Question:
Given $X = x$, $Y$ is defined the following way:
$$ Y = x + 2W $$
where $ W \sim N(0,1) $. What is the conditional mean and variance of $Y$ given $X = x$? Write down the conditional density of $Y$ given $X = x$.
My work:
- Finding the conditional mean :
\begin{align}{ E[ Y | X = x] = E [x + 2W | X = x] \\ E[Y|X=x] =E[x | X = x] + 2E[W | X = x] \\ E[Y| X = x] = x + 2*0 = x } \end{align}
- Finding the conditional variance: \begin{align}{ \text{Var}(Y | X = x) = \text{Var}(x + 2W | X = x) \\\text{Var}(Y |X =x) = \text{Var}(x | X=x) + 4\text{Var}(W|X = x) \\ \text{Var}(Y|X=x) = 0 + 4*1 = 4 } \end{align}
- Finding the pdf. $Y$ is defined in terms of $W$, which is approx. norm. We calculated the conditional mean and conditional variance to be $x$ and $4$, respectively. Therefore, the pdf for this variable will be approx. norm with $\mu = x$ and $\sigma^2 = 4$.
\begin{align}{ f_{Y|X=x}(Y|X=x) = f_{x + 2W|X=x}(x + 2W|X=x) \\ f_{Y|X=x}(Y|X=x) = \frac{1}{\sqrt{2\pi\sigma^2}}*e^{\frac{-(y-\mu)^2}{2*\sigma^2}} \\ f_{Y|X=x}(Y|X=x) = \frac{1}{\sqrt{8\pi}}*e^{\frac{-(y-x)^2}{8}} } \end{align}
Does this seem right? Thanks in advance!