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Assume that $X_1$, and $X_2$ are i.i.d. normal random variables with mean $0$ and variance $1$. Let $Y_1$ and $Y_2$ be defined as $Y_1 =8X_1+6X_2$ and $Y_2 = X_1$.

  1. $E[Y_1]= 0$ correct? because it's just $8 \times 0 + 6 \times 0$?
  2. $Var(Y_1) = 100$? I'm getting this because $Var(x) = a^2Var(X)$ So each equal $64($variance of $X_1 = 1) + 36($variance of $X_2 = 1) = 100$.
  3. $P(Y_1 \ge 12)= 1 - P(Y_1 < 12) = $Do I use the standard normal table for this and shift the standard normal over or something like that?
  4. $Cov(Y_1,Y_2) =? $. I know $$Cov(X,Y) = E[(X-E[X])·(Y- E[Y])] = E[XY ]- E[X]E[Y ]$$ but I'm not sure how I apply it in this situation.

Any help solving these problems and offering suggestions would be greatly appreciated. Thank you all very much! I'm happy to also offer more clarification.

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1 Answer 1

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  1. Correct.

  2. Correct.

  3. If you use tables for your normal CDF calculations then yes, recast to a standard normal with $Z=\frac{Y_1-\mu}{\sigma}$ and read off the number you need.

  4. You're a fair bit of the way there. The next step is to write $E[Y_1 Y_2]$ in terms of expectations involving $X_1,X_2$, specifically: $E[Y_1 Y_2]=E[(8X_1+6X_2)X_1]=E[8X_1^2+6X_2X_1]$. Any ideas for the next step?

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  • $\begingroup$ I distribute X1 and plug in al the parts where I know mean for the variables? Do I need the formula Var(X) = E[X2]−(E[X])2 ? Also for part 3 do I say z = 12-0/1? That doesn't seem right $\endgroup$ Mar 29, 2016 at 11:49
  • $\begingroup$ @user2598152 In part 3 you use the mean and variance of $Y_1$, the normal variable of interest. As for part 4, technically you will need that formula, though you also know $E[X_1]=0$ so... $\endgroup$
    – Ian
    Mar 29, 2016 at 12:41
  • $\begingroup$ Is the answer to four 0? lol $\endgroup$ Mar 29, 2016 at 12:45
  • $\begingroup$ @user2598152 Nope. $\endgroup$
    – Ian
    Mar 29, 2016 at 12:56
  • $\begingroup$ I figured it out haha thank you so much! $\endgroup$ Mar 29, 2016 at 12:59

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