Question

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. Determine the bias and the mean squared error for this estimator of the mean. Bias: Mean Square Error:

Answer 1

Bias for the estimator = -0.56

Mean Square Error for the estimator = 6.6311

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

`\mu = (3X_(1) + 4X_(2) )/(8)`

Now

Bias for the estimator = E(μ bar) - μ

=
`E( (3X_(1) + 4X_(2) )/(8)) - 4.5`

=
`(3E(X_(1)) + 4E(X_(2)))/(8) - 4.5`

=
`(3(4.5) + 4(4.5))/(8) - 4.5`

=
`(13.5 + 18)/(8) - 4.5`

=
`(31.5)/(8) - 4.5`

= 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

=
`Var( (3X_(1) + 4X_(2) )/(8)) + 0.3136`

=
`(1)/(64) Var( {3X_(1) + 4X_(2) }) + 0.3136`

=
`(1)/(64) ( [{3Var(X_(1)) + 4Var(X_(2))] }) + 0.3136`

=
`(1)/(64) [{3(57.76) + 4(57.76)}] } + 0.3136`

=
`(1)/(64) [7(57.76)}] } + 0.3136`

=
`(1)/(64) [404.32] } + 0.3136`

=
`6.3175 + 0.3136`

= 6.6311

∴ we get

Mean Square Error for the estimator = 6.6311