Let $X$ and $Y$ i.i.d standardized normally distributed random variables.
Calculate the conditional expectation of : $$ \mathbb{E}[(X+Y)^{3} | \mathscr{G}] $$
where $\mathscr{G} = \sigma(X)$ ($\sigma$-field generated from $X$)
Proposal
$$X,Y \sim N(0,1)$$
$$\mathbb{E}[(X+Y)|X] = \int_{-\infty}^{+\infty} (X+Y) f_{X+Y|X}(X+Y|X)dx$$
$X$ and $Y$ are independent continuous random variables with density functions $f_X$ and $f_Y$, respectively. First i find the density function of $X + Y$.
Secondly I use the first calculation in order to find the density of the sum of two independent standard normal random variables.
Conditioning on $X$:
\begin{align*} \mathbb{E}[X+Y|X]= P(X+Y \leq t) &= \int_{-\infty}^{+\infty} P(X+Y \leq t |X=x)f_X(X)dx \\ &= \int_{-\infty}^{+\infty} P(Y \leq t-x |X=x)f_X(X)dx \\ &= \int_{-\infty}^{+\infty} P(Y \leq t-x)f_X(X)dx \\ \end{align*} Differentiating with respect to $t$ gives
$$\int_{-\infty}^{+\infty} f_y( t-x)f_X(X)dx (1)$$
Now For $X$ and $Y$ independent standard normal random variables, by (1), the sum $X + Y$ has density
\begin{align*} f_{X+Y}(t) &= \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2\pi}}e^{-(t-x)^2/2} \cdot \frac{1}{\sqrt{2\pi}}e^{-x^2/2} dx \\ &= \frac{1}{\sqrt{4\pi}}e^{-t^2/4} \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi(1/2)}}e^{-(x-t/2)^2/2(1/2)}dx\\ &= \frac{1}{\sqrt{4\pi}}e^{-t^2/4} \end{align*}
