# conditional expectation of iid $X,Y$ cubic sum

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*}

• $$(Y+X)|X\sim\mathcal N(X,1)$$ Feb 5, 2021 at 14:20
• how?can you explain ?@Xi'an
– user310375
Feb 5, 2021 at 15:12
• Drawing a picture helps. If you would like the details, see stats.stackexchange.com/a/71303/919.
– whuber
Feb 5, 2021 at 15:33
• @whuber I edited my question any help?
– user310375
Feb 5, 2021 at 22:32

"Taking out what is known" is the basic property

$$E[X\mid \sigma(X)] = X$$

and therefore when $$f$$ is a measurable function of $$X$$ and $$g$$ is a measurable function of $$Y,$$

$$E[f(X)g(Y)\mid \sigma(X)] = f(X)E[g(Y)\mid \sigma(X)].$$

When $$X$$ and $$Y$$ are independent this further simplifies to $$f(X)E[g(Y)].$$

Use this fact along with linearity of expectation and the independence of $$(X,Y)$$ to compute

\begin{aligned} E[(X+Y)^3\mid \sigma(X)] &= E[X^3+3X^2Y+3XY^2+Y^3\mid \sigma(X)] \\ &= X^3 + 3X^2E[Y] + 3XE[Y^2] + E[Y^3]. \end{aligned}

You don't have to integrate to find these moments of $$Y,$$ because you already have the information you need:

• Because $$Y$$ and $$-Y$$ have the same distribution and their first and third moments are finite, $$E[Y]=E[-Y]=-E[Y]$$ and $$E[Y^3]=E[(-Y)^3]=-E[Y^3]$$ show these expectations are zero.

• $$E[Y^2] = \operatorname{Var}(Y)+E[Y]^2 = 1+0=1.$$

• so I have to integrate only $\mathbb{E}[-Y^3]$?
– user310375
Feb 5, 2021 at 22:47
• or just the $$\int_{-\infty}^{+\infty} yf_{Y|X}(Y|X)dy$$
– user310375
Feb 5, 2021 at 23:09
• So according to these I will have $x^3+3x$
– user310375
Feb 5, 2021 at 23:35
• For $\mathbb{E}[Y]$ we have : \begin{align*} \int_{-\infty}^{+\infty} yf(y)dy &= \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2\pi}}y e^{-y^2/2}dy \\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}y e^{-y^2/2}dy \\ &= \frac{1}{\sqrt{2\pi}}-y e^{-y^2/2} \\ &= \left[-\frac{1}{\sqrt{2\pi}}y e^{-y^2/2}\right]_{-\infty}^{+\infty} \\ &=0 \end{align*} Therefore : \begin{aligned} \mathbb{E}[(X+Y)^3\mid \sigma(X)] &= X^3 + 3X^2\mathbb{E}[Y] + 3X\mathbb{E}[Y^2] + \mathbb{E}[Y^3]\\ &= X^3 + 3X^2\cdot 0 + 3X \cdot 1 + 0 \\ &= X^3 + 3X \\ & = X(X^2+3) \end{aligned}
– user310375
Feb 6, 2021 at 13:24