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Is is possible to calculate the expected value for a continuous variable using the sum of discrete distribution?
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I would like to know if it is possible to calculate the expected value for a continuous random variable as:
$ mathopmathbbE(x)=int xf(x)d(x)=frac1Nsum_i=1^NX_i $
If that is true, how can we convert the integral to the sum? Also, is the sum part $frac1Nsum_i=1^NX_i$ used to calculate the discrete uniform distribution?
probability expectation
asked Jul 19 at 17:38
Adam Amin
11
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favorite
I would like to know if it is possible to calculate the expected value for a continuous random variable as:
$ mathopmathbbE(x)=int xf(x)d(x)=frac1Nsum_i=1^NX_i $
If that is true, how can we convert the integral to the sum? Also, is the sum part $frac1Nsum_i=1^NX_i$ used to calculate the discrete uniform distribution?
probability expectation
asked Jul 19 at 17:38
Adam Amin
11
Can you explain what is the random variable (or random variables)? What is $x$? What is $X_i$? What is $N$? If $X$ is a random variable with PDF $f_X(x)$ then indeed $E[X] = int_-infty^infty xf_X(x)dx$.
– Michael
Jul 19 at 17:49
@Michael I have a series of different values that I want to calculate the expected value for each of them. An example of the series is 5.6, 3.2, 1.9, 6.1, 7.2. $N$ in this case is 5, $X_i$ is each value in the series.
– Adam Amin
Jul 19 at 17:56
@AdamAmin And where is the continuous distribution?
– callculus
Jul 19 at 18:03
@callculus I am not sure. It is confusing me how to calculate the expected value for each value in the series. I need this because I am calculating the autocorrelation for the series I have and part of the equation is to get the expected value.
– Adam Amin
Jul 19 at 18:09
1
This is called monte carlo integration
– RHowe
Jul 19 at 18:13
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I would like to know if it is possible to calculate the expected value for a continuous random variable as:
$ mathopmathbbE(x)=int xf(x)d(x)=frac1Nsum_i=1^NX_i $
If that is true, how can we convert the integral to the sum? Also, is the sum part $frac1Nsum_i=1^NX_i$ used to calculate the discrete uniform distribution?
probability expectation
asked Jul 19 at 17:38
Adam Amin
11
I would like to know if it is possible to calculate the expected value for a continuous random variable as:
$ mathopmathbbE(x)=int xf(x)d(x)=frac1Nsum_i=1^NX_i $
If that is true, how can we convert the integral to the sum? Also, is the sum part $frac1Nsum_i=1^NX_i$ used to calculate the discrete uniform distribution?
probability expectation
asked Jul 19 at 17:38
Adam Amin
11
asked Jul 19 at 17:38
Adam Amin
11
asked Jul 19 at 17:38
Adam Amin
11
asked Jul 19 at 17:38
Adam Amin
11
11
Can you explain what is the random variable (or random variables)? What is $x$? What is $X_i$? What is $N$? If $X$ is a random variable with PDF $f_X(x)$ then indeed $E[X] = int_-infty^infty xf_X(x)dx$.
– Michael
Jul 19 at 17:49
@Michael I have a series of different values that I want to calculate the expected value for each of them. An example of the series is 5.6, 3.2, 1.9, 6.1, 7.2. $N$ in this case is 5, $X_i$ is each value in the series.
– Adam Amin
Jul 19 at 17:56
@AdamAmin And where is the continuous distribution?
– callculus
Jul 19 at 18:03
@callculus I am not sure. It is confusing me how to calculate the expected value for each value in the series. I need this because I am calculating the autocorrelation for the series I have and part of the equation is to get the expected value.
– Adam Amin
Jul 19 at 18:09
1
This is called monte carlo integration
– RHowe
Jul 19 at 18:13
add a comment |Â
Can you explain what is the random variable (or random variables)? What is $x$? What is $X_i$? What is $N$? If $X$ is a random variable with PDF $f_X(x)$ then indeed $E[X] = int_-infty^infty xf_X(x)dx$.
– Michael
Jul 19 at 17:49
@Michael I have a series of different values that I want to calculate the expected value for each of them. An example of the series is 5.6, 3.2, 1.9, 6.1, 7.2. $N$ in this case is 5, $X_i$ is each value in the series.
– Adam Amin
Jul 19 at 17:56
@AdamAmin And where is the continuous distribution?
– callculus
Jul 19 at 18:03
@callculus I am not sure. It is confusing me how to calculate the expected value for each value in the series. I need this because I am calculating the autocorrelation for the series I have and part of the equation is to get the expected value.
– Adam Amin
Jul 19 at 18:09
1
This is called monte carlo integration
– RHowe
Jul 19 at 18:13
Can you explain what is the random variable (or random variables)? What is $x$? What is $X_i$? What is $N$? If $X$ is a random variable with PDF $f_X(x)$ then indeed $E[X] = int_-infty^infty xf_X(x)dx$.
– Michael
Jul 19 at 17:49
Can you explain what is the random variable (or random variables)? What is $x$? What is $X_i$? What is $N$? If $X$ is a random variable with PDF $f_X(x)$ then indeed $E[X] = int_-infty^infty xf_X(x)dx$.
– Michael
Jul 19 at 17:49
@Michael I have a series of different values that I want to calculate the expected value for each of them. An example of the series is 5.6, 3.2, 1.9, 6.1, 7.2. $N$ in this case is 5, $X_i$ is each value in the series.
– Adam Amin
Jul 19 at 17:56
@Michael I have a series of different values that I want to calculate the expected value for each of them. An example of the series is 5.6, 3.2, 1.9, 6.1, 7.2. $N$ in this case is 5, $X_i$ is each value in the series.
– Adam Amin
Jul 19 at 17:56
@AdamAmin And where is the continuous distribution?
– callculus
Jul 19 at 18:03
@AdamAmin And where is the continuous distribution?
– callculus
Jul 19 at 18:03
@callculus I am not sure. It is confusing me how to calculate the expected value for each value in the series. I need this because I am calculating the autocorrelation for the series I have and part of the equation is to get the expected value.
– Adam Amin
Jul 19 at 18:09
@callculus I am not sure. It is confusing me how to calculate the expected value for each value in the series. I need this because I am calculating the autocorrelation for the series I have and part of the equation is to get the expected value.
– Adam Amin
Jul 19 at 18:09
1
1
This is called monte carlo integration
– RHowe
Jul 19 at 18:13
This is called monte carlo integration
– RHowe
Jul 19 at 18:13
add a comment |Â
1 Answer
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I think you are asking if it is possible to estimate the mean of a continuous distribution by taking the sample mean of a large sample (of size $N$) from that distribution. If so, the answer is Yes, but be aware that the result is an estimate and that the error of the estimate depends on the sample size $N$ (typically, smaller error for larger $N$). Here are some examples in which
we sample at random from a few known distributions. (Sampling done using
R statistical software.)
1) $N=50$ observations from the distributon $mathsfNorm(mu = 75,, sigma=3).$ The R statement rnorm samples at random from a specified normal population. You will get a somewhat different estimate each time you take
a sample and find its mean. In 95% of runs the answer will be within
x = rnorm(50, 75, 3); mean(x); 2*sd(x)/sqrt(50)
## 74.84001 # aprx E(X) = 75
## 0.7656806 # 95% margin of simulation error
2) $N = 5000$ observations from $mathsfNorm(mu = 75,, sigma=3).$
x = rnorm(5000, 75, 3); mean(x); 2*sd(x)/sqrt(5000)
## 75.00454
## 0.08390667
3) $N = 1000$ observations from $mathsfUnif(a = 50, b = 100).$ The mean of this population distribution is $mu = 75.$
y = runif(1000, 50, 100); mean(y); 2*sd(y)/sqrt(1000)
## 75.54673
## 0.900703
4) $N = 10^6$ observations from an exponential distribution with rate 1/5.
This distribution has mean $mu = 5.$
w = rexp(10^6, 1/5); mean(w); 2*sd(w)/sqrt(10^6)
[1] 4.997817
[1] 0.00998948
5) $n = 10,000$ observations from the beta distribution with shape parameters $alpha = 1.2$ and $beta = 2.8.$ You can look in your text or read about the beta distribution on Wikipedia, if you have not encountered it yet. Its mean is $$mu = int_0^1 x,fracGamma(alpha + beta)Gamma(alpha)+Gamma(beta),x^alpha-1(1-x)^beta-1,dx = fracalphaalpha+beta = 0.3.$$
v = rbeta(10000, 1.2, 2.8); mean(v); 2*sd(v)/sqrt(10000)
[1] 0.3017991
[1] 0.004096549
Notes: (a) The Law of Large Numbers (LLN) guarantees that the mean of a sample
converges 'in probability' to its population mean as $N rightarrow infty.$
By definition, that is $lim_N rightarrow inftyP(|bar X - mu| < epsilon) = 0,$ for any $epsilon > 0.$
(b) The margins of simulation error depend on the dispersion of the population and the sample size. For most non-normal distributions the same
formula I have used in the examples above works because of the Central Limit Theorem (CLT).
(c) As in @Geronimo's Comment (+1), this is one form of 'Monte Carlo Integration.' It is a method often used in applications, but you don't literally "convert the integral into a sum." As remarked above you use the LLN and the CLT to get an approximate answer and to have some confidence how accurate that answer is. (Of course, it's only a 95% confidence interval, so there are some occasions on which the estimate is farther from the target than hoped for.)
answered Jul 20 at 20:30
BruceET
33.2k61440
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
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I think you are asking if it is possible to estimate the mean of a continuous distribution by taking the sample mean of a large sample (of size $N$) from that distribution. If so, the answer is Yes, but be aware that the result is an estimate and that the error of the estimate depends on the sample size $N$ (typically, smaller error for larger $N$). Here are some examples in which
we sample at random from a few known distributions. (Sampling done using
R statistical software.)
1) $N=50$ observations from the distributon $mathsfNorm(mu = 75,, sigma=3).$ The R statement rnorm samples at random from a specified normal population. You will get a somewhat different estimate each time you take
a sample and find its mean. In 95% of runs the answer will be within
x = rnorm(50, 75, 3); mean(x); 2*sd(x)/sqrt(50)
## 74.84001 # aprx E(X) = 75
## 0.7656806 # 95% margin of simulation error
2) $N = 5000$ observations from $mathsfNorm(mu = 75,, sigma=3).$
x = rnorm(5000, 75, 3); mean(x); 2*sd(x)/sqrt(5000)
## 75.00454
## 0.08390667
3) $N = 1000$ observations from $mathsfUnif(a = 50, b = 100).$ The mean of this population distribution is $mu = 75.$
y = runif(1000, 50, 100); mean(y); 2*sd(y)/sqrt(1000)
## 75.54673
## 0.900703
4) $N = 10^6$ observations from an exponential distribution with rate 1/5.
This distribution has mean $mu = 5.$
w = rexp(10^6, 1/5); mean(w); 2*sd(w)/sqrt(10^6)
[1] 4.997817
[1] 0.00998948
5) $n = 10,000$ observations from the beta distribution with shape parameters $alpha = 1.2$ and $beta = 2.8.$ You can look in your text or read about the beta distribution on Wikipedia, if you have not encountered it yet. Its mean is $$mu = int_0^1 x,fracGamma(alpha + beta)Gamma(alpha)+Gamma(beta),x^alpha-1(1-x)^beta-1,dx = fracalphaalpha+beta = 0.3.$$
v = rbeta(10000, 1.2, 2.8); mean(v); 2*sd(v)/sqrt(10000)
[1] 0.3017991
[1] 0.004096549
Notes: (a) The Law of Large Numbers (LLN) guarantees that the mean of a sample
converges 'in probability' to its population mean as $N rightarrow infty.$
By definition, that is $lim_N rightarrow inftyP(|bar X - mu| < epsilon) = 0,$ for any $epsilon > 0.$
(b) The margins of simulation error depend on the dispersion of the population and the sample size. For most non-normal distributions the same
formula I have used in the examples above works because of the Central Limit Theorem (CLT).
(c) As in @Geronimo's Comment (+1), this is one form of 'Monte Carlo Integration.' It is a method often used in applications, but you don't literally "convert the integral into a sum." As remarked above you use the LLN and the CLT to get an approximate answer and to have some confidence how accurate that answer is. (Of course, it's only a 95% confidence interval, so there are some occasions on which the estimate is farther from the target than hoped for.)
answered Jul 20 at 20:30
BruceET
33.2k61440
add a comment |Â
up vote
0
down vote
I think you are asking if it is possible to estimate the mean of a continuous distribution by taking the sample mean of a large sample (of size $N$) from that distribution. If so, the answer is Yes, but be aware that the result is an estimate and that the error of the estimate depends on the sample size $N$ (typically, smaller error for larger $N$). Here are some examples in which
we sample at random from a few known distributions. (Sampling done using
R statistical software.)
1) $N=50$ observations from the distributon $mathsfNorm(mu = 75,, sigma=3).$ The R statement rnorm samples at random from a specified normal population. You will get a somewhat different estimate each time you take
a sample and find its mean. In 95% of runs the answer will be within
x = rnorm(50, 75, 3); mean(x); 2*sd(x)/sqrt(50)
## 74.84001 # aprx E(X) = 75
## 0.7656806 # 95% margin of simulation error
2) $N = 5000$ observations from $mathsfNorm(mu = 75,, sigma=3).$
x = rnorm(5000, 75, 3); mean(x); 2*sd(x)/sqrt(5000)
## 75.00454
## 0.08390667
3) $N = 1000$ observations from $mathsfUnif(a = 50, b = 100).$ The mean of this population distribution is $mu = 75.$
y = runif(1000, 50, 100); mean(y); 2*sd(y)/sqrt(1000)
## 75.54673
## 0.900703
4) $N = 10^6$ observations from an exponential distribution with rate 1/5.
This distribution has mean $mu = 5.$
w = rexp(10^6, 1/5); mean(w); 2*sd(w)/sqrt(10^6)
[1] 4.997817
[1] 0.00998948
5) $n = 10,000$ observations from the beta distribution with shape parameters $alpha = 1.2$ and $beta = 2.8.$ You can look in your text or read about the beta distribution on Wikipedia, if you have not encountered it yet. Its mean is $$mu = int_0^1 x,fracGamma(alpha + beta)Gamma(alpha)+Gamma(beta),x^alpha-1(1-x)^beta-1,dx = fracalphaalpha+beta = 0.3.$$
v = rbeta(10000, 1.2, 2.8); mean(v); 2*sd(v)/sqrt(10000)
[1] 0.3017991
[1] 0.004096549
Notes: (a) The Law of Large Numbers (LLN) guarantees that the mean of a sample
converges 'in probability' to its population mean as $N rightarrow infty.$
By definition, that is $lim_N rightarrow inftyP(|bar X - mu| < epsilon) = 0,$ for any $epsilon > 0.$
(b) The margins of simulation error depend on the dispersion of the population and the sample size. For most non-normal distributions the same
formula I have used in the examples above works because of the Central Limit Theorem (CLT).
(c) As in @Geronimo's Comment (+1), this is one form of 'Monte Carlo Integration.' It is a method often used in applications, but you don't literally "convert the integral into a sum." As remarked above you use the LLN and the CLT to get an approximate answer and to have some confidence how accurate that answer is. (Of course, it's only a 95% confidence interval, so there are some occasions on which the estimate is farther from the target than hoped for.)
answered Jul 20 at 20:30
BruceET
33.2k61440
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think you are asking if it is possible to estimate the mean of a continuous distribution by taking the sample mean of a large sample (of size $N$) from that distribution. If so, the answer is Yes, but be aware that the result is an estimate and that the error of the estimate depends on the sample size $N$ (typically, smaller error for larger $N$). Here are some examples in which
we sample at random from a few known distributions. (Sampling done using
R statistical software.)
1) $N=50$ observations from the distributon $mathsfNorm(mu = 75,, sigma=3).$ The R statement rnorm samples at random from a specified normal population. You will get a somewhat different estimate each time you take
a sample and find its mean. In 95% of runs the answer will be within
x = rnorm(50, 75, 3); mean(x); 2*sd(x)/sqrt(50)
## 74.84001 # aprx E(X) = 75
## 0.7656806 # 95% margin of simulation error
2) $N = 5000$ observations from $mathsfNorm(mu = 75,, sigma=3).$
x = rnorm(5000, 75, 3); mean(x); 2*sd(x)/sqrt(5000)
## 75.00454
## 0.08390667
3) $N = 1000$ observations from $mathsfUnif(a = 50, b = 100).$ The mean of this population distribution is $mu = 75.$
y = runif(1000, 50, 100); mean(y); 2*sd(y)/sqrt(1000)
## 75.54673
## 0.900703
4) $N = 10^6$ observations from an exponential distribution with rate 1/5.
This distribution has mean $mu = 5.$
w = rexp(10^6, 1/5); mean(w); 2*sd(w)/sqrt(10^6)
[1] 4.997817
[1] 0.00998948
5) $n = 10,000$ observations from the beta distribution with shape parameters $alpha = 1.2$ and $beta = 2.8.$ You can look in your text or read about the beta distribution on Wikipedia, if you have not encountered it yet. Its mean is $$mu = int_0^1 x,fracGamma(alpha + beta)Gamma(alpha)+Gamma(beta),x^alpha-1(1-x)^beta-1,dx = fracalphaalpha+beta = 0.3.$$
v = rbeta(10000, 1.2, 2.8); mean(v); 2*sd(v)/sqrt(10000)
[1] 0.3017991
[1] 0.004096549
Notes: (a) The Law of Large Numbers (LLN) guarantees that the mean of a sample
converges 'in probability' to its population mean as $N rightarrow infty.$
By definition, that is $lim_N rightarrow inftyP(|bar X - mu| < epsilon) = 0,$ for any $epsilon > 0.$
(b) The margins of simulation error depend on the dispersion of the population and the sample size. For most non-normal distributions the same
formula I have used in the examples above works because of the Central Limit Theorem (CLT).
(c) As in @Geronimo's Comment (+1), this is one form of 'Monte Carlo Integration.' It is a method often used in applications, but you don't literally "convert the integral into a sum." As remarked above you use the LLN and the CLT to get an approximate answer and to have some confidence how accurate that answer is. (Of course, it's only a 95% confidence interval, so there are some occasions on which the estimate is farther from the target than hoped for.)
answered Jul 20 at 20:30
BruceET
33.2k61440
I think you are asking if it is possible to estimate the mean of a continuous distribution by taking the sample mean of a large sample (of size $N$) from that distribution. If so, the answer is Yes, but be aware that the result is an estimate and that the error of the estimate depends on the sample size $N$ (typically, smaller error for larger $N$). Here are some examples in which
we sample at random from a few known distributions. (Sampling done using
R statistical software.)
1) $N=50$ observations from the distributon $mathsfNorm(mu = 75,, sigma=3).$ The R statement rnorm samples at random from a specified normal population. You will get a somewhat different estimate each time you take
a sample and find its mean. In 95% of runs the answer will be within
x = rnorm(50, 75, 3); mean(x); 2*sd(x)/sqrt(50)
## 74.84001 # aprx E(X) = 75
## 0.7656806 # 95% margin of simulation error
2) $N = 5000$ observations from $mathsfNorm(mu = 75,, sigma=3).$
x = rnorm(5000, 75, 3); mean(x); 2*sd(x)/sqrt(5000)
## 75.00454
## 0.08390667
3) $N = 1000$ observations from $mathsfUnif(a = 50, b = 100).$ The mean of this population distribution is $mu = 75.$
y = runif(1000, 50, 100); mean(y); 2*sd(y)/sqrt(1000)
## 75.54673
## 0.900703
4) $N = 10^6$ observations from an exponential distribution with rate 1/5.
This distribution has mean $mu = 5.$
w = rexp(10^6, 1/5); mean(w); 2*sd(w)/sqrt(10^6)
[1] 4.997817
[1] 0.00998948
5) $n = 10,000$ observations from the beta distribution with shape parameters $alpha = 1.2$ and $beta = 2.8.$ You can look in your text or read about the beta distribution on Wikipedia, if you have not encountered it yet. Its mean is $$mu = int_0^1 x,fracGamma(alpha + beta)Gamma(alpha)+Gamma(beta),x^alpha-1(1-x)^beta-1,dx = fracalphaalpha+beta = 0.3.$$
v = rbeta(10000, 1.2, 2.8); mean(v); 2*sd(v)/sqrt(10000)
[1] 0.3017991
[1] 0.004096549
Notes: (a) The Law of Large Numbers (LLN) guarantees that the mean of a sample
converges 'in probability' to its population mean as $N rightarrow infty.$
By definition, that is $lim_N rightarrow inftyP(|bar X - mu| < epsilon) = 0,$ for any $epsilon > 0.$
(b) The margins of simulation error depend on the dispersion of the population and the sample size. For most non-normal distributions the same
formula I have used in the examples above works because of the Central Limit Theorem (CLT).
(c) As in @Geronimo's Comment (+1), this is one form of 'Monte Carlo Integration.' It is a method often used in applications, but you don't literally "convert the integral into a sum." As remarked above you use the LLN and the CLT to get an approximate answer and to have some confidence how accurate that answer is. (Of course, it's only a 95% confidence interval, so there are some occasions on which the estimate is farther from the target than hoped for.)
answered Jul 20 at 20:30
BruceET
33.2k61440
edited Jul 20 at 20:58
answered Jul 20 at 20:30
BruceET
33.2k61440
answered Jul 20 at 20:30
BruceET
33.2k61440
answered Jul 20 at 20:30
BruceET
33.2k61440
33.2k61440
add a comment |Â
add a comment |Â
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Can you explain what is the random variable (or random variables)? What is $x$? What is $X_i$? What is $N$? If $X$ is a random variable with PDF $f_X(x)$ then indeed $E[X] = int_-infty^infty xf_X(x)dx$.
– Michael
Jul 19 at 17:49
@Michael I have a series of different values that I want to calculate the expected value for each of them. An example of the series is 5.6, 3.2, 1.9, 6.1, 7.2. $N$ in this case is 5, $X_i$ is each value in the series.
– Adam Amin
Jul 19 at 17:56
@AdamAmin And where is the continuous distribution?
– callculus
Jul 19 at 18:03
@callculus I am not sure. It is confusing me how to calculate the expected value for each value in the series. I need this because I am calculating the autocorrelation for the series I have and part of the equation is to get the expected value.
– Adam Amin
Jul 19 at 18:09
1
This is called monte carlo integration
– RHowe
Jul 19 at 18:13