Using Random Sample to Find Estimate
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0
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I have to use the Inversion Sampling Method to generate a random sample of 100 from the function $f(x)=theta x^{theta - 1}$ if $theta =5$. Here is my function so far:
X = function(n)
{
U = runif(n)
sample = (U/5)^{1/4}
return(sample)
}
So I get $100$ random values when I input $X(100)$. Now I have to use this data to find an estimate for $theta = 5$. Where do I go from here? Do I make a histogram?
statistics sampling sampling-theory
asked Nov 28 at 22:09
numericalorange
1,719311
add a comment |
up vote
0
down vote
favorite
I have to use the Inversion Sampling Method to generate a random sample of 100 from the function $f(x)=theta x^{theta - 1}$ if $theta =5$. Here is my function so far:
X = function(n)
{
U = runif(n)
sample = (U/5)^{1/4}
return(sample)
}
So I get $100$ random values when I input $X(100)$. Now I have to use this data to find an estimate for $theta = 5$. Where do I go from here? Do I make a histogram?
statistics sampling sampling-theory
asked Nov 28 at 22:09
numericalorange
1,719311
If you know $theta$ equals $5$, why estimate it?
– StubbornAtom
Nov 29 at 15:41
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have to use the Inversion Sampling Method to generate a random sample of 100 from the function $f(x)=theta x^{theta - 1}$ if $theta =5$. Here is my function so far:
X = function(n)
{
U = runif(n)
sample = (U/5)^{1/4}
return(sample)
}
So I get $100$ random values when I input $X(100)$. Now I have to use this data to find an estimate for $theta = 5$. Where do I go from here? Do I make a histogram?
statistics sampling sampling-theory
asked Nov 28 at 22:09
numericalorange
1,719311
I have to use the Inversion Sampling Method to generate a random sample of 100 from the function $f(x)=theta x^{theta - 1}$ if $theta =5$. Here is my function so far:
X = function(n)
{
U = runif(n)
sample = (U/5)^{1/4}
return(sample)
}
So I get $100$ random values when I input $X(100)$. Now I have to use this data to find an estimate for $theta = 5$. Where do I go from here? Do I make a histogram?
statistics sampling sampling-theory
statistics sampling sampling-theory
asked Nov 28 at 22:09
numericalorange
1,719311
asked Nov 28 at 22:09
numericalorange
1,719311
asked Nov 28 at 22:09
numericalorange
1,719311
asked Nov 28 at 22:09
numericalorange
1,719311
asked Nov 28 at 22:09
numericalorange
1,719311
1,719311
If you know $theta$ equals $5$, why estimate it?
– StubbornAtom
Nov 29 at 15:41
add a comment |
If you know $theta$ equals $5$, why estimate it?
– StubbornAtom
Nov 29 at 15:41
If you know $theta$ equals $5$, why estimate it?
– StubbornAtom
Nov 29 at 15:41
If you know $theta$ equals $5$, why estimate it?
– StubbornAtom
Nov 29 at 15:41
add a comment |
1 Answer
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votes
up vote
1
down vote
accepted
You can derive the MLE estimator, i.e.,
$$
L(theta; X) = theta ^n ( prod x_i )^{theta - 1 } ,
$$
the log-likelihood is
$$
l(theta) = n ln theta+(theta - 1)sumln x_i,
$$
$$
l'(theta) = frac{n}{theta} + sumln x_i = 0,
$$
hence the MLE is
$$
hat{theta}_{ML}=- frac{n}{sum ln x_i}.
$$
Verifying that it indeed maximizes the likelihood,
$$
l''(hat{theta}) = - frac{ n }{hat{theta}_{ML}} < 0.
$$
answered Nov 30 at 23:27
V. Vancak
10.8k2926
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You can derive the MLE estimator, i.e.,
$$
L(theta; X) = theta ^n ( prod x_i )^{theta - 1 } ,
$$
the log-likelihood is
$$
l(theta) = n ln theta+(theta - 1)sumln x_i,
$$
$$
l'(theta) = frac{n}{theta} + sumln x_i = 0,
$$
hence the MLE is
$$
hat{theta}_{ML}=- frac{n}{sum ln x_i}.
$$
Verifying that it indeed maximizes the likelihood,
$$
l''(hat{theta}) = - frac{ n }{hat{theta}_{ML}} < 0.
$$
answered Nov 30 at 23:27
V. Vancak
10.8k2926
add a comment |
up vote
1
down vote
accepted
You can derive the MLE estimator, i.e.,
$$
L(theta; X) = theta ^n ( prod x_i )^{theta - 1 } ,
$$
the log-likelihood is
$$
l(theta) = n ln theta+(theta - 1)sumln x_i,
$$
$$
l'(theta) = frac{n}{theta} + sumln x_i = 0,
$$
hence the MLE is
$$
hat{theta}_{ML}=- frac{n}{sum ln x_i}.
$$
Verifying that it indeed maximizes the likelihood,
$$
l''(hat{theta}) = - frac{ n }{hat{theta}_{ML}} < 0.
$$
answered Nov 30 at 23:27
V. Vancak
10.8k2926
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can derive the MLE estimator, i.e.,
$$
L(theta; X) = theta ^n ( prod x_i )^{theta - 1 } ,
$$
the log-likelihood is
$$
l(theta) = n ln theta+(theta - 1)sumln x_i,
$$
$$
l'(theta) = frac{n}{theta} + sumln x_i = 0,
$$
hence the MLE is
$$
hat{theta}_{ML}=- frac{n}{sum ln x_i}.
$$
Verifying that it indeed maximizes the likelihood,
$$
l''(hat{theta}) = - frac{ n }{hat{theta}_{ML}} < 0.
$$
answered Nov 30 at 23:27
V. Vancak
10.8k2926
You can derive the MLE estimator, i.e.,
$$
L(theta; X) = theta ^n ( prod x_i )^{theta - 1 } ,
$$
the log-likelihood is
$$
l(theta) = n ln theta+(theta - 1)sumln x_i,
$$
$$
l'(theta) = frac{n}{theta} + sumln x_i = 0,
$$
hence the MLE is
$$
hat{theta}_{ML}=- frac{n}{sum ln x_i}.
$$
Verifying that it indeed maximizes the likelihood,
$$
l''(hat{theta}) = - frac{ n }{hat{theta}_{ML}} < 0.
$$
answered Nov 30 at 23:27
V. Vancak
10.8k2926
answered Nov 30 at 23:27
V. Vancak
10.8k2926
answered Nov 30 at 23:27
V. Vancak
10.8k2926
answered Nov 30 at 23:27
V. Vancak
10.8k2926
10.8k2926
add a comment |
add a comment |
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If you know $theta$ equals $5$, why estimate it?
– StubbornAtom
Nov 29 at 15:41