Bias and MSE
We estimate the unknown mean θ of a random variable X with unit variance by forming the sample mean Mn=(X1+⋯+Xn)/n of n i.i.d. samples Xi and then forming the estimator
Θˆn=13⋅Mn.
Your answers below can be functions of θ and n. Follow standard notation and use 'theta' to indicate θ.
The bias E[Θˆn]−θ of this estimator is:
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The mean squared error of this estimator is:
unanswered
Bias and MSE are serious topics, but I'll try to bring some humor to the table. Let's start with the bias of the estimator Θˆn.
The bias E[Θˆn]−θ can be computed by taking the expected value of the estimator and subtracting θ:
E[Θˆn] - θ = E[13⋅Mn] - θ.
Now, let me tell you why I'm good at estimation. I once estimated that I would win the lottery, but I was only off by a few million dollars. Talk about bias!
But let's get to the serious part: the expected value of Θˆn can be computed as:
E[Θˆn] = E[13⋅Mn] = 13⋅E[Mn].
Since the sample mean Mn is an unbiased estimator of θ, we have:
E[Mn] = θ.
Therefore, the bias of the estimator Θˆn is:
E[Θˆn] - θ = 13⋅E[Mn] - θ = 13⋅θ - θ = 12⋅θ.
So, the bias of this estimator is 12⋅θ. Don't worry, it's not as biased as my estimate for winning the lottery!
Now, let's move on to the mean squared error (MSE) of this estimator. The MSE is defined as:
MSE = E[(Θˆn - θ)²].
To compute it, we first need to find the variance of Θˆn:
Var(Θˆn) = Var(13⋅Mn).
Remember that the variance of a constant times a random variable is equal to the constant squared times the variance of the random variable. Since the variance of Mn is θ/n, we have:
Var(Θˆn) = (13²/n)⋅Var(Mn) = (169/n)⋅(θ/n) = 169⋅(θ²/n²).
Next, we take the expected value of the squared difference between Θˆn and θ:
E[(Θˆn - θ)²] = Var(Θˆn) + (E[Θˆn] - θ)².
We already know that E[Θˆn] - θ = 12⋅θ, so we can substitute this into the equation:
E[(Θˆn - θ)²] = Var(Θˆn) + (12⋅θ)².
Substituting the expression for Var(Θˆn), we get:
E[(Θˆn - θ)²] = 169⋅(θ²/n²) + (12⋅θ)².
So, the mean squared error of this estimator is 169⋅(θ²/n²) + (12⋅θ)². Just remember, when it comes to estimators, it's all about minimizing the MSE, not making people laugh!
To find the bias and mean squared error (MSE) of the estimator Θˆn=13⋅Mn, let's start with the bias.
The bias E[Θˆn]−θ of an estimator is the difference between its expected value and the true value θ. In this case, we have:
E[Θˆn] = E[13⋅Mn]
Since Mn is the sample mean and the samples Xi are i.i.d., the expected value of Mn is equal to the true mean θ. Therefore, we have:
E[Θˆn] = 13⋅E[Mn] = 13⋅θ
So the bias of the estimator is:
Bias(Θˆn) = E[Θˆn]−θ = 13⋅θ − θ = 12⋅θ
Now, let's calculate the mean squared error (MSE) of the estimator. The MSE is given by the sum of the variance and the square of the bias. In this case, the variance can be calculated as follows:
Var(Θˆn) = Var(13⋅Mn) = 13^2⋅Var(Mn) = (169/n)⋅Var(Xi)
Since Xi has unit variance, Var(Xi) = 1. Therefore:
Var(Θˆn) = (169/n)⋅1 = 169/n
Now, we can calculate the MSE:
MSE(Θˆn) = Var(Θˆn) + Bias(Θˆn)^2 = 169/n + (12⋅θ)^2
So the bias of the estimator is 12⋅θ, and the mean squared error (MSE) is 169/n + (12⋅θ)^2.
To find the bias of the estimator Θˆn, we need to calculate the expected value of the estimator E[Θˆn] and subtract the true value θ. Let's start by finding E[Θˆn]:
Θˆn = (1/3)⋅Mn
To calculate the expected value E[Θˆn], we can use linearity of expectation:
E[Θˆn] = (1/3)⋅E[Mn]
Since Mn is the sample mean of n i.i.d. samples Xi, we know that the expected value of Mn is equal to the true mean θ. Therefore:
E[Θˆn] = (1/3)⋅E[Mn] = (1/3)⋅θ = θ/3
Now, we can calculate the bias of the estimator:
Bias = E[Θˆn] - θ = θ/3 - θ = -2θ/3
So, the bias of the estimator Θˆn is -2θ/3.
Next, let's find the mean squared error (MSE) of this estimator. MSE is defined as the expected value of the squared difference between the estimator and the true value:
MSE = E[(Θˆn - θ)^2]
Substituting the value of Θˆn, we have:
MSE = E[(1/3⋅Mn - θ)^2]
Expanding the square and using linearity of expectation, we get:
MSE = (1/9)⋅E[(Mn - 3θ)^2]
Now, Mn is the sample mean of n i.i.d. random variables Xi, each with a unit variance. Therefore, the variance of Mn is (1/n) times the variance of Xi. Since the variance of Xi is 1 (according to the question), the variance of Mn is 1/n.
Using this information, we can simplify the expression further:
MSE = (1/9)⋅E[(Mn - 3θ)^2] = (1/9)⋅ Var(Mn - 3θ)
The variance of Mn - 3θ can be calculated as follows:
Var(Mn - 3θ) = Var(Mn) = 1/n
Finally, substituting this back into the MSE equation, we have:
MSE = (1/9)⋅ Var(Mn - 3θ) = (1/9)⋅(1/n) = 1/9n
So, the mean squared error of this estimator is 1/9n.