Mixing
Tossing a fair coin $N$ times, what is the variance?
$begingroup$
$X$ is a random variable that equals the number of heads.
$E(X)$ is the expected value of the number of heads when a coin is tossed $n$ times, so since $P(X)= 0.5, E(X) = frac n2$
I know that the variance is calculated by $V(X) = E(X^2) - E(X)^2$, but my question is how would I calculate $E(X^2)$?
probability
edited Jan 22 at 10:54
EdOverflow
25119
asked Jan 22 at 10:44
2000mroliver2000mroliver
644
$endgroup$
add a comment |
$begingroup$
$X$ is a random variable that equals the number of heads.
$E(X)$ is the expected value of the number of heads when a coin is tossed $n$ times, so since $P(X)= 0.5, E(X) = frac n2$
I know that the variance is calculated by $V(X) = E(X^2) - E(X)^2$, but my question is how would I calculate $E(X^2)$?
probability
edited Jan 22 at 10:54
EdOverflow
25119
asked Jan 22 at 10:44
2000mroliver2000mroliver
644
$endgroup$
add a comment |
$begingroup$
$X$ is a random variable that equals the number of heads.
$E(X)$ is the expected value of the number of heads when a coin is tossed $n$ times, so since $P(X)= 0.5, E(X) = frac n2$
I know that the variance is calculated by $V(X) = E(X^2) - E(X)^2$, but my question is how would I calculate $E(X^2)$?
probability
edited Jan 22 at 10:54
EdOverflow
25119
asked Jan 22 at 10:44
2000mroliver2000mroliver
644
$endgroup$
$X$ is a random variable that equals the number of heads.
$E(X)$ is the expected value of the number of heads when a coin is tossed $n$ times, so since $P(X)= 0.5, E(X) = frac n2$
I know that the variance is calculated by $V(X) = E(X^2) - E(X)^2$, but my question is how would I calculate $E(X^2)$?
probability
probability
edited Jan 22 at 10:54
EdOverflow
25119
asked Jan 22 at 10:44
2000mroliver2000mroliver
644
edited Jan 22 at 10:54
EdOverflow
25119
asked Jan 22 at 10:44
2000mroliver2000mroliver
644
edited Jan 22 at 10:54
EdOverflow
25119
edited Jan 22 at 10:54
EdOverflow
25119
edited Jan 22 at 10:54
EdOverflow
25119
25119
asked Jan 22 at 10:44
2000mroliver2000mroliver
644
asked Jan 22 at 10:44
2000mroliver2000mroliver
644
asked Jan 22 at 10:44
2000mroliver2000mroliver
644
644
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
This is a classical example of a binomial experiment, in short the probability distribution of the variable $X$ can be written as
$$
P(X = x) = {n choose x}p^x (1 - p)^{n-x} tag{1}
$$
in your case $p = 1/2$ is the probability of getting heads when the coin is tossed. And from here you can do the math to show
$$
mathbb{E}[X] = sum_x x P(X = x) = np
$$
and
$$
mathbb{V}{rm ar}[X] = sum_x (x - mathbb{E}[X])^2 P(X = x) = np(1 - p)
$$
answered Jan 22 at 11:05
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
okay thank you, is the formula np(1-p) for the variance something that I can use in any problem?
$endgroup$
– 2000mroliver
Jan 22 at 11:09
$begingroup$
@OliverBeck If the problem is binomial, then yes. Otherwise you cannot start applying this result to every problem you find.
$endgroup$
– caverac
Jan 22 at 11:10
add a comment |
$begingroup$
You can calculate $mathbb EX^2$ purely on base of the equality:$$mathbb EX^2=sum_{k=0}^nk^2P(X=k)$$
If you take that (cumbersome) route then it is handsome to make use of: $$cdots=sum_{k=0}^nk(k-1)P(X=k)+sum_{k=0}^nkP(X=k)=sum_{k=0}^nk(k-1)P(X=k)+mathbb EX$$
However you are dealing with a random variable $X$ that has binomial distribution with parameters $n$ and $p=frac12$. Then it is much more handsome to write $X=B_1+cdots+B_n$ where the $B_i$ are iid random variables having Bernoulli-distribution with parameter $p=frac12$.
Then with linearity of expectation, symmetry and independence we find:$$mathbb EX^2=mathbb Eleft[sum_{i=1}^nsum_{j=1}^nB_iB_jright]=nmathbb EB_1^2+n(n-1)mathbb EB_1mathbb EB_2=ncdotfrac12+n(n-1)cdotfrac12frac12=frac14n+frac14n^2$$leading to $mathsf{Var}X=frac14n+frac14n^2-left(frac12nright)^2=frac14n$.
If you are aiming only at variance and not specifically $mathbb EX^2$ then there is even a faster route:$$mathsf{Var}
X=sum_{k=1}^nmathsf{Var}B_k=nmathsf{Var}B_1=frac14n$$
This because the variance is also linear under the extra condition that the terms are independent.
answered Jan 22 at 11:15
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
HINT
Given the random variable $X$ and its corresponding probability mass function $f_{X}(x)$, we have
$$textbf{E}(g(X)) = sum_{x}g(x)f(x)$$
answered Jan 22 at 11:01
user1337user1337
46110
$endgroup$
$begingroup$
sorry but I am unfamiliar with the probability mass function, is it necessary here to find the variance?
$endgroup$
– 2000mroliver
Jan 22 at 11:06
add a comment |
$begingroup$
There is some misunderstanding in your model, because you mixed up one single toss with the whole experiment.
The right model:
You have $n$ random variables $X_i, i=1,dots,n$ which are identically and independently distributed. Each $X_i$ returns $1$ if you toss head and 0 else. Therefore the distribution is given by
$$
P(X_i = 1) = 0.5 quadtext{and} quad P(X_i = 0) = 0.5 quadtext{for all} quad i in {1,dots,n}.
$$
The random variable $X$ which counts the number of heads is $X = sum_{i=1}^{n} X_i$. Therefore, you have $P(X=k) = binom{n}{k}cdot 0.5^{n}$.
Now remember that $mathbb{E}(Y) := sum_{iin Y^{-1}(mathbb{R})} iP(Y=i)$ (for discrete random variables)
answered Jan 22 at 11:17
Nathanael SkrepekNathanael Skrepek
1,7111615
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a classical example of a binomial experiment, in short the probability distribution of the variable $X$ can be written as
$$
P(X = x) = {n choose x}p^x (1 - p)^{n-x} tag{1}
$$
in your case $p = 1/2$ is the probability of getting heads when the coin is tossed. And from here you can do the math to show
$$
mathbb{E}[X] = sum_x x P(X = x) = np
$$
and
$$
mathbb{V}{rm ar}[X] = sum_x (x - mathbb{E}[X])^2 P(X = x) = np(1 - p)
$$
answered Jan 22 at 11:05
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
okay thank you, is the formula np(1-p) for the variance something that I can use in any problem?
$endgroup$
– 2000mroliver
Jan 22 at 11:09
$begingroup$
@OliverBeck If the problem is binomial, then yes. Otherwise you cannot start applying this result to every problem you find.
$endgroup$
– caverac
Jan 22 at 11:10
add a comment |
$begingroup$
This is a classical example of a binomial experiment, in short the probability distribution of the variable $X$ can be written as
$$
P(X = x) = {n choose x}p^x (1 - p)^{n-x} tag{1}
$$
in your case $p = 1/2$ is the probability of getting heads when the coin is tossed. And from here you can do the math to show
$$
mathbb{E}[X] = sum_x x P(X = x) = np
$$
and
$$
mathbb{V}{rm ar}[X] = sum_x (x - mathbb{E}[X])^2 P(X = x) = np(1 - p)
$$
answered Jan 22 at 11:05
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
okay thank you, is the formula np(1-p) for the variance something that I can use in any problem?
$endgroup$
– 2000mroliver
Jan 22 at 11:09
$begingroup$
@OliverBeck If the problem is binomial, then yes. Otherwise you cannot start applying this result to every problem you find.
$endgroup$
– caverac
Jan 22 at 11:10
add a comment |
$begingroup$
This is a classical example of a binomial experiment, in short the probability distribution of the variable $X$ can be written as
$$
P(X = x) = {n choose x}p^x (1 - p)^{n-x} tag{1}
$$
in your case $p = 1/2$ is the probability of getting heads when the coin is tossed. And from here you can do the math to show
$$
mathbb{E}[X] = sum_x x P(X = x) = np
$$
and
$$
mathbb{V}{rm ar}[X] = sum_x (x - mathbb{E}[X])^2 P(X = x) = np(1 - p)
$$
answered Jan 22 at 11:05
caveraccaverac
14.8k31130
$endgroup$
This is a classical example of a binomial experiment, in short the probability distribution of the variable $X$ can be written as
$$
P(X = x) = {n choose x}p^x (1 - p)^{n-x} tag{1}
$$
in your case $p = 1/2$ is the probability of getting heads when the coin is tossed. And from here you can do the math to show
$$
mathbb{E}[X] = sum_x x P(X = x) = np
$$
and
$$
mathbb{V}{rm ar}[X] = sum_x (x - mathbb{E}[X])^2 P(X = x) = np(1 - p)
$$
answered Jan 22 at 11:05
caveraccaverac
14.8k31130
answered Jan 22 at 11:05
caveraccaverac
14.8k31130
answered Jan 22 at 11:05
caveraccaverac
14.8k31130
answered Jan 22 at 11:05
caveraccaverac
14.8k31130
14.8k31130
$begingroup$
okay thank you, is the formula np(1-p) for the variance something that I can use in any problem?
$endgroup$
– 2000mroliver
Jan 22 at 11:09
$begingroup$
@OliverBeck If the problem is binomial, then yes. Otherwise you cannot start applying this result to every problem you find.
$endgroup$
– caverac
Jan 22 at 11:10
add a comment |
$begingroup$
okay thank you, is the formula np(1-p) for the variance something that I can use in any problem?
$endgroup$
– 2000mroliver
Jan 22 at 11:09
$begingroup$
@OliverBeck If the problem is binomial, then yes. Otherwise you cannot start applying this result to every problem you find.
$endgroup$
– caverac
Jan 22 at 11:10
$begingroup$
okay thank you, is the formula np(1-p) for the variance something that I can use in any problem?
$endgroup$
– 2000mroliver
Jan 22 at 11:09
$begingroup$
okay thank you, is the formula np(1-p) for the variance something that I can use in any problem?
$endgroup$
– 2000mroliver
Jan 22 at 11:09
$begingroup$
@OliverBeck If the problem is binomial, then yes. Otherwise you cannot start applying this result to every problem you find.
$endgroup$
– caverac
Jan 22 at 11:10
$begingroup$
@OliverBeck If the problem is binomial, then yes. Otherwise you cannot start applying this result to every problem you find.
$endgroup$
– caverac
Jan 22 at 11:10
add a comment |
$begingroup$
You can calculate $mathbb EX^2$ purely on base of the equality:$$mathbb EX^2=sum_{k=0}^nk^2P(X=k)$$
If you take that (cumbersome) route then it is handsome to make use of: $$cdots=sum_{k=0}^nk(k-1)P(X=k)+sum_{k=0}^nkP(X=k)=sum_{k=0}^nk(k-1)P(X=k)+mathbb EX$$
However you are dealing with a random variable $X$ that has binomial distribution with parameters $n$ and $p=frac12$. Then it is much more handsome to write $X=B_1+cdots+B_n$ where the $B_i$ are iid random variables having Bernoulli-distribution with parameter $p=frac12$.
Then with linearity of expectation, symmetry and independence we find:$$mathbb EX^2=mathbb Eleft[sum_{i=1}^nsum_{j=1}^nB_iB_jright]=nmathbb EB_1^2+n(n-1)mathbb EB_1mathbb EB_2=ncdotfrac12+n(n-1)cdotfrac12frac12=frac14n+frac14n^2$$leading to $mathsf{Var}X=frac14n+frac14n^2-left(frac12nright)^2=frac14n$.
If you are aiming only at variance and not specifically $mathbb EX^2$ then there is even a faster route:$$mathsf{Var}
X=sum_{k=1}^nmathsf{Var}B_k=nmathsf{Var}B_1=frac14n$$
This because the variance is also linear under the extra condition that the terms are independent.
answered Jan 22 at 11:15
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
You can calculate $mathbb EX^2$ purely on base of the equality:$$mathbb EX^2=sum_{k=0}^nk^2P(X=k)$$
If you take that (cumbersome) route then it is handsome to make use of: $$cdots=sum_{k=0}^nk(k-1)P(X=k)+sum_{k=0}^nkP(X=k)=sum_{k=0}^nk(k-1)P(X=k)+mathbb EX$$
However you are dealing with a random variable $X$ that has binomial distribution with parameters $n$ and $p=frac12$. Then it is much more handsome to write $X=B_1+cdots+B_n$ where the $B_i$ are iid random variables having Bernoulli-distribution with parameter $p=frac12$.
Then with linearity of expectation, symmetry and independence we find:$$mathbb EX^2=mathbb Eleft[sum_{i=1}^nsum_{j=1}^nB_iB_jright]=nmathbb EB_1^2+n(n-1)mathbb EB_1mathbb EB_2=ncdotfrac12+n(n-1)cdotfrac12frac12=frac14n+frac14n^2$$leading to $mathsf{Var}X=frac14n+frac14n^2-left(frac12nright)^2=frac14n$.
If you are aiming only at variance and not specifically $mathbb EX^2$ then there is even a faster route:$$mathsf{Var}
X=sum_{k=1}^nmathsf{Var}B_k=nmathsf{Var}B_1=frac14n$$
This because the variance is also linear under the extra condition that the terms are independent.
answered Jan 22 at 11:15
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
You can calculate $mathbb EX^2$ purely on base of the equality:$$mathbb EX^2=sum_{k=0}^nk^2P(X=k)$$
If you take that (cumbersome) route then it is handsome to make use of: $$cdots=sum_{k=0}^nk(k-1)P(X=k)+sum_{k=0}^nkP(X=k)=sum_{k=0}^nk(k-1)P(X=k)+mathbb EX$$
However you are dealing with a random variable $X$ that has binomial distribution with parameters $n$ and $p=frac12$. Then it is much more handsome to write $X=B_1+cdots+B_n$ where the $B_i$ are iid random variables having Bernoulli-distribution with parameter $p=frac12$.
Then with linearity of expectation, symmetry and independence we find:$$mathbb EX^2=mathbb Eleft[sum_{i=1}^nsum_{j=1}^nB_iB_jright]=nmathbb EB_1^2+n(n-1)mathbb EB_1mathbb EB_2=ncdotfrac12+n(n-1)cdotfrac12frac12=frac14n+frac14n^2$$leading to $mathsf{Var}X=frac14n+frac14n^2-left(frac12nright)^2=frac14n$.
If you are aiming only at variance and not specifically $mathbb EX^2$ then there is even a faster route:$$mathsf{Var}
X=sum_{k=1}^nmathsf{Var}B_k=nmathsf{Var}B_1=frac14n$$
This because the variance is also linear under the extra condition that the terms are independent.
answered Jan 22 at 11:15
drhabdrhab
103k545136
$endgroup$
You can calculate $mathbb EX^2$ purely on base of the equality:$$mathbb EX^2=sum_{k=0}^nk^2P(X=k)$$
If you take that (cumbersome) route then it is handsome to make use of: $$cdots=sum_{k=0}^nk(k-1)P(X=k)+sum_{k=0}^nkP(X=k)=sum_{k=0}^nk(k-1)P(X=k)+mathbb EX$$
However you are dealing with a random variable $X$ that has binomial distribution with parameters $n$ and $p=frac12$. Then it is much more handsome to write $X=B_1+cdots+B_n$ where the $B_i$ are iid random variables having Bernoulli-distribution with parameter $p=frac12$.
Then with linearity of expectation, symmetry and independence we find:$$mathbb EX^2=mathbb Eleft[sum_{i=1}^nsum_{j=1}^nB_iB_jright]=nmathbb EB_1^2+n(n-1)mathbb EB_1mathbb EB_2=ncdotfrac12+n(n-1)cdotfrac12frac12=frac14n+frac14n^2$$leading to $mathsf{Var}X=frac14n+frac14n^2-left(frac12nright)^2=frac14n$.
If you are aiming only at variance and not specifically $mathbb EX^2$ then there is even a faster route:$$mathsf{Var}
X=sum_{k=1}^nmathsf{Var}B_k=nmathsf{Var}B_1=frac14n$$
This because the variance is also linear under the extra condition that the terms are independent.
answered Jan 22 at 11:15
drhabdrhab
103k545136
answered Jan 22 at 11:15
drhabdrhab
103k545136
answered Jan 22 at 11:15
drhabdrhab
103k545136
answered Jan 22 at 11:15
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
$begingroup$
HINT
Given the random variable $X$ and its corresponding probability mass function $f_{X}(x)$, we have
$$textbf{E}(g(X)) = sum_{x}g(x)f(x)$$
answered Jan 22 at 11:01
user1337user1337
46110
$endgroup$
$begingroup$
sorry but I am unfamiliar with the probability mass function, is it necessary here to find the variance?
$endgroup$
– 2000mroliver
Jan 22 at 11:06
add a comment |
$begingroup$
HINT
Given the random variable $X$ and its corresponding probability mass function $f_{X}(x)$, we have
$$textbf{E}(g(X)) = sum_{x}g(x)f(x)$$
answered Jan 22 at 11:01
user1337user1337
46110
$endgroup$
$begingroup$
sorry but I am unfamiliar with the probability mass function, is it necessary here to find the variance?
$endgroup$
– 2000mroliver
Jan 22 at 11:06
add a comment |
$begingroup$
HINT
Given the random variable $X$ and its corresponding probability mass function $f_{X}(x)$, we have
$$textbf{E}(g(X)) = sum_{x}g(x)f(x)$$
answered Jan 22 at 11:01
user1337user1337
46110
$endgroup$
HINT
Given the random variable $X$ and its corresponding probability mass function $f_{X}(x)$, we have
$$textbf{E}(g(X)) = sum_{x}g(x)f(x)$$
answered Jan 22 at 11:01
user1337user1337
46110
answered Jan 22 at 11:01
user1337user1337
46110
answered Jan 22 at 11:01
user1337user1337
46110
answered Jan 22 at 11:01
user1337user1337
46110
46110
$begingroup$
sorry but I am unfamiliar with the probability mass function, is it necessary here to find the variance?
$endgroup$
– 2000mroliver
Jan 22 at 11:06
add a comment |
$begingroup$
sorry but I am unfamiliar with the probability mass function, is it necessary here to find the variance?
$endgroup$
– 2000mroliver
Jan 22 at 11:06
$begingroup$
sorry but I am unfamiliar with the probability mass function, is it necessary here to find the variance?
$endgroup$
– 2000mroliver
Jan 22 at 11:06
$begingroup$
sorry but I am unfamiliar with the probability mass function, is it necessary here to find the variance?
$endgroup$
– 2000mroliver
Jan 22 at 11:06
add a comment |
$begingroup$
There is some misunderstanding in your model, because you mixed up one single toss with the whole experiment.
The right model:
You have $n$ random variables $X_i, i=1,dots,n$ which are identically and independently distributed. Each $X_i$ returns $1$ if you toss head and 0 else. Therefore the distribution is given by
$$
P(X_i = 1) = 0.5 quadtext{and} quad P(X_i = 0) = 0.5 quadtext{for all} quad i in {1,dots,n}.
$$
The random variable $X$ which counts the number of heads is $X = sum_{i=1}^{n} X_i$. Therefore, you have $P(X=k) = binom{n}{k}cdot 0.5^{n}$.
Now remember that $mathbb{E}(Y) := sum_{iin Y^{-1}(mathbb{R})} iP(Y=i)$ (for discrete random variables)
answered Jan 22 at 11:17
Nathanael SkrepekNathanael Skrepek
1,7111615
$endgroup$
add a comment |
$begingroup$
There is some misunderstanding in your model, because you mixed up one single toss with the whole experiment.
The right model:
You have $n$ random variables $X_i, i=1,dots,n$ which are identically and independently distributed. Each $X_i$ returns $1$ if you toss head and 0 else. Therefore the distribution is given by
$$
P(X_i = 1) = 0.5 quadtext{and} quad P(X_i = 0) = 0.5 quadtext{for all} quad i in {1,dots,n}.
$$
The random variable $X$ which counts the number of heads is $X = sum_{i=1}^{n} X_i$. Therefore, you have $P(X=k) = binom{n}{k}cdot 0.5^{n}$.
Now remember that $mathbb{E}(Y) := sum_{iin Y^{-1}(mathbb{R})} iP(Y=i)$ (for discrete random variables)
answered Jan 22 at 11:17
Nathanael SkrepekNathanael Skrepek
1,7111615
$endgroup$
add a comment |
$begingroup$
There is some misunderstanding in your model, because you mixed up one single toss with the whole experiment.
The right model:
You have $n$ random variables $X_i, i=1,dots,n$ which are identically and independently distributed. Each $X_i$ returns $1$ if you toss head and 0 else. Therefore the distribution is given by
$$
P(X_i = 1) = 0.5 quadtext{and} quad P(X_i = 0) = 0.5 quadtext{for all} quad i in {1,dots,n}.
$$
The random variable $X$ which counts the number of heads is $X = sum_{i=1}^{n} X_i$. Therefore, you have $P(X=k) = binom{n}{k}cdot 0.5^{n}$.
Now remember that $mathbb{E}(Y) := sum_{iin Y^{-1}(mathbb{R})} iP(Y=i)$ (for discrete random variables)
answered Jan 22 at 11:17
Nathanael SkrepekNathanael Skrepek
1,7111615
$endgroup$
There is some misunderstanding in your model, because you mixed up one single toss with the whole experiment.
The right model:
You have $n$ random variables $X_i, i=1,dots,n$ which are identically and independently distributed. Each $X_i$ returns $1$ if you toss head and 0 else. Therefore the distribution is given by
$$
P(X_i = 1) = 0.5 quadtext{and} quad P(X_i = 0) = 0.5 quadtext{for all} quad i in {1,dots,n}.
$$
The random variable $X$ which counts the number of heads is $X = sum_{i=1}^{n} X_i$. Therefore, you have $P(X=k) = binom{n}{k}cdot 0.5^{n}$.
Now remember that $mathbb{E}(Y) := sum_{iin Y^{-1}(mathbb{R})} iP(Y=i)$ (for discrete random variables)
answered Jan 22 at 11:17
Nathanael SkrepekNathanael Skrepek
1,7111615
edited Jan 22 at 11:22
answered Jan 22 at 11:17
Nathanael SkrepekNathanael Skrepek
1,7111615
answered Jan 22 at 11:17
Nathanael SkrepekNathanael Skrepek
1,7111615
answered Jan 22 at 11:17
Nathanael SkrepekNathanael Skrepek
1,7111615
1,7111615
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