Respuesta :
Answer:
Bias for the estimator = -0.56
Mean Square Error for the estimator = 6.6311
Step-by-step explanation:
Given - A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and X2. The mean is estimated using the formula (3X1 + 4X2)/8.
To find - Determine the bias and the mean squared error for this estimator of the mean.
Proof -
Let us denote
X be a random variable such that X ~ N(mean = 4.5, SD = 7.6)
Now,
An estimate of mean, μ is suggested as
[tex]\mu = \frac{3X_{1} + 4X_{2} }{8}[/tex]
Now
Bias for the estimator = E(μ bar) - μ
= [tex]E( \frac{3X_{1} + 4X_{2} }{8}) - 4.5[/tex]
= [tex]\frac{3E(X_{1}) + 4E(X_{2})}{8} - 4.5[/tex]
= [tex]\frac{3(4.5) + 4(4.5)}{8} - 4.5[/tex]
= [tex]\frac{13.5 + 18}{8} - 4.5[/tex]
= [tex]\frac{31.5}{8} - 4.5[/tex]
= 3.9375 - 4.5
= - 0.5625 ≈ -0.56
∴ we get
Bias for the estimator = -0.56
Now,
Mean Square Error for the estimator = E[(μ bar - μ)²]
= Var(μ bar) + [Bias(μ bar, μ)]²
= [tex]Var( \frac{3X_{1} + 4X_{2} }{8}) + 0.3136[/tex]
= [tex]\frac{1}{64} Var( {3X_{1} + 4X_{2} }) + 0.3136[/tex]
= [tex]\frac{1}{64} ( [{3Var(X_{1}) + 4Var(X_{2})] }) + 0.3136[/tex]
= [tex]\frac{1}{64} [{3(57.76) + 4(57.76)}] } + 0.3136[/tex]
= [tex]\frac{1}{64} [7(57.76)}] } + 0.3136[/tex]
= [tex]\frac{1}{64} [404.32] } + 0.3136[/tex]
= [tex]6.3175 + 0.3136[/tex]
= 6.6311
∴ we get
Mean Square Error for the estimator = 6.6311
