Question

if (X+Y) variance Z when Y constant and (X+Z) variance Y when Z constant then prove that (X+Y+Z) variance YZ when Y and Z both variable.​

Answer 1

Given data:

• $\mathsf{X+Y\propto Z}$ when $\mathsf{Y}$ is constant

• $\mathsf{X+Z\propto Y}$ when $\mathsf{Z}$ is constant

To prove:

$\mathsf{X+Y+Z\propto YZ}$ when both $\mathsf{Y}$ and $\mathsf{Z}$ are variables

Step-by-step explanation:

Step 1.

Given, $\mathsf{X+Y\propto Z}$ when $\mathsf{Y}$ is constant

$\mathsf{\Rightarrow X+Y=k_{1}Z}$ where $\mathsf{Y}$ and $\mathsf{k_{1}}$ are constants

Adding $\mathsf{Z}$ in both sides, we get

$\quad \mathsf{X+Y+Z=k_{1}Z+Z}$

$\mathsf{\Rightarrow X+Y+Z=(1+k_{1})Z}$

$\mathsf{\Rightarrow X+Y+Z\propto Z}$ . . . (i), since $\mathsf{1+k_{1}}$ is a constant

Step 2.

Given, $\mathsf{X+Z\propto Y}$ when $\mathsf{Z}$ is constant

$\Rightarrow \mathsf{X+Z=k_{2}Y}$ where $\mathsf{Z}$ and $\mathsf{k_{2}}$ are constants

Adding $\mathsf{Y}$ in both sides, we get

$\quad \mathsf{X+Y+Z=k_{2}Y+Y}$

$\mathsf{\Rightarrow X+Y+Z=(1+k_{2})Y}$

$\mathsf{\Rightarrow X+Y+Z\propto Y}$ . . . (ii), since $\mathsf{1+k_{2}}$ is a constant

Step 3.

From (i) and (ii), using the compound theorem of variation, we get

$\quad \mathsf{X+Y+Z\propto ZY}$ since both $\mathsf{Z}$ and $\mathsf{Y}$ are variables

$\mathsf{\Rightarrow X+Y+Z\propto YZ}$ when both $\mathsf{Y}$ and $\mathsf{Z}$ are variables

Conclusion:

$\mathsf{X+Y+Z\propto YZ}$ when both $\mathsf{Y}$ and $\mathsf{Z}$ are variables.

Hence proved.