# Prüfungsvorbereitung 05.05.2015

• Hi,

is someone else taking the exam on 5th and wants to meet up for studying?

Straight to the topic:
I'm currently solving the previous exams and compare it with the solutions from the VOWI.
(https://vowi.fsinf.at/images/4…r%C3%BCfungsbeispiele.zip)

At some parts I think there are errors in the VOWI solution and at some parts my solution differs from it, but I don't know what is correct. So I'd like to get the solutions correct..maybe others join here...
I just post my thoughts so you can comment on it and next generations have better conditions.

At the moment I'm trying to solve the exam from January 2008.
For 1) (The triangular pdf)

b:
I think it should be

${\sigma_x}^2 = P_x - {\mu_x}^2 = \frac{1}{18}$

. He flipped the sign.

d:
For calculating

${\sigma_y}^2 = P_y - {\mu_y}^2$

, we need

$P_y = E\left\{ (x_1 + x_2)^2 \right\} = E\left\{ {x_1}^2 \right\} + E\left\{ {x_2}^2 \right\} + 2*E\left\{x_1*x_2 \right\}$

In the VOWI solution

$E\left\{x_1 x_2\right\}$

has been directly set equal to zero...why? :O I guess he confused the properties of

$C_{x1,x2}$

and

$R_{x1,x2}$

.

$X1$

and

$X2$

are statistically independent. For statistically independent random variables it holds that

$R_{x_1, x_2} = E\left\{ x_1 x_2 \right\} = E\left\{ x_1 \right\} E\left\{ x_2 \right\}$

and

$C_{x1,x2} = 0$

. Probably he assumed it the other way around..
Though I wouldn't know how to get

$E\left\{x_1 x_2\right\}$

for not independent random variables without using the

$pdf_y$

. (They did not ask for it, but still..maybe someone has an idea!?)

So we get

$P_y = \frac{1}{6} + \frac{1}{6} + 2*E\left\{x_1 \right\}*E\left\{x_2 \right\}= \frac{1}{3} + \frac{2}{9} \rightarrow P_y = \frac{5}{9}$

$\sigma(y)^2 = P_y - {\mu_y}^2 = \frac{5}{9} - \frac{4}{9}
\rightarrow \sigma(y)^2 =\frac{1}{9}$

More to follow...