Expected Number of edges in the graph
$begingroup$
G(V,E) is a simple graph with 8 vertices. The edges of G are decided by tossing the coin for each 2 vertices combination. Edge is added between any two vertices iff head is turned up. Expected number of edges in the Graph G(V,E) is.
I thought of doing like this
Let X be a random variable denoting the number of edges in the graph.
Total possible edges->$_8C_2=28$
Now, for each of those 28 edges, we tossed a coin and if it turned out to be heads,that edge was included.
(1)X=0,P(X)=$_{28}C_0 frac{1}{2^{28}}$. All those 28 tosses of coin are independent with probability of heads=Probability of getting tails=$frac{1}{2}$
(2)X=1(One edge), P(X)=$_{28}C_1 timesfrac{1}{2} times frac{1}{2^{27}}$-Means in those 28 trials, exactly 1 heads and rest tails.
(3)Similarly for all 28 edges-> $X=28,P(X)=_{28}C_{28} times frac{1}{2^{28}}$
And then we add all cases of X from 0 to 28 because $E[X]=sum x.p(x)$
But this all together seems to be a very huge number and answer is given to be 14, where I am wrong in my reasoning?
probability graph-theory
edited Dec 17 '18 at 7:04
bof
52.1k558121
asked Dec 17 '18 at 6:20
user3767495user3767495
4078
$endgroup$
add a comment |
$begingroup$
G(V,E) is a simple graph with 8 vertices. The edges of G are decided by tossing the coin for each 2 vertices combination. Edge is added between any two vertices iff head is turned up. Expected number of edges in the Graph G(V,E) is.
I thought of doing like this
Let X be a random variable denoting the number of edges in the graph.
Total possible edges->$_8C_2=28$
Now, for each of those 28 edges, we tossed a coin and if it turned out to be heads,that edge was included.
(1)X=0,P(X)=$_{28}C_0 frac{1}{2^{28}}$. All those 28 tosses of coin are independent with probability of heads=Probability of getting tails=$frac{1}{2}$
(2)X=1(One edge), P(X)=$_{28}C_1 timesfrac{1}{2} times frac{1}{2^{27}}$-Means in those 28 trials, exactly 1 heads and rest tails.
(3)Similarly for all 28 edges-> $X=28,P(X)=_{28}C_{28} times frac{1}{2^{28}}$
And then we add all cases of X from 0 to 28 because $E[X]=sum x.p(x)$
But this all together seems to be a very huge number and answer is given to be 14, where I am wrong in my reasoning?
probability graph-theory
edited Dec 17 '18 at 7:04
bof
52.1k558121
asked Dec 17 '18 at 6:20
user3767495user3767495
4078
$endgroup$
$begingroup$
Two remarks: (1) What you tried to write, is not very clear. You do not define what the $C_0,C_1,C_2,...$ are. (2) The problem becomes much simpler if you were to use the linearity of the expectation. By finding a collection of 'simple' random variables for which the calculation of expectation is simple, and that sum up to $X$, you will have simplified the problem many times over.
$endgroup$
– Keen-ameteur
Dec 17 '18 at 6:31
$begingroup$
It is not such a "very huge number" because the factor $frac1{2^{28}}$ is pretty small.
$endgroup$
– bof
Dec 17 '18 at 7:06
add a comment |
$begingroup$
G(V,E) is a simple graph with 8 vertices. The edges of G are decided by tossing the coin for each 2 vertices combination. Edge is added between any two vertices iff head is turned up. Expected number of edges in the Graph G(V,E) is.
I thought of doing like this
Let X be a random variable denoting the number of edges in the graph.
Total possible edges->$_8C_2=28$
Now, for each of those 28 edges, we tossed a coin and if it turned out to be heads,that edge was included.
(1)X=0,P(X)=$_{28}C_0 frac{1}{2^{28}}$. All those 28 tosses of coin are independent with probability of heads=Probability of getting tails=$frac{1}{2}$
(2)X=1(One edge), P(X)=$_{28}C_1 timesfrac{1}{2} times frac{1}{2^{27}}$-Means in those 28 trials, exactly 1 heads and rest tails.
(3)Similarly for all 28 edges-> $X=28,P(X)=_{28}C_{28} times frac{1}{2^{28}}$
And then we add all cases of X from 0 to 28 because $E[X]=sum x.p(x)$
But this all together seems to be a very huge number and answer is given to be 14, where I am wrong in my reasoning?
probability graph-theory
edited Dec 17 '18 at 7:04
bof
52.1k558121
asked Dec 17 '18 at 6:20
user3767495user3767495
4078
$endgroup$
G(V,E) is a simple graph with 8 vertices. The edges of G are decided by tossing the coin for each 2 vertices combination. Edge is added between any two vertices iff head is turned up. Expected number of edges in the Graph G(V,E) is.
I thought of doing like this
Let X be a random variable denoting the number of edges in the graph.
Total possible edges->$_8C_2=28$
Now, for each of those 28 edges, we tossed a coin and if it turned out to be heads,that edge was included.
(1)X=0,P(X)=$_{28}C_0 frac{1}{2^{28}}$. All those 28 tosses of coin are independent with probability of heads=Probability of getting tails=$frac{1}{2}$
(2)X=1(One edge), P(X)=$_{28}C_1 timesfrac{1}{2} times frac{1}{2^{27}}$-Means in those 28 trials, exactly 1 heads and rest tails.
(3)Similarly for all 28 edges-> $X=28,P(X)=_{28}C_{28} times frac{1}{2^{28}}$
And then we add all cases of X from 0 to 28 because $E[X]=sum x.p(x)$
But this all together seems to be a very huge number and answer is given to be 14, where I am wrong in my reasoning?
probability graph-theory
probability graph-theory
edited Dec 17 '18 at 7:04
bof
52.1k558121
asked Dec 17 '18 at 6:20
user3767495user3767495
4078
edited Dec 17 '18 at 7:04
bof
52.1k558121
asked Dec 17 '18 at 6:20
user3767495user3767495
4078
edited Dec 17 '18 at 7:04
bof
52.1k558121
edited Dec 17 '18 at 7:04
bof
52.1k558121
edited Dec 17 '18 at 7:04
bof
52.1k558121
52.1k558121
asked Dec 17 '18 at 6:20
user3767495user3767495
4078
asked Dec 17 '18 at 6:20
user3767495user3767495
4078
asked Dec 17 '18 at 6:20
user3767495user3767495
4078
4078
$begingroup$
Two remarks: (1) What you tried to write, is not very clear. You do not define what the $C_0,C_1,C_2,...$ are. (2) The problem becomes much simpler if you were to use the linearity of the expectation. By finding a collection of 'simple' random variables for which the calculation of expectation is simple, and that sum up to $X$, you will have simplified the problem many times over.
$endgroup$
– Keen-ameteur
Dec 17 '18 at 6:31
$begingroup$
It is not such a "very huge number" because the factor $frac1{2^{28}}$ is pretty small.
$endgroup$
– bof
Dec 17 '18 at 7:06
add a comment |
$begingroup$
Two remarks: (1) What you tried to write, is not very clear. You do not define what the $C_0,C_1,C_2,...$ are. (2) The problem becomes much simpler if you were to use the linearity of the expectation. By finding a collection of 'simple' random variables for which the calculation of expectation is simple, and that sum up to $X$, you will have simplified the problem many times over.
$endgroup$
– Keen-ameteur
Dec 17 '18 at 6:31
$begingroup$
It is not such a "very huge number" because the factor $frac1{2^{28}}$ is pretty small.
$endgroup$
– bof
Dec 17 '18 at 7:06
$begingroup$
Two remarks: (1) What you tried to write, is not very clear. You do not define what the $C_0,C_1,C_2,...$ are. (2) The problem becomes much simpler if you were to use the linearity of the expectation. By finding a collection of 'simple' random variables for which the calculation of expectation is simple, and that sum up to $X$, you will have simplified the problem many times over.
$endgroup$
– Keen-ameteur
Dec 17 '18 at 6:31
$begingroup$
Two remarks: (1) What you tried to write, is not very clear. You do not define what the $C_0,C_1,C_2,...$ are. (2) The problem becomes much simpler if you were to use the linearity of the expectation. By finding a collection of 'simple' random variables for which the calculation of expectation is simple, and that sum up to $X$, you will have simplified the problem many times over.
$endgroup$
– Keen-ameteur
Dec 17 '18 at 6:31
$begingroup$
It is not such a "very huge number" because the factor $frac1{2^{28}}$ is pretty small.
$endgroup$
– bof
Dec 17 '18 at 7:06
$begingroup$
It is not such a "very huge number" because the factor $frac1{2^{28}}$ is pretty small.
$endgroup$
– bof
Dec 17 '18 at 7:06
add a comment |
1 Answer
1
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$begingroup$
In other words, you want to know the expected number of times you're going to get heads in $28$ independent tosses of a fair coin. To do it your way, you can add up the $29$ terms by hand, or you can use the identity
$$sum_{k=0}^nkbinom nk=n2^{n-1}tag1$$
which is obtained by differentiating the binomial identity
$$(1+x)^n=sum_{k=0}^nbinom nkx^n$$
with respect to $x$ and then setting $x=1$. Using $(1)$ we get
$$E(X)=sum_{k=0}^{28}kbinom{28}kleft(frac12right)^kleft(frac12right)^{28-k}=left(frac12right)^{28}sum_{k=0}^{28}kbinom{28}k=left(frac12right)^{28}cdot28cdot2^{27}=14.$$
This shows that, if you toss a coin $28$ times, on average you're going to get $14$ heads. You can get the same result more easily by using the additivity of expectations: if $X$ is the sum of $28$ random variables, each of which has an expected value of $frac12$, then the expected value of $X$ is $28cdotfrac12=14$.
answered Dec 17 '18 at 7:28
bofbof
52.1k558121
$endgroup$
1
$begingroup$
Another way to find $sum_{k=0}^n k binom nk$ is to substitute $binom nk = frac nk binom{n-1}{k-1}$ in all except the $k=0$ term, getting $sum_{k=1}^n n binom{n-1}{k-1} = n sum_{j=0}^{n-1}binom{n-1}{j} = n2^{n-1}$.
$endgroup$
– Misha Lavrov
Dec 18 '18 at 4:18
add a comment |
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$begingroup$
In other words, you want to know the expected number of times you're going to get heads in $28$ independent tosses of a fair coin. To do it your way, you can add up the $29$ terms by hand, or you can use the identity
$$sum_{k=0}^nkbinom nk=n2^{n-1}tag1$$
which is obtained by differentiating the binomial identity
$$(1+x)^n=sum_{k=0}^nbinom nkx^n$$
with respect to $x$ and then setting $x=1$. Using $(1)$ we get
$$E(X)=sum_{k=0}^{28}kbinom{28}kleft(frac12right)^kleft(frac12right)^{28-k}=left(frac12right)^{28}sum_{k=0}^{28}kbinom{28}k=left(frac12right)^{28}cdot28cdot2^{27}=14.$$
This shows that, if you toss a coin $28$ times, on average you're going to get $14$ heads. You can get the same result more easily by using the additivity of expectations: if $X$ is the sum of $28$ random variables, each of which has an expected value of $frac12$, then the expected value of $X$ is $28cdotfrac12=14$.
answered Dec 17 '18 at 7:28
bofbof
52.1k558121
$endgroup$
1
$begingroup$
Another way to find $sum_{k=0}^n k binom nk$ is to substitute $binom nk = frac nk binom{n-1}{k-1}$ in all except the $k=0$ term, getting $sum_{k=1}^n n binom{n-1}{k-1} = n sum_{j=0}^{n-1}binom{n-1}{j} = n2^{n-1}$.
$endgroup$
– Misha Lavrov
Dec 18 '18 at 4:18
add a comment |
$begingroup$
In other words, you want to know the expected number of times you're going to get heads in $28$ independent tosses of a fair coin. To do it your way, you can add up the $29$ terms by hand, or you can use the identity
$$sum_{k=0}^nkbinom nk=n2^{n-1}tag1$$
which is obtained by differentiating the binomial identity
$$(1+x)^n=sum_{k=0}^nbinom nkx^n$$
with respect to $x$ and then setting $x=1$. Using $(1)$ we get
$$E(X)=sum_{k=0}^{28}kbinom{28}kleft(frac12right)^kleft(frac12right)^{28-k}=left(frac12right)^{28}sum_{k=0}^{28}kbinom{28}k=left(frac12right)^{28}cdot28cdot2^{27}=14.$$
This shows that, if you toss a coin $28$ times, on average you're going to get $14$ heads. You can get the same result more easily by using the additivity of expectations: if $X$ is the sum of $28$ random variables, each of which has an expected value of $frac12$, then the expected value of $X$ is $28cdotfrac12=14$.
answered Dec 17 '18 at 7:28
bofbof
52.1k558121
$endgroup$
1
$begingroup$
Another way to find $sum_{k=0}^n k binom nk$ is to substitute $binom nk = frac nk binom{n-1}{k-1}$ in all except the $k=0$ term, getting $sum_{k=1}^n n binom{n-1}{k-1} = n sum_{j=0}^{n-1}binom{n-1}{j} = n2^{n-1}$.
$endgroup$
– Misha Lavrov
Dec 18 '18 at 4:18
add a comment |
$begingroup$
In other words, you want to know the expected number of times you're going to get heads in $28$ independent tosses of a fair coin. To do it your way, you can add up the $29$ terms by hand, or you can use the identity
$$sum_{k=0}^nkbinom nk=n2^{n-1}tag1$$
which is obtained by differentiating the binomial identity
$$(1+x)^n=sum_{k=0}^nbinom nkx^n$$
with respect to $x$ and then setting $x=1$. Using $(1)$ we get
$$E(X)=sum_{k=0}^{28}kbinom{28}kleft(frac12right)^kleft(frac12right)^{28-k}=left(frac12right)^{28}sum_{k=0}^{28}kbinom{28}k=left(frac12right)^{28}cdot28cdot2^{27}=14.$$
This shows that, if you toss a coin $28$ times, on average you're going to get $14$ heads. You can get the same result more easily by using the additivity of expectations: if $X$ is the sum of $28$ random variables, each of which has an expected value of $frac12$, then the expected value of $X$ is $28cdotfrac12=14$.
answered Dec 17 '18 at 7:28
bofbof
52.1k558121
$endgroup$
In other words, you want to know the expected number of times you're going to get heads in $28$ independent tosses of a fair coin. To do it your way, you can add up the $29$ terms by hand, or you can use the identity
$$sum_{k=0}^nkbinom nk=n2^{n-1}tag1$$
which is obtained by differentiating the binomial identity
$$(1+x)^n=sum_{k=0}^nbinom nkx^n$$
with respect to $x$ and then setting $x=1$. Using $(1)$ we get
$$E(X)=sum_{k=0}^{28}kbinom{28}kleft(frac12right)^kleft(frac12right)^{28-k}=left(frac12right)^{28}sum_{k=0}^{28}kbinom{28}k=left(frac12right)^{28}cdot28cdot2^{27}=14.$$
This shows that, if you toss a coin $28$ times, on average you're going to get $14$ heads. You can get the same result more easily by using the additivity of expectations: if $X$ is the sum of $28$ random variables, each of which has an expected value of $frac12$, then the expected value of $X$ is $28cdotfrac12=14$.
answered Dec 17 '18 at 7:28
bofbof
52.1k558121
edited Dec 18 '18 at 2:49
answered Dec 17 '18 at 7:28
bofbof
52.1k558121
answered Dec 17 '18 at 7:28
bofbof
52.1k558121
answered Dec 17 '18 at 7:28
bofbof
52.1k558121
52.1k558121
1
$begingroup$
Another way to find $sum_{k=0}^n k binom nk$ is to substitute $binom nk = frac nk binom{n-1}{k-1}$ in all except the $k=0$ term, getting $sum_{k=1}^n n binom{n-1}{k-1} = n sum_{j=0}^{n-1}binom{n-1}{j} = n2^{n-1}$.
$endgroup$
– Misha Lavrov
Dec 18 '18 at 4:18
add a comment |
1
$begingroup$
Another way to find $sum_{k=0}^n k binom nk$ is to substitute $binom nk = frac nk binom{n-1}{k-1}$ in all except the $k=0$ term, getting $sum_{k=1}^n n binom{n-1}{k-1} = n sum_{j=0}^{n-1}binom{n-1}{j} = n2^{n-1}$.
$endgroup$
– Misha Lavrov
Dec 18 '18 at 4:18
1
1
$begingroup$
Another way to find $sum_{k=0}^n k binom nk$ is to substitute $binom nk = frac nk binom{n-1}{k-1}$ in all except the $k=0$ term, getting $sum_{k=1}^n n binom{n-1}{k-1} = n sum_{j=0}^{n-1}binom{n-1}{j} = n2^{n-1}$.
$endgroup$
– Misha Lavrov
Dec 18 '18 at 4:18
$begingroup$
Another way to find $sum_{k=0}^n k binom nk$ is to substitute $binom nk = frac nk binom{n-1}{k-1}$ in all except the $k=0$ term, getting $sum_{k=1}^n n binom{n-1}{k-1} = n sum_{j=0}^{n-1}binom{n-1}{j} = n2^{n-1}$.
$endgroup$
– Misha Lavrov
Dec 18 '18 at 4:18
add a comment |
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$begingroup$
Two remarks: (1) What you tried to write, is not very clear. You do not define what the $C_0,C_1,C_2,...$ are. (2) The problem becomes much simpler if you were to use the linearity of the expectation. By finding a collection of 'simple' random variables for which the calculation of expectation is simple, and that sum up to $X$, you will have simplified the problem many times over.
$endgroup$
– Keen-ameteur
Dec 17 '18 at 6:31
$begingroup$
It is not such a "very huge number" because the factor $frac1{2^{28}}$ is pretty small.
$endgroup$
– bof
Dec 17 '18 at 7:06