LetX1,...,Xn be a random sample of size n from the geometric distribution for which p is the probability of success.
(a) Use the method of moments to find a point estimator for p.
(b) Use the following data (simulated from geometric distribution) to find
the moment estimator for p:
2 5 7 43 18 19 16 11 22
4 34 19 21 23 6 21 7 12
The pdf of a geometric distribution is f(x)= p(1-p)^x-1, for x,.... Also population mean (mew) = 1/p.]
LetX1,...,Xn be a random sample of size n from the geometric distribution for which p is...
5.2.1. Let XX be a random sample of size n from the geometric distribution for which p is the probability of success. (a) Use the method of moments to find a point estimator for p. (b) Use the following data (simulated from geometric distribution) to find the moment estimator for p: 2 5743 18 19 16 11 22 4 34 19 21 23 6 21 7 12
5. Let X1, X2,. , Xn be a random sample from a distribution with pdf of f(x) (0+1)x,0< x<1 a. What is the moment estimator for 0 using the method of moments technique? b. What is the MLE for 0?
5. Let X1, X2, ..., Xn be a random sample from a distribution with pdf of f(x) = (@+1)xº,0<x<1. a. What is the moment estimator for 0 using the method of moments technique? b. What is the MLE for @ ?
MoM stands for Method of Moments.
4. (a) If X Geometric(p), prove that the moment-generating function for X is Mx(t) pe 1-(1-p)e' (b) Use your result of part (a) to show that E(X) = p and V(X) = Now, we have X1, X2,... X, d Geometricíp). (c) Find a MoM estimator for p based on the first moment. (d) Explain why your estimator makes sense intuitively. (e) Use the following data to give a point estimate of p: XnGeometric(p 3,...
1. Let X1, X2,... .Xn be a random sample of size n from a Bernoulli distribution for which p is the probability of success. We know the maximum likelihood estimator for p is p = 1 Σ_i Xi. ·Show that p is an unbiased estimator of p.
Let X1, X2, ..., Xn be a random sample of size n from the distribution with probability density function f(x1) = 2 Æ e-dz?, x > 0, 1 > 0. a. Obtain the maximum likelihood estimator of 1 . Enter a formula below. Use * for multiplication, / for divison, ^ for power. Use m1 for the sample mean X, m2 for the second moment and pi for the constant n. That is, m1 = * = *Šxi, m2 =...
Let X1, X2, ..., Xn be a random sample of size n from the
distribution with probability density function
To answer this question, enter you answer as a formula. In
addition to the usual guidelines, two more instructions for this
problem only : write
as single variable p and
as m. and these can be used as inputs of functions as usual
variables e.g log(p), m^2, exp(m) etc. Remember p represents the
product of
s only, but will not work...
Number 2 only PLEASE
1. [40] 6.4-5. Let Xi, X2..,Xn be a random sample from dis- tributions with the given probability density functions. In each case, find the maximum likelihood estimator . 6.4-10. Let X1, X2,... ,Xn be a random sample of size n from a geometric distribution for which p is the probabil- ity of success. (a) Use the method of moments to find a point estimate 2. [20] for p. 100] 6.5-3. The midterm and final exam scores...
Let X1, X2, .. , Xn be a random sample of size n from a geometric distribution with pmf =0.75 . 0.25z-1, f(x) X-1.2.3. ) Let Zn 3 n n-2ућ. Find Mz, (t), the mgf of Žn. Then find the limiting mgf limn→oo MZm (t). What is the limiting distribution of Z,'?
Let X1, X2, .. , Xn be a random sample of size n from a geometric distribution with pmf =0.75 . 0.25z-1, f(x) X-1.2.3.
) Let Zn 3...
Let X1 Xn be a random sample from a distribution with the pdf f(x(9) = θ(1 +0)-r(0-1) (1-2), 0 < x < 1, θ > 0. the estimator T-4 is a method of moments estimator for θ. It can be shown that the asymptotic distribution of T is Normal with ETT θ and Var(T) 0042)2 Apply the integral transform method (provide an equation that should be solved to obtain random observations from the distribution) to generate a sam ple of...