Likelihood function for $\theta = 0$ (Calculate $\sum_{i=1}^4x_i ^2$ given $\sum_{i=1}^4 x_i$)

Question

I need to find $L_x(\theta)$ for $X_1, X_2, X_3, X_4$ c.i.i.d random variables such that $X_i | \theta \sim N(\theta,1)$ when:

$\theta = 0$

$\bar{x} = -0.7$ ($\bar{x} = \Large\frac{\sum_{i=1}^n x_i}{n}$)

Since my $L_x(\theta) = \left(\large\frac{1}{\sqrt{2 \pi}}\right)^n exp\left(-\large\frac{1}{2} \large\sum_{i=1}^n (x_i - \theta)^2 \right)$, for this data I will need $\sum_{i=1}^4 x_i^2$ . Is that a eay to obtain it using that I have $\bar{x}$? I've tried to find it without sucess.

Thanks in advance!

Answer 1

No, you cannot. For example, $$ X_1=X_2=X_3=X_4=-0.7 $$ gives $\bar{X}=-0.7$ and $\sum_{i=1}^4X_i^2=4\times0.7^2=1.96$. But, $$ X_1=100, \quad X_2=-101.4, \quad X_3=X_4=-0.7 $$ also gives $\bar{X}=-0.7$, but $\sum_{i=1}^4 X_i^2=20282.4$. There is no way you can distinguish between them. Essentially, what you are doing is that you are trying to solve a system with four unknowns using only two equations, which gives you an infinite number of solutions.