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Consider the function $g(W)=-e^{-W}$, where $W$ is some random variable s.t.$W=X+YZ$. Furthermore, it holds that all the random variables $X,Y,Z$ follow the normal distribution with the following properties:

  • $X\sim N(x,\sigma_x^{2})$, where $x$ denotes the constant mean of the r.v. $X$ and its variance is $\sigma_x^{2}>0$
  • $Y\sim N(0,\sigma_y^{2})$ and $Z\sim N(0,\sigma_z^{2})$
  • $X$ and $Z$ are correlated and we denote their covariance as $\sigma_{x,z}\neq 0$, while $\sigma_{y,z}=0$, $\sigma_{x,y}=0$.
  • furthermore, $I=\{S,Y\}$ is an information set, where $S=X-x+U$, where $U$ is independently normally distributed with mean zero and variance $\sigma_u^{2}$, i.e. $U\sim N(0,\sigma_u^{2})$

If $(X,Y,Z,S)$ is jointly normally distributed, what is the conditional expectation$$E(g(W)|I)=E[-e^{-(X+YZ)}|\{S,Y\}]$$

Nav89
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    This is the third incarnation of your question. You aren't getting answers because there aren't any: you need to evaluate this expectation numerically. – whuber Oct 30 '19 at 22:08
  • This can't be! What do you mean by saying numerically? I can not understand you...The problem can be solved – Nav89 Oct 30 '19 at 22:26
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    I appreciate you challenging like that, because I see now that the conditioning on $\tilde{y}$ removes the principal complication. However, that basic insight is so deeply hidden behind distracting and irrelevant details that readers should be excused for glancing at your question and just moving on. You can help yourself by doing some preliminary analysis to simplify the question. – whuber Oct 30 '19 at 22:35
  • Ok i will do this, but can you help me to edit the question? I am searching for a precise answer and it is very signifficant to me. How should I edit it? Differently, in such a way not to make thsi distracting? – Nav89 Oct 30 '19 at 22:40
  • whuber please help me edit my question! – Nav89 Oct 30 '19 at 22:41
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    $\delta,$ $\alpha,$ $\beta,$ and some of the variances are superfluous. The tildes everywhere are off-putting and generally unnecessary. With some analysis it looks like you could eliminate the variables $s_1$ (why the subscript?) and $\tilde{y}$ altogether, turning this into a question about the expectation of a Lognormal variable. Even if this simplification cannot be attained, it's worth aiming for it. – whuber Oct 30 '19 at 22:43
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    You say " $Y\sim N(0,\sigma_y^{2})$ and $Z\sim N(0,\sigma_z^{2})$, and they are identically and normally distributed". If the distributions are identical, why did you give them distinct variances? – Glen_b Oct 31 '19 at 00:59
  • Glen_b You are right, they follow different normal distributions. I stated that wrong, I have changed it. – Nav89 Oct 31 '19 at 06:05
  • If this is part of assignment, please add the `self-study` tag. And have a look at the recommendations on the [wiki](https://stats.stackexchange.com/tags/self-study/info) page. – Xi'an Nov 01 '19 at 10:20

2 Answers2

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Reproducing this earlier question on the forum,

For ${\boldsymbol Y} \sim \mathcal{N}(\boldsymbol\mu, \Sigma)$, consider partitioning $\boldsymbol\mu$ and ${\boldsymbol Y}$ into $$\boldsymbol\mu = \begin{bmatrix} \boldsymbol\mu_1 \\ \boldsymbol\mu_2 \end{bmatrix} $$ $${\boldsymbol Y}=\begin{bmatrix}{\boldsymbol y}_1 \\ {\boldsymbol y}_2 \end{bmatrix}$$

with a similar partition of $\Sigma$ into $$ \begin{bmatrix} \Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22} \end{bmatrix} $$ Then, $f({\boldsymbol y}_1|{\boldsymbol y}_2={\boldsymbol a})$, the conditional distribution of the first partition given the second, is $\mathcal{N}(\overline{\boldsymbol\mu},\overline{\Sigma})$, with mean $$ \overline{\boldsymbol\mu}=\boldsymbol\mu_1+\Sigma_{12}{\Sigma_{22}}^{-1}({\boldsymbol a}-\boldsymbol\mu_2) $$ and covariance matrix $$ \overline{\Sigma}=\Sigma_{11}-\Sigma_{12}{\Sigma_{22}}^{-1}\Sigma_{21}$$

Therefore, since the vector $(X,Z,S)$ is Gaussian with zero mean, without loss of generality since $$\mathbb E[\exp\{X\}]=\mathbb E[\exp\{X-\mu+\mu\}]=\mathbb E[\exp\{X-\mu\}]\exp\{\mu\},\tag{1}$$ and arbitrary covariance matrix $\Sigma$, with the vector being independent from $Y$, the conditional distribution of $(X,Z)$ conditional on $(S,Y)$ is Gaussian with mean $\Sigma_{12}{\Sigma_{22}}^{-1}S$ and variance $\overline{\Sigma}$ (see above). This implies that the expectation of interest which is also the MGF of $X+YZ$ at $t=-1$ is \begin{align*}\mathbb E[\exp\{W\}]&=\mathbb E[\mathbb E[\exp\{-\underbrace{(X+YZ)}_{(1\ Y)\,(X\ Z)^\text{T}}\}|S,Y]]\\ &=\exp\left\{-(1\ Y)\Sigma_{12}{\Sigma_{22}}^{-1}S + \frac{1}{2}(1\ Y) \overline{\Sigma} (1\ Y)^\text{T} \right\} \end{align*} since $$(1\ Y)\,(X\ Z)^\text{T} | S,Y \sim \mathcal N\left((1\ Y)\Sigma_{12}{\Sigma_{22}}^{-1}S,(1\ Y) \overline{\Sigma} (1\ Y)^\text{T}\right)$$

Xi'an
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  • Thank you Mr Xi'an! This is exactly what i was searching for! – Nav89 Oct 31 '19 at 19:04
  • @Xi'an I miss something in the solution. Why is the vector $(X,Z,S)$ Gaussian with zero mean? If we see them individually, $X$ has mean $x$, while $Z$ and $S$ have mean zero. If we combine them, why is this happening? Does this come from the formula of the projection theorem you mention above? Could you explain it? – Nav89 Nov 03 '19 at 10:43
  • Well, I do not know if you can help me claryfing this but, looking at the answer there is something that it is confusing. Specifically, you claim that $X+YZ=(1\ Y)\,(X\ Z)^\text{T}$. Is this some kind of an inner product and why does this hold? I understand that we need this because on the information set, these two random variables, $(1\ Y)$,$(X\ Z)^\text{T}$ are indepedent...but i do not get it... – Nav89 Apr 18 '20 at 05:30
  • So, by writing $(1Y)(XZ)^T=(1Y)\begin{pmatrix} X \\ Z \end{pmatrix}$ you decompose somehow the sum beacasu $Y|I$ is gaussian and $XZ|I$ is also gaussian normal, but for instance $X=(x_1,x_2)$ and $Z=(z_1,z_2)$, $1=(1,1)$ and $Y=(y_1,y_2)$ by making the calculation the LHS=RHS of the equation? – Nav89 Apr 18 '20 at 07:31
  • It is not obvious to me how it can be proven that the sum is equivalent to this product of vectors, if it is called product...why is it $1Y$? and how this product is the sumw$X+ZY=W$? – Nav89 Apr 18 '20 at 08:02
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A little more analytical I believe this is the case, following the solution of @Xi'an...Suppose that $W=X+ZY\sim N(\mu,\Sigma)$ is a partition where $$\mu = \begin{bmatrix} \mu_X \\ \mu_Z \\ \mu_Y \end{bmatrix} = \begin{bmatrix} x \\ 0 \\ 0 \end{bmatrix}$$ and $$\Sigma=\begin{bmatrix} \sigma_x^2 & \sigma_{xz} & 0 \\ \sigma_{xz} & \sigma_z^2 & 0 \\ 0 & 0 & \sigma_{y}^2 \end{bmatrix}$$ Since, $(X,Z,S)$ is jointly normal vector, then we can claim that $\begin{bmatrix} X \\ Z \end{bmatrix}|I=\{S,Y\}$ is also normal since $X\in I$ and $Z\sim N(0,\sigma_z^2)$, where the lemma gives you the $\mu_{\begin{bmatrix} X \\ Z\end{bmatrix}|I}=\mu_{xz}+(\dfrac{\sigma_{x}^2}{\sigma_{x}^2+\sigma_{u}^2},\dfrac{\sigma_{z}^2}{\sigma_{x}^2+\sigma_{u}^2})S=\begin{pmatrix}x + \dfrac{\sigma_{x}^2}{\sigma_{x}^2+\sigma_{u}^2}S & \dfrac{\sigma_{z}^2}{\sigma_{x}^2+\sigma_{u}^2}S \end{pmatrix}^T$ and $\Sigma_{\begin{bmatrix} X \\ Z \end{bmatrix}|I}=\begin{pmatrix}(1-\dfrac{\sigma_{x}^2}{\sigma_{x}^2+\sigma_{u}^2})\sigma_{x}^2 & (1-\dfrac{\sigma_{x}^2}{\sigma_{x}^2+\sigma_{u}^2})\sigma_{xz}\\ (1-\dfrac{\sigma_{x}^2}{\sigma_{x}^2+\sigma_{u}^2})\sigma_{xz} & \sigma_{z}^2-\dfrac{\sigma_{xz}^2}{\sigma_{x}^2+\sigma_{u}^2}\end{pmatrix}$. Moreover $Y|I$ is also a normally distributed, since $Y\in I$ and in particular $\mu_{Y|I}=Y$ and $\Sigma_{Y|I}=0$. As a consequence, $W|I=X+ZY|I=1X+ZY|I=\begin{pmatrix}1 & Y \end{pmatrix}\begin{pmatrix}X \\Z\end{pmatrix}|I=\begin{pmatrix}1 & Y \end{pmatrix}\begin{pmatrix}X & Z\end{pmatrix}^T|I\sim N(\mathbb{E}[W|I], \mathbb{V}ar[W|I])$

Specifically, the conditional expectaion of the r.v. $W|I$ is also a mgf of a Gaussian normal distribution for $t=-1$, that it: $$M_W(t)=\mathbb{E}[\exp\{-W\}]=\mathbb{E}\left(\mathbb{E}[\exp\{-(X+YZ)\}|I]\right)=\mathbb{E}[\exp\{-\begin{pmatrix}1 & Y \end{pmatrix}\begin{pmatrix}X & Z\end{pmatrix}^T\}|I]$$ Since, $Y \in I$ and $\begin{pmatrix}Z & X\end{pmatrix}^T|I\sim N\left(\mu_{\begin{pmatrix}X & Z\end{pmatrix}^T|I}, \Sigma_{\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)$ we obtain that $$\mathbb{E}\left(\exp\{-\begin{pmatrix}1 & Y \end{pmatrix}\}\mathbb{E}[\exp\{-\begin{pmatrix}X & Z\end{pmatrix}^T\}|I]\right)=\\ =\mathbb{E}\left(\exp\{-\begin{pmatrix}1 & Y \end{pmatrix}\}\exp\{\underbrace{-\mu_{\begin{pmatrix}X & Z\end{pmatrix}^T|I}-\dfrac{1}{2}\Sigma_{\begin{pmatrix}X & Z\end{pmatrix}^T|I}}\}_{mgf\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)=\\ =\exp\{-\begin{pmatrix}1 & Y \end{pmatrix}\}\exp\{-\mathbb{E}\left({\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)-\dfrac{1}{2}\mathbb{V}ar\left({\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)\}=\\ =\exp\{-\begin{pmatrix}1 & Y \end{pmatrix}\mathbb{E}\left({\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)-\dfrac{1}{2}\begin{pmatrix}1 & Y \end{pmatrix}\mathbb{V}ar\left({\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)\begin{pmatrix}1 & Y \end{pmatrix}^T\}=\\ =\exp\{-\mathbb{E}\left({\begin{pmatrix}1 & Y \end{pmatrix}\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)-\dfrac{1}{2}\mathbb{V}ar\left({\begin{pmatrix}1 & Y \end{pmatrix}\begin{pmatrix}X & Z\end{pmatrix}^T|I}\right)\}$$ Thus, $W|I$ is the Gasussian normal with $\mu_{W|I}=\mathbb{E}[W|I]$ and $\Sigma_{W|I}=\mathbb{V}ar\left[W|I\right]$ I believe this is the full solution...it took me some hours to understand, but there is nothing complex here...

Hunger Learn
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