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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*}

1 Answers1

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"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.

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

whuber
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  • so I have to integrate only $\mathbb{E}[-Y^3]$? –  Feb 05 '21 at 22:47
  • or just the $$\int_{-\infty}^{+\infty} yf_{Y|X}(Y|X)dy$$ –  Feb 05 '21 at 23:09
  • So according to these I will have $x^3+3x$ –  Feb 05 '21 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 : $$\begin{aligned} \mathbb{E}[(X+Y)^3\mid \sigma(X)] &= X^3 + 3X^2\mathbb{E}[Y] + 3X\mathbb{E}[Y^2] + \mathbb{E}[Y^3]\\ &= X^3 + 3X^2\cdot 0 + 3X \cdot 1 + 0 \\ &= X^3 + 3X \\ & = X(X^2+3) \end{aligned}$$ –  Feb 06 '21 at 13:24