Probability Distribution Of Random Variable Defined By Gaussian Variables I don't have answers to this question
Please let me know if there are mistakes


Problem:



  $X$, $Y$, and $Z$ are independent Gaussian random variables with:
  $E(X) = 2$$E(Y) = −1$$E(Z) = 4$$Var(X) = 2$ $Var(Y) = 1$  $Var(Z) = 2$ 
   If the random variable $W$ is given by $W = 3X + 3Y − Z$
  
  i) Find the mean and variance of $W$ ii) Write down the probability
  distribution for $pW(w)$ iii) What is the probability that a sample
  of $W$ will be between 0 and 5?



Attempt:

i)

\begin{align}
E(W) &amp;= E(3X + 3Y - Z)\\
\ &amp;= E(3X) + E(3Y) - E(Z) \\
\ &amp;= 3E(X) + 3E(Y) - E(Z) \\ 
\ &amp;= 3(2) + 3(-1) - 4\\
\ &amp;= -1 \\
\end{align}



\begin{align}
V(W) &amp;= V(3X + 3Y - Z)\\
\ &amp;= V(3X) + V(3Y) - V(Z) \\
\ &amp;= 9V(X) + 9V(Y) - V(Z) \\ 
\ &amp;= 9(2) + 9(1) - 2\\
\ &amp;= 25 \\
\end{align}



ii)

I am not sure how to find $pW(w)$ knowing $W$, $E(W)$ and $V(W)$.



iii)

Given $pW(w)$ from part ii I assume the answer would be:
$P(0 ≤ W ≤ 5) = \int_{0}^5 pW(w) dW$


Thank you
In part a, you mistakenly subtract the third variance rather than adding it. You know how when you multiply a variable by a constant, the variance gets multiplied by the constant squared? Well, the same goes for $(-1).$ So the variance should be $29,$ not $25$

For part $b$ any sum of independent Gaussians is a Gaussian and a Gaussian is fully characterized by its mean and variance. So we have $W\sim N(-1,29)$ and it has PDF $$f_W(w) = \frac{1}{\sqrt{58\pi}}e^{-\frac{(w+1)^2}{58}}.$$

Part c looks good too, and now you know what to plug in. It might be easier to write it in terms of the $N(0,1)$ quantity $\frac{W+1}{\sqrt{29}}$. We have $$0&lt;W&lt;5 \iff \frac{1}{\sqrt{29}}&lt; \frac{W+1}{\sqrt{29}} &lt;\frac{6}{\sqrt{29}} $$ so the answer should be $$P(1/\sqrt{29} &lt; N(0,1) &lt; 6/\sqrt{29}) = \Phi(6/\sqrt{29})-\Phi(1/\sqrt{29})$$ where $\Phi$ is the standardard normal CDF.

Note A previous version mistakenly gave OP an 'all clear' on part (a).
In part a, you mistakenly subtract the third variance rather than adding it. You know how when you multiply a variable by a constant, the variance gets multiplied by the constant squared? Well, the same goes for $(-1).$ So the variance should be $29,$ not $25$

For part $b$ any sum of independent Gaussians is a Gaussian and a Gaussian is fully characterized by its mean and variance. So we have $W\sim N(-1,29)$ and it has PDF $$f_W(w) = \frac{1}{\sqrt{58\pi}}e^{-\frac{(w+1)^2}{58}}.$$

Part c looks good too, and now you know what to plug in. It might be easier to write it in terms of the $N(0,1)$ quantity $\frac{W+1}{\sqrt{29}}$. We have $$0&lt;W&lt;5 \iff \frac{1}{\sqrt{29}}&lt; \frac{W+1}{\sqrt{29}} &lt;\frac{6}{\sqrt{29}} $$ so the answer should be $$P(1/\sqrt{29} &lt; N(0,1) &lt; 6/\sqrt{29}) = \Phi(6/\sqrt{29})-\Phi(1/\sqrt{29})$$ where $\Phi$ is the standardard normal CDF.

Note A previous version mistakenly gave OP an 'all clear' on part (a).