{What is the regression line $y=K(x)$? At what point $y_0$ does it intersect the $y$ axis?
{What is the regression line $\rm (RL)$? At what point $y_0$ does it intersect the $y$–axis?
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$y_0\ = \ $ { 2.5 3% }
$y_0\ = \ $ { 2.5 3% }
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{Calculate the marginal probability density function $f_y(y)$. What is the probability that the random variableö&aerospace;e $y >3$ ?
{Calculate the marginal probability density function $f_y(y)$. What is the probability that the random variablee $y >3$?
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${\rm Pr}(y > 3)\ = \ $ { 0.167 3% }
${\rm Pr}(y > 3)\ = \ $ { 0.167 3% }
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*The area of the triangle $(1,0)\ (1,4)\ (-1,3)$ gives $0.5 · 4 · 2 = 4$.
*The area of the triangle $(1,0)\ (1,4)\ (-1,3)$ gives $0.5 · 4 · 2 = 4$.
*The total area is double: $F = 8$.
*The total area is double: $F = 8$.
*Since the PDF–volume is always $1$ , then $H= 1/F\hspace{0.15cm}\underline{ = 0.125}$.
*Since the PDF volume is always $1$ , then $H= 1/F\hspace{0.15cm}\underline{ = 0.125}$.
'''(2)''' The minimum value of $x$ is obtained for $\underline{ u=0}$ and $\underline{ v=1}$.
'''(2)''' The minimum value of $x$ is obtained for $\underline{ u=0}$ and $\underline{ v=1}$.
*From the above equations, the results $x= -1$ and $y= +3$ follow.
*From the above equations, the results $x= -1$ and $y= +3$ follow.
'''(3)''' The equation given in the theory section is valid in general, i.e., for any PDF of the two statistically independent variables $u$ and $v$,
*as long as they have equal standard deviations $(\sigma_u = \sigma_v)$.
'''(3)''' The equation given in the theory section is valid in general, i.e., for any PDF of the two statistically independent variables $u$ and $v$, as long as they have equal standard deviations $(\sigma_u = \sigma_v)$.
*From the linear means $m_u = m_v = 0.5$ and the equations given in the problem statement, we obtain $m_x = 1$ and $m_y = 2$.
*From the linear means $m_u = m_v = 0.5$ and the equations given in the problem statement, we obtain $m_x = 1$ and $m_y = 2$.
*The variances of $u$ and $v$ are respectively $\sigma_u^2 = \sigma_v^2 =1/12$. It follows:
*The variances of $u$ and $v$ are respectively $\sigma_u^2 = \sigma_v^2 =1/12$. It follows:
*From this follows the value $y_0=K(x=0)\hspace{0.15cm}\underline{ = 2.5}$
*From this follows with $x=0$ the value $y_0=\hspace{0.15cm}\underline{ = 2.5}$
'''(5)''' With the auxiliary quantities $q= 2u$, $r= -2v$ and $s= x-1$ the relation holds: $s= q+r$.
'''(5)''' With the auxiliary quantities $q= 2u$, $r= -2v$ and $s= x-1$: $s= q+r$.
*Since $u$ and $v$ are each uniformly distributed between $0$ and $1$ , $q$ has a uniform distribution in the range from $0$ to $2$ and $r$ is uniformly distributed between $-2$ and $0$.
*In addition, since $q$ and $r$ are not statistically dependent on each other, the PDF of the sum is:
[[File:P_ID414__Sto_A_4_8_e.png|right|frame|Triangular PDF $f_x(x)$]]
[[File:P_ID414__Sto_A_4_8_e.png|right|frame|Triangular PDF $f_x(x)$]]
*Since $u$ and $v$ are each uniformly distributed between $0$ and $1$, $q$ has a uniform distribution in the range from $0$ to $2$ and $r$ is uniformly distributed between $-2$ and $0$.
*In addition, since $q$ and $r$ are not statistically dependent on each other, the PDF of the sum is:
:$$f_s(s) = f_q(q) \star f_r(r).$$
:$$f_s(s) = f_q(q) \star f_r(r).$$
*The addition $x = s+1$ leads to a shift of the triangular–PDF by $1$ to the right.
*The addition $x = s+1$ leads to a shift of the triangular–PDF by $1$ to the right.
*For the sought probability (highlighted in green in the following image) therefore holds: ${\rm Pr}(x < 0)\hspace{0.15cm}\underline{ = 0.125}$.
*For the sought probability (highlighted in green in the graphic) therefore holds:
We consider a two-dimensional random variable $(x,\hspace{0.08cm} y)$ whose components arise as linear combinations of two random variables $u$ and $v$:
$$x=2u-2v+1,$$
$$y=u+3v.$$
Further, note:
The two statistically independent random variables $u$ and $v$ are each uniformly distributed between $0$ and $1$.
In the figure you can see the joint PDF. Within the parallelogram drawn in blue holds:
$$f_{xy}(x,\hspace{0.08cm} y) = H = {\rm const.}$$
Outside the parallelogram no values are possible: $f_{xy}(x,\hspace{0.08cm} y) = 0$.
(1) The area of the parallelogram can be composed of two triangles of equal size.
The area of the triangle $(1,0)\ (1,4)\ (-1,3)$ gives $0.5 · 4 · 2 = 4$.
The total area is double: $F = 8$.
Since the PDF volume is always $1$ , then $H= 1/F\hspace{0.15cm}\underline{ = 0.125}$.
(2) The minimum value of $x$ is obtained for $\underline{ u=0}$ and $\underline{ v=1}$.
From the above equations, the results $x= -1$ and $y= +3$ follow.
(3) The equation given in the theory section is valid in general, i.e., for any PDF of the two statistically independent variables $u$ and $v$, as long as they have equal standard deviations $(\sigma_u = \sigma_v)$.
With $A = 2$, $B = -2$, $D = 1$ and $E = 3$ we obtain:
From this follows with $x=0$ the value $y_0=\hspace{0.15cm}\underline{ = 2.5}$
(5) With the auxiliary quantities $q= 2u$, $r= -2v$ and $s= x-1$: $s= q+r$.
Triangular PDF $f_x(x)$
Since $u$ and $v$ are each uniformly distributed between $0$ and $1$, $q$ has a uniform distribution in the range from $0$ to $2$ and $r$ is uniformly distributed between $-2$ and $0$.
In addition, since $q$ and $r$ are not statistically dependent on each other, the PDF of the sum is:
$$f_s(s) = f_q(q) \star f_r(r).$$
The addition $x = s+1$ leads to a shift of the triangular–PDF by $1$ to the right.
For the sought probability (highlighted in green in the graphic) therefore holds:
