Find Maximum Likelihood Estimator of Gamma Distribution
Given:
f(x; β) = [ 1/( β^2) ] * x * e^(-x/ β) for 0 < x < infinity
EX = 2β and VarX = 2(β^2)
Questions:
1/First, I believe this is a gamma distribution with alpha = 2. Is that right?
2/Find Maximum likelihood estimator of β
-First, I get the likelihood fn L(β) = product of all the f(xi; β)
-After doing all arithmetic, I get L(β) = β^(-2n) * (x1 * x2 * … *xn) * e^[(-1/ β) * (x1 + x2 +…+xn)]
-Then I take the log function of L(β), I call it l(β), find derivative of this l(β), equate the derivative to 0 and solve.
I get the MLE is (1/n * (x1 + x2+ … + xn)) / 2 = (sample mean )/2.
** Am I correct till this point?
3/If the given said EX = 2 β, can I assume that X must be 2 β, and thus for the population, β is X / 2 ??
4/I think this MLE is not bias, but I get confused when I try to find the Variance of the MLE. If I let this MLE of β being β’, then for variance, I believe that I should find E(β’ ^ 2) – [ E(β’) ] ^2. But from here, my question is that should I use the fact that β’ is (sample mean) / 2, then just plug in and solve ? Or should I do integration?
Because I think since this distribution is continuous, isn’t it that E(anything) is integration of that “anything” with f(x)dx ??
Please help me. Thanks in advance.
1/ Yes, you are correct. The given probability density function (pdf) represents a gamma distribution with α = 2.
2/ To find the maximum likelihood estimator (MLE) of β, you correctly start by forming the likelihood function, L(β), which is the product of the pdf for each data point. In this case, it will be L(β) = β^(-2n) * (x1 * x2 * … * xn) * e^[(-1/β) * (x1 + x2 + … + xn)], where n is the number of data points.
Next, you take the logarithm of the likelihood function, l(β), to simplify calculations. You then differentiate l(β) with respect to β, set it equal to zero, and solve for β to find the MLE.
You proceed correctly up to this point, and the MLE you obtained, (sample mean)/2, is correct as well.
3/ Given that E(X) = 2β, you cannot assume that X must be 2β. Instead, it means that the expected value of the random variable X, denoted by E(X), is equal to 2β. This implies that 2β is the expected value of the population, not the value of X. Therefore, you cannot conclude that β is X/2.
4/ To determine whether the MLE is biased, you need to calculate the bias of the estimator, which is the difference between the expected value of the estimator and the true value of the parameter being estimated. In this case, you need to find E(β') - β^2, where β' is the MLE of β. To find the expectation, you can use the fact that E(aX) = aE(X), where a is a constant and X is a random variable.
In this case, β' = (sample mean)/2, so you can rewrite it as β' = (1/n * (x1 + x2 + … + xn))/2. Now you can find E(β') by substituting the expression for β'. Since the sample mean is an unbiased estimator of the population mean, E(β') will be equal to β. Therefore, the MLE of β is unbiased.
Regarding the variance, you need to find Var(β'). To do this, you can use the following formula: Var(aX) = a^2Var(X), where a is a constant and X is a random variable. In this case, β' = (sample mean)/2, so you can rewrite it as β' = (1/n * (x1 + x2 + … + xn))/2. Now, using the formula for variance, you can calculate Var(β') = (1/4n^2)Var(sample mean). The variance of the sample mean is Var(sample mean) = σ^2/n, where σ^2 is the population variance. Since VarX = 2(β^2), you can substitute it in the expression to get Var(β') = (1/4n^2)(2(β^2)/n). Simplifying this expression will give you the variance of the MLE of β.
Hope this helps! Let me know if you have any further questions.
1/ Yes, you are correct. The given density function corresponds to the gamma distribution with shape parameter α = 2.
2/ To find the maximum likelihood estimator (MLE) of β:
First, let's write the likelihood function based on the given data: L(β) = ∏[1/( β^2)] * x * e^(-x/ β)
Taking the logarithm of the likelihood function, we have: l(β) = -2n * ln(β) + (x1 + x2 + ... + xn)/β
To find the MLE, we need to differentiate l(β) with respect to β and set it equal to 0:
dl(β)/dβ = -2n/β + (x1 + x2 + ... + xn)/β^2 = 0
Multiply through by β^2 to simplify and solve:
-2nβ + (x1 + x2 + ... + xn) = 0
Multiplying by -1 and dividing by 2n, we get:
β = (x1 + x2 + ... + xn)/(2n)
So, the MLE of β is (sample mean) / 2.
You are correct up to this point.
3/ No, you cannot assume that X must be 2 β. The given EX = 2β states that the expected value of X is equal to 2β, not that X itself is equal to 2β. So, β should not be assumed as X/2 based on this statement.
4/ To determine the bias and variance of the MLE, β', let's calculate them separately.
Bias of the MLE:
Bias(β') = E(β') - β
Since β' = (sample mean)/2, we can substitute it into the equation:
Bias(β') = E((sample mean)/2) - β
= (1/2) * E(sample mean) - β
= β/2 - β
= -β/2
Therefore, the MLE is biased since the bias is not zero.
Variance of the MLE:
To find the variance of the MLE, Var(β'), we can use the property of linearity of expectation: Var(β') = Var((sample mean)/2)
Since Var(cX) = c^2 * Var(X), we have:
Var(β') = (1/4) * Var(sample mean)
For a gamma distribution, Var(X) = 2(β^2), as given.
Therefore, Var(β') = (1/4) * (2(β^2)/n)
= (β^2)/(2n)
So, the variance of the MLE, Var(β'), is (β^2)/(2n).