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I am trying to solve the following question and I am not sure if I am thinking of the correct solution.
Given X that could take any of the following values with the corresponding probabilities:

\begin{equation} X= \begin{cases} 0, & p =0.3 \\ 1, & p=0.4 \\ 2, & p=0.3 \end{cases} \end{equation}

Let \begin{equation} Y = 5 - X^{2} + \varepsilon \end{equation} Where \begin{equation} X ~and ~ \varepsilon \end{equation} are independent and

\begin{equation} \varepsilon= \begin{cases} -1, & p =0.5 \\ 1, & p=0.5 \end{cases} \end{equation}

find

\begin{equation} g(x) = E[Y|X=x]\\ \end{equation} and the least squares linear predictor of Y.

Since the variables are independent can I say that

\begin{equation} Y= \begin{cases} 4-X^{2}, & p =0.5 \\ 6-X^{2}, & p=0.5 \end{cases} \end{equation}

and would

\begin{equation} h(X) = \mu_Y + b\mu_X \end{equation}

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The third and fourth line in @nafizh's answer are incorrect. In the third line the value of $X = x$ is given and therfore no expectation is needed. In the fourth line the values of $\epsilon$ do not need to be squared. \begin{eqnarray*} g(x) = E[Y|X=x] &=& E[5 - X^2 + \epsilon|X =x]\\ &=& E[5] - E[X^2|X=x] + E[\epsilon|X=x]) \\ &=& 5 - x^2 + E[\epsilon] \\ &=& 5 - x^2 + [-1(0.5) + 1(0.5)] \\ &=& 5 - x^2 \end{eqnarray*}

The same result occurs from start at the last line of your working: \begin{eqnarray*} g(x) = E[Y|X=x] &=& 0.5 \times (4 - x^2) + 0.5 \times (6 - x^2)\\ &=& 5 - x^2 \end{eqnarray*}

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