciated!
probability statistics random-variables covariance expected-value
asked Nov 26 at 8:28
s0ulr3aper07
625
Question:
A die is thrown $n+2$ times. After each throw a '$+$' is recorded for $4$, $5$, or $6$ and '$-$' for $1$,$2$, or $3$, the signs forming an ordered sequence. To each, except the first and the last sign, is attached a characteristic random variable which takes the value $1$ if both the neighboring signs differ from the one between them and $0$ otherwise. If $X_1, X_2, ldots , X_n$ are the characteristic random variables, find the mean and variance of $X = sum_{i=1}^n X_i$
Problematic part:
$$ V(X) = V(X_1+X_2+cdots+X_n) = sum_{i=1}^n V(X_i) +2sum_{i=1}^nsum_{j=1, j>i}^n Cov(X_i,X_j) $$
Calculating the variance of $X$ requires calculating the covariance of two arbitrarily chosen random variables $X_i$ and $X_j$.
The formula then used is $Cov(X_i,X_j) = E(X_iX_j) - E(X_i)E(X_j)$
Which brings us to the essence of my problem -- How to find $E(X_iX_j)$?
It is certain that $X_iX_j$ can take only two values, namely, $0$ and $1$. Therefore $E(X_iX_j) = 1.P(X_iX_j=1) + 0.P(X_iX_j=0)= P(X_iX_j=1)$
But at this point I'm not sure how to compute the probability. The book I'm using says the probability is $1/8$, but I can't seem to wrap my head around the reasoning. An intuitive explanation would be highly appreciated!
probability statistics random-variables covariance expected-value
probability statistics random-variables covariance expected-value
asked Nov 26 at 8:28
s0ulr3aper07
625
asked Nov 26 at 8:28
s0ulr3aper07
625
asked Nov 26 at 8:28
s0ulr3aper07
625
asked Nov 26 at 8:28
s0ulr3aper07
625
asked Nov 26 at 8:28
s0ulr3aper07
625
625
add a comment |
add a comment |
1 Answer
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Assuming the die is thrown independently, $X_i$ and $X_j$ are independent if $j geq i + 3$. So in that case the convariance is $0$. So you only have to calculate $E[X_iX_{i+1}] = E[X_1X_2]$ and $E[X_iX_{i+2}]=E[X_1X_3]$ which should be easy by just looking at all the possible outcomes.
answered Nov 26 at 9:30
Tki Deneb
26710
Thanks! The textbook I'm using did not talk about the case where j = i + 2. This lead to some confusion.
– s0ulr3aper07
Nov 29 at 9:30
add a comment |
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1 Answer
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Assuming the die is thrown independently, $X_i$ and $X_j$ are independent if $j geq i + 3$. So in that case the convariance is $0$. So you only have to calculate $E[X_iX_{i+1}] = E[X_1X_2]$ and $E[X_iX_{i+2}]=E[X_1X_3]$ which should be easy by just looking at all the possible outcomes.
answered Nov 26 at 9:30
Tki Deneb
26710
Thanks! The textbook I'm using did not talk about the case where j = i + 2. This lead to some confusion.
– s0ulr3aper07
Nov 29 at 9:30
add a comment |
Assuming the die is thrown independently, $X_i$ and $X_j$ are independent if $j geq i + 3$. So in that case the convariance is $0$. So you only have to calculate $E[X_iX_{i+1}] = E[X_1X_2]$ and $E[X_iX_{i+2}]=E[X_1X_3]$ which should be easy by just looking at all the possible outcomes.
answered Nov 26 at 9:30
Tki Deneb
26710
Thanks! The textbook I'm using did not talk about the case where j = i + 2. This lead to some confusion.
– s0ulr3aper07
Nov 29 at 9:30
add a comment |
Assuming the die is thrown independently, $X_i$ and $X_j$ are independent if $j geq i + 3$. So in that case the convariance is $0$. So you only have to calculate $E[X_iX_{i+1}] = E[X_1X_2]$ and $E[X_iX_{i+2}]=E[X_1X_3]$ which should be easy by just looking at all the possible outcomes.
answered Nov 26 at 9:30
Tki Deneb
26710
Assuming the die is thrown independently, $X_i$ and $X_j$ are independent if $j geq i + 3$. So in that case the convariance is $0$. So you only have to calculate $E[X_iX_{i+1}] = E[X_1X_2]$ and $E[X_iX_{i+2}]=E[X_1X_3]$ which should be easy by just looking at all the possible outcomes.
answered Nov 26 at 9:30
Tki Deneb
26710
edited Nov 26 at 9:36
answered Nov 26 at 9:30
Tki Deneb
26710
answered Nov 26 at 9:30
Tki Deneb
26710
answered Nov 26 at 9:30
Tki Deneb
26710
26710
Thanks! The textbook I'm using did not talk about the case where j = i + 2. This lead to some confusion.
– s0ulr3aper07
Nov 29 at 9:30
add a comment |
Thanks! The textbook I'm using did not talk about the case where j = i + 2. This lead to some confusion.
– s0ulr3aper07
Nov 29 at 9:30
Thanks! The textbook I'm using did not talk about the case where j = i + 2. This lead to some confusion.
– s0ulr3aper07
Nov 29 at 9:30
Thanks! The textbook I'm using did not talk about the case where j = i + 2. This lead to some confusion.
– s0ulr3aper07
Nov 29 at 9:30
add a comment |
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