Mixing
Finding the mean and variance of the number of successes of a sequence of independent trials.
$begingroup$
In a sequence of $n$ independent trials the probability of a success at the $i^{mathrm{th}}$ trial is $p_i$. Find the mean and variance of the total number of successes.
My problem is should I let $X_i$ be the event that the $i^{mathrm{th}}$ is a success or that $i$ trials have been successful, where $X=X_1+X_2+cdots +X_n$.
probability
edited Feb 24 '15 at 20:17
Math1000
19.3k31745
asked Feb 24 '15 at 17:08
guest10923guest10923
2418
$endgroup$
add a comment |
$begingroup$
In a sequence of $n$ independent trials the probability of a success at the $i^{mathrm{th}}$ trial is $p_i$. Find the mean and variance of the total number of successes.
My problem is should I let $X_i$ be the event that the $i^{mathrm{th}}$ is a success or that $i$ trials have been successful, where $X=X_1+X_2+cdots +X_n$.
probability
edited Feb 24 '15 at 20:17
Math1000
19.3k31745
asked Feb 24 '15 at 17:08
guest10923guest10923
2418
$endgroup$
$begingroup$
Your idea is a useful one. We define the random variable $X_i$ by $X_i=1$ if we have a success on the $i$-th trial, and by $X_i=0$ otherwise. Then the number $X$ of successes is given by $X=X_1+cdots+X_n$. Now $E(X)$ is immediate by the linearity of expectation. For the variance, it will be a good idea to expand $(X_1+cdots+X_n)^2$.
$endgroup$
– André Nicolas
Feb 24 '15 at 17:14
$begingroup$
thank you, I think I can see where to go from here!
$endgroup$
– guest10923
Feb 24 '15 at 17:36
$begingroup$
You are welcome. I thought it best to outline things only, so that you could do the rest. Note that there is a simpler way to get at the variance, since we are dealing with an independent sum.
$endgroup$
– André Nicolas
Feb 24 '15 at 17:40
$begingroup$
Hmm, I can't find a way to tidy up the (X1+...+Xn)^2 expression. I tried to use the fact the events are independent therefore E(XY)=E(X)E(Y). I am not sure I know a simpler formula for the variance.
$endgroup$
– guest10923
Feb 24 '15 at 18:00
$begingroup$
The simple way is to use $text{Var}(X)=sum text{Var}(X_i)$. An easy computation (or standard fact) shows that $text{Var}(X_i)=p_i(1-p_i)$. The harder way is to expand. The mean of $X_i^2$ is $p_i$ since $X_i^2=X_i$. The cross terms have expectation $2sum_{ilt j}p_ip_j$. So the expectation of $X^2$ is $sum p_i+2sum_{ilt j}p_ip_j$. Subtract $(E(X))^2$. We get a messy expression that simplifies a lot.
$endgroup$
– André Nicolas
Feb 24 '15 at 18:12
add a comment |
$begingroup$
In a sequence of $n$ independent trials the probability of a success at the $i^{mathrm{th}}$ trial is $p_i$. Find the mean and variance of the total number of successes.
My problem is should I let $X_i$ be the event that the $i^{mathrm{th}}$ is a success or that $i$ trials have been successful, where $X=X_1+X_2+cdots +X_n$.
probability
edited Feb 24 '15 at 20:17
Math1000
19.3k31745
asked Feb 24 '15 at 17:08
guest10923guest10923
2418
$endgroup$
In a sequence of $n$ independent trials the probability of a success at the $i^{mathrm{th}}$ trial is $p_i$. Find the mean and variance of the total number of successes.
My problem is should I let $X_i$ be the event that the $i^{mathrm{th}}$ is a success or that $i$ trials have been successful, where $X=X_1+X_2+cdots +X_n$.
probability
probability
edited Feb 24 '15 at 20:17
Math1000
19.3k31745
asked Feb 24 '15 at 17:08
guest10923guest10923
2418
edited Feb 24 '15 at 20:17
Math1000
19.3k31745
asked Feb 24 '15 at 17:08
guest10923guest10923
2418
edited Feb 24 '15 at 20:17
Math1000
19.3k31745
edited Feb 24 '15 at 20:17
Math1000
19.3k31745
edited Feb 24 '15 at 20:17
Math1000
19.3k31745
19.3k31745
asked Feb 24 '15 at 17:08
guest10923guest10923
2418
asked Feb 24 '15 at 17:08
guest10923guest10923
2418
asked Feb 24 '15 at 17:08
guest10923guest10923
2418
2418
$begingroup$
Your idea is a useful one. We define the random variable $X_i$ by $X_i=1$ if we have a success on the $i$-th trial, and by $X_i=0$ otherwise. Then the number $X$ of successes is given by $X=X_1+cdots+X_n$. Now $E(X)$ is immediate by the linearity of expectation. For the variance, it will be a good idea to expand $(X_1+cdots+X_n)^2$.
$endgroup$
– André Nicolas
Feb 24 '15 at 17:14
$begingroup$
thank you, I think I can see where to go from here!
$endgroup$
– guest10923
Feb 24 '15 at 17:36
$begingroup$
You are welcome. I thought it best to outline things only, so that you could do the rest. Note that there is a simpler way to get at the variance, since we are dealing with an independent sum.
$endgroup$
– André Nicolas
Feb 24 '15 at 17:40
$begingroup$
Hmm, I can't find a way to tidy up the (X1+...+Xn)^2 expression. I tried to use the fact the events are independent therefore E(XY)=E(X)E(Y). I am not sure I know a simpler formula for the variance.
$endgroup$
– guest10923
Feb 24 '15 at 18:00
$begingroup$
The simple way is to use $text{Var}(X)=sum text{Var}(X_i)$. An easy computation (or standard fact) shows that $text{Var}(X_i)=p_i(1-p_i)$. The harder way is to expand. The mean of $X_i^2$ is $p_i$ since $X_i^2=X_i$. The cross terms have expectation $2sum_{ilt j}p_ip_j$. So the expectation of $X^2$ is $sum p_i+2sum_{ilt j}p_ip_j$. Subtract $(E(X))^2$. We get a messy expression that simplifies a lot.
$endgroup$
– André Nicolas
Feb 24 '15 at 18:12
add a comment |
$begingroup$
Your idea is a useful one. We define the random variable $X_i$ by $X_i=1$ if we have a success on the $i$-th trial, and by $X_i=0$ otherwise. Then the number $X$ of successes is given by $X=X_1+cdots+X_n$. Now $E(X)$ is immediate by the linearity of expectation. For the variance, it will be a good idea to expand $(X_1+cdots+X_n)^2$.
$endgroup$
– André Nicolas
Feb 24 '15 at 17:14
$begingroup$
thank you, I think I can see where to go from here!
$endgroup$
– guest10923
Feb 24 '15 at 17:36
$begingroup$
You are welcome. I thought it best to outline things only, so that you could do the rest. Note that there is a simpler way to get at the variance, since we are dealing with an independent sum.
$endgroup$
– André Nicolas
Feb 24 '15 at 17:40
$begingroup$
Hmm, I can't find a way to tidy up the (X1+...+Xn)^2 expression. I tried to use the fact the events are independent therefore E(XY)=E(X)E(Y). I am not sure I know a simpler formula for the variance.
$endgroup$
– guest10923
Feb 24 '15 at 18:00
$begingroup$
The simple way is to use $text{Var}(X)=sum text{Var}(X_i)$. An easy computation (or standard fact) shows that $text{Var}(X_i)=p_i(1-p_i)$. The harder way is to expand. The mean of $X_i^2$ is $p_i$ since $X_i^2=X_i$. The cross terms have expectation $2sum_{ilt j}p_ip_j$. So the expectation of $X^2$ is $sum p_i+2sum_{ilt j}p_ip_j$. Subtract $(E(X))^2$. We get a messy expression that simplifies a lot.
$endgroup$
– André Nicolas
Feb 24 '15 at 18:12
$begingroup$
Your idea is a useful one. We define the random variable $X_i$ by $X_i=1$ if we have a success on the $i$-th trial, and by $X_i=0$ otherwise. Then the number $X$ of successes is given by $X=X_1+cdots+X_n$. Now $E(X)$ is immediate by the linearity of expectation. For the variance, it will be a good idea to expand $(X_1+cdots+X_n)^2$.
$endgroup$
– André Nicolas
Feb 24 '15 at 17:14
$begingroup$
Your idea is a useful one. We define the random variable $X_i$ by $X_i=1$ if we have a success on the $i$-th trial, and by $X_i=0$ otherwise. Then the number $X$ of successes is given by $X=X_1+cdots+X_n$. Now $E(X)$ is immediate by the linearity of expectation. For the variance, it will be a good idea to expand $(X_1+cdots+X_n)^2$.
$endgroup$
– André Nicolas
Feb 24 '15 at 17:14
$begingroup$
thank you, I think I can see where to go from here!
$endgroup$
– guest10923
Feb 24 '15 at 17:36
$begingroup$
thank you, I think I can see where to go from here!
$endgroup$
– guest10923
Feb 24 '15 at 17:36
$begingroup$
You are welcome. I thought it best to outline things only, so that you could do the rest. Note that there is a simpler way to get at the variance, since we are dealing with an independent sum.
$endgroup$
– André Nicolas
Feb 24 '15 at 17:40
$begingroup$
You are welcome. I thought it best to outline things only, so that you could do the rest. Note that there is a simpler way to get at the variance, since we are dealing with an independent sum.
$endgroup$
– André Nicolas
Feb 24 '15 at 17:40
$begingroup$
Hmm, I can't find a way to tidy up the (X1+...+Xn)^2 expression. I tried to use the fact the events are independent therefore E(XY)=E(X)E(Y). I am not sure I know a simpler formula for the variance.
$endgroup$
– guest10923
Feb 24 '15 at 18:00
$begingroup$
Hmm, I can't find a way to tidy up the (X1+...+Xn)^2 expression. I tried to use the fact the events are independent therefore E(XY)=E(X)E(Y). I am not sure I know a simpler formula for the variance.
$endgroup$
– guest10923
Feb 24 '15 at 18:00
$begingroup$
The simple way is to use $text{Var}(X)=sum text{Var}(X_i)$. An easy computation (or standard fact) shows that $text{Var}(X_i)=p_i(1-p_i)$. The harder way is to expand. The mean of $X_i^2$ is $p_i$ since $X_i^2=X_i$. The cross terms have expectation $2sum_{ilt j}p_ip_j$. So the expectation of $X^2$ is $sum p_i+2sum_{ilt j}p_ip_j$. Subtract $(E(X))^2$. We get a messy expression that simplifies a lot.
$endgroup$
– André Nicolas
Feb 24 '15 at 18:12
$begingroup$
The simple way is to use $text{Var}(X)=sum text{Var}(X_i)$. An easy computation (or standard fact) shows that $text{Var}(X_i)=p_i(1-p_i)$. The harder way is to expand. The mean of $X_i^2$ is $p_i$ since $X_i^2=X_i$. The cross terms have expectation $2sum_{ilt j}p_ip_j$. So the expectation of $X^2$ is $sum p_i+2sum_{ilt j}p_ip_j$. Subtract $(E(X))^2$. We get a messy expression that simplifies a lot.
$endgroup$
– André Nicolas
Feb 24 '15 at 18:12
add a comment |
1 Answer
1
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$begingroup$
Recall that for independent random variables $X_i$:
$$mathbb Eleft[sum_{i=1}^n X_iright] = sum_{i=1}^nmathbb E[X_i] $$
and
$$operatorname{Var}left(sum_{i=1}^n X_iright) = sum_{i=1}^n operatorname{Var}(X_i). $$
Applying these with $mathbb E[X_i]=p_i$ and $operatorname{Var}(X_i)=p_i(1-p_i)$ we get that the mean and variance of the sum are
$$sum_{i=1}^n p_i$$
and
$$ sum_{i=1}^n p_i(1-p_i),$$
respectively.
answered Feb 24 '15 at 20:41
Math1000Math1000
19.3k31745
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Recall that for independent random variables $X_i$:
$$mathbb Eleft[sum_{i=1}^n X_iright] = sum_{i=1}^nmathbb E[X_i] $$
and
$$operatorname{Var}left(sum_{i=1}^n X_iright) = sum_{i=1}^n operatorname{Var}(X_i). $$
Applying these with $mathbb E[X_i]=p_i$ and $operatorname{Var}(X_i)=p_i(1-p_i)$ we get that the mean and variance of the sum are
$$sum_{i=1}^n p_i$$
and
$$ sum_{i=1}^n p_i(1-p_i),$$
respectively.
answered Feb 24 '15 at 20:41
Math1000Math1000
19.3k31745
$endgroup$
add a comment |
$begingroup$
Recall that for independent random variables $X_i$:
$$mathbb Eleft[sum_{i=1}^n X_iright] = sum_{i=1}^nmathbb E[X_i] $$
and
$$operatorname{Var}left(sum_{i=1}^n X_iright) = sum_{i=1}^n operatorname{Var}(X_i). $$
Applying these with $mathbb E[X_i]=p_i$ and $operatorname{Var}(X_i)=p_i(1-p_i)$ we get that the mean and variance of the sum are
$$sum_{i=1}^n p_i$$
and
$$ sum_{i=1}^n p_i(1-p_i),$$
respectively.
answered Feb 24 '15 at 20:41
Math1000Math1000
19.3k31745
$endgroup$
add a comment |
$begingroup$
Recall that for independent random variables $X_i$:
$$mathbb Eleft[sum_{i=1}^n X_iright] = sum_{i=1}^nmathbb E[X_i] $$
and
$$operatorname{Var}left(sum_{i=1}^n X_iright) = sum_{i=1}^n operatorname{Var}(X_i). $$
Applying these with $mathbb E[X_i]=p_i$ and $operatorname{Var}(X_i)=p_i(1-p_i)$ we get that the mean and variance of the sum are
$$sum_{i=1}^n p_i$$
and
$$ sum_{i=1}^n p_i(1-p_i),$$
respectively.
answered Feb 24 '15 at 20:41
Math1000Math1000
19.3k31745
$endgroup$
Recall that for independent random variables $X_i$:
$$mathbb Eleft[sum_{i=1}^n X_iright] = sum_{i=1}^nmathbb E[X_i] $$
and
$$operatorname{Var}left(sum_{i=1}^n X_iright) = sum_{i=1}^n operatorname{Var}(X_i). $$
Applying these with $mathbb E[X_i]=p_i$ and $operatorname{Var}(X_i)=p_i(1-p_i)$ we get that the mean and variance of the sum are
$$sum_{i=1}^n p_i$$
and
$$ sum_{i=1}^n p_i(1-p_i),$$
respectively.
answered Feb 24 '15 at 20:41
Math1000Math1000
19.3k31745
answered Feb 24 '15 at 20:41
Math1000Math1000
19.3k31745
answered Feb 24 '15 at 20:41
Math1000Math1000
19.3k31745
answered Feb 24 '15 at 20:41
Math1000Math1000
19.3k31745
19.3k31745
add a comment |
add a comment |
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$begingroup$
Your idea is a useful one. We define the random variable $X_i$ by $X_i=1$ if we have a success on the $i$-th trial, and by $X_i=0$ otherwise. Then the number $X$ of successes is given by $X=X_1+cdots+X_n$. Now $E(X)$ is immediate by the linearity of expectation. For the variance, it will be a good idea to expand $(X_1+cdots+X_n)^2$.
$endgroup$
– André Nicolas
Feb 24 '15 at 17:14
$begingroup$
thank you, I think I can see where to go from here!
$endgroup$
– guest10923
Feb 24 '15 at 17:36
$begingroup$
You are welcome. I thought it best to outline things only, so that you could do the rest. Note that there is a simpler way to get at the variance, since we are dealing with an independent sum.
$endgroup$
– André Nicolas
Feb 24 '15 at 17:40
$begingroup$
Hmm, I can't find a way to tidy up the (X1+...+Xn)^2 expression. I tried to use the fact the events are independent therefore E(XY)=E(X)E(Y). I am not sure I know a simpler formula for the variance.
$endgroup$
– guest10923
Feb 24 '15 at 18:00
$begingroup$
The simple way is to use $text{Var}(X)=sum text{Var}(X_i)$. An easy computation (or standard fact) shows that $text{Var}(X_i)=p_i(1-p_i)$. The harder way is to expand. The mean of $X_i^2$ is $p_i$ since $X_i^2=X_i$. The cross terms have expectation $2sum_{ilt j}p_ip_j$. So the expectation of $X^2$ is $sum p_i+2sum_{ilt j}p_ip_j$. Subtract $(E(X))^2$. We get a messy expression that simplifies a lot.
$endgroup$
– André Nicolas
Feb 24 '15 at 18:12