Question

し.folloir _Atalisti.OM. ancillang: (s) Te X+-X1.5 Caution:Vefurclain acl cass

Answer 1

We know that for given two normal random variables X and Y, independent of each other s.t.

$X \sim N(\mu_X,\sigma^2_X) \ \ \ ; Y \sim N(\mu_Y,\sigma^2_Y) \\ \Rightarrow X \pm Y \sim N(\mu_X \pm \mu_Y , \sigma^2_X + \sigma^2_Y)$

(i)

Since, X4 and X₁ are independent of each other

TN(0, 4)

Clearly, the distribution of T is independent of $\theta$ .

Thus, T is ancillary.

(ii)
$X_1 \sim N(\theta,2) \\ \Rightarrow 2X_1 \sim N(2\theta,2^2*2 = 8)$

Now, since, X₂, X₃ and 2X₁ are independent of each other

TN(0, 12)

Clearly, the distribution of T is independent of $\theta$ .

Thus, T is ancillary.

(iii)

$X_4 \sim N(\theta,2) \\ \Rightarrow 2X_4 \sim N(2\theta,2^2*2 = 8)$

Now, since, X₁, X₂, X₃ and 2X₄ are independent of each other

$T = X_1+ X_2 + X_3 + 2X_4 \sim N(\theta + \theta + \theta + 2\theta, 2+2+2+8) \\ \Rightarrow T \sim N(5\theta, 14)$

Clearly, the distribution of T is not independent of $\theta$ .

Thus, T is not ancillary.

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