In the algebra of random variables, is the following identity correct? $$X + X \neq 2X$$
The underlying source of this questions is a mathematical text that considers a MLE estimator for mean of a Gaussian as an equation of random variables (rather than variates as is typically used in descriptive statistics).
So $$\hat{\mu} = \frac{x_{1} + x_{2} + ... + x_{n}}{n}$$ becomes $$\hat{M} = \frac{X_{1} + X_{2} + ... + X_{n}}{n}$$.
I know that samples are taken from the same distribution so effectively $$X_{1} + X_{2} + ... + X_{n} = X$$, wich leads to $$\hat{M} = \frac{X_{1} + X_{2} + ... + X_{n}}{n} = \frac{nX}{n}=X$$ which seems incorrect.
This made me think, that perhaps my assumption that $X + X = 2X$ is incorrect.
Consider the random variable $X$ with the following pmf: $$\mathbb{P}(X=k)=\begin{cases}1/2&\text{for }k=1\\ 1/2&\text{for }k=0.\end{cases}$$
If $X_{1},X_{2},X$ are i.i.d., then let's compute the pmf of $Y=X_{1}+X_{2}$: $$\mathbb{P}(Y=k)=\begin{cases}1/4&\text{for }k=2\\1/2&\text{for }k=1\\ 1/4&\text{for }k=0.\end{cases}$$
On the other hand, the pmf of $Z=2X$ is given by $$\mathbb{P}(Z=k)=\begin{cases}1/2&\text{for }k=2\\ 1/2&\text{for }k=0.\end{cases}$$
So we can see that these are not the same distribution. The difference is coming about because in the case of $X_{1}+X_{2},$ $X_{1}$ and $X_{2}$ are allowed to vary independently, where in the case of $2X,$ you are essentially forcing $X_{1}=X_{2},$ violating the independence assumption.