6
$\begingroup$

Let $X = N(0,\frac{1}{\alpha})$, $Y = 2X + 8 + N_{y}$, and $N_{y}$ be a noise $N_{y} = N(0,1)$. Then, $P(y|x) = \frac{1}{\sqrt{2\pi}}exp\{ -\frac{1}{2}(y - 2x - 8)^{2} \}$ and $P(x) = \sqrt{\frac{\alpha}{2\pi}}exp\{-\frac{\alpha x^{2}}{2}\} $.

The mean vector is:

$$\mathbf{\mu} = \left( \begin{array}{c} \mu_{x}\\ \mu_{y}\end{array} \right)= \left( \begin{array}{c} 0\\ 8\end{array} \right).$$

The question is how to calculate the variance of Y.

I know that the correct answer is

$$\frac{4}{\alpha} + 1, $$

but don't know how to get from $$var(Y) = E[(Y-\mu_{y})^{2}] = E[(2X+N_{y})^{2}] $$

to

$$\frac{4}{\alpha} + 1. $$

Can anybody help? UPDATE: Thank you All for answers

$\endgroup$
8
$\begingroup$

The law of iterated expectations can help here. We have:

$$Var[Y]=E(Var[Y|X])+Var[E(Y|X)]$$

Now conditional on $X$ the expected value of $Y$ is $2X+8$, and its variance is $1$. So we have:

$$Var[Y]=E(1)+Var[2X+8]=1+4 Var[X]=1+\frac{4}{\alpha}$$

$\endgroup$
  • $\begingroup$ (+1) not so straightforward algebra, nice alternative solution. $\endgroup$ – Dmitrij Celov Apr 2 '11 at 16:14
6
$\begingroup$

Solution to this homework is straightforward application of simple algebra and independence of $X$ and $N_y$: $\mathbb{E} (2 X + N_y)^2 = 4 \mathbb{E} X^2 + 4 \mathbb{E} X \mathbb{E} N_y + \mathbb{E} N_y^2 = 4 Var X + 0 + Var N_y = \frac{4}{\alpha} + 1$.

$\endgroup$
  • $\begingroup$ Thanks for a swift answer, not a homework though. Would have tagged it as such if it really was. $\endgroup$ – matcheek Apr 2 '11 at 14:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.