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While $$ \mathrm{E}(\tilde{\mathrm{y}})=\alpha+\beta \tilde{\mathrm{x}} $$
Subject is Regression Analysis and this formula is from the "Features of Estimation ". and y is a neutral variable.
Formula for the variance of "y".
Where this plus 1 ( the one that has red underline) came from.
In linear regression we have $$y = \alpha + \beta x + \varepsilon$$ where $\varepsilon \sim N(0, \sigma^2)$ are iid (independent and identically distributed) error terms.
For a sufficiently large sample size, the maximum likelihood estimators for $\alpha$ and $\beta$ are jointly Normal: $$\begin{pmatrix} \hat{\alpha} \\ \hat{\beta} \end{pmatrix} \sim N \left\{ \begin{pmatrix} \alpha \\ \beta \end{pmatrix}, \begin{pmatrix} \sigma^2(1/n + \bar{x}^2/S_{xx}) &-\sigma^2 \bar{x}/S_{xx} \\-\sigma^2 \bar{x}/S_{xx} & \sigma^2/S_{xx}\end{pmatrix}\right\}$$ Where $S_{xx} - \sum_{i=1}^n(\bar{x} - x_i)^2$. Now if I have a new value of $x$ denotes $\tilde{x}$ we have $\tilde{y} = \alpha + \beta\tilde{x} + \varepsilon$. The obvious (and probably best) predictor of $\tilde{y}$ is $\hat{\tilde{y}} = \hat{\alpha} + \hat{\beta}\tilde{x}$. Now suppose we want an idea of the variability of $\tilde{y}$. To do this we need to consider the uncertainty in our estimated of $\alpha$ and $\beta$. So we have, using standard properties of $Var()$, $Cov()$ and the assumption that $(\hat{\alpha}, \hat{\beta})$ is independent of $\varepsilon$:
$$\begin{align} Var(\tilde{y}) &= Var(\hat{\alpha} + \hat{\beta}\tilde{x} + \varepsilon)\\ & = Var(\hat{\alpha} + \hat{\beta}\tilde{x}) + Var(\varepsilon) \\ & = Var(\hat{\alpha} + \hat{\alpha}\tilde{x}) + \sigma^2 \\ &=Var(\hat{\alpha}) + \tilde{x}^2Var(\hat{\beta}) + 2\tilde{x}Cov(\hat{\alpha}, \hat{\beta}) + \sigma^2\end{align}$$
Then extracting the $Var$ and $Cov$ terms from the variance matrix of the Normal distribution above we have
\begin{align}Var(\tilde{y}) &= \sigma^2(1/n + \bar{x}^2/S_{xx}) + \tilde{x}^2\sigma^2/S_{xx} - 2\tilde{x}\bar{x}\sigma^2/S_{xx} + \sigma^2\\ &=\sigma^2 \left\{1/n + \bar{x}^2/S_{xx} + \tilde{x}^2/S_{xx} - 2\tilde{x}\bar{x}/S_{xx} + 1 \right\} \text{*notice the $+1$ from factoring out $\sigma^2$}\\ &=\sigma^2 \left\{\frac{1}{n} + \frac{(\bar{x}-\tilde{x})^2}{S_{xx}} + 1 \right\}\end{align} Now simply rearranging the additive terms leads to $$Var(\tilde{y}) = \sigma^2 \left\{1 + \frac{1}{n} + \frac{(\bar{x}-\tilde{x})^2}{S_{xx}} \right\}$$
Now my answer has slightly different notation to yours because you haven't define $KT_x$, however, this should show you where $+1$ come from.
