Solutions - 5220-04

(a)

(1) Suppose $X,Y\sim 𝒩(\mu ,{\sigma }^2)$ are IID. If $X+X\sim X+Y$, then $E[X+X]=E[X+Y]$ and $\text{Var}(X+X)=\text{Var}(X+Y)$.

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(2) Check the first condition:

$$E[X+X]=2\mu =E[X]+E[Y]=E[X+Y]$$

The first condition holds for any $\mu \in ℝ$.

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(3) Check the second condition:

$$\begin{array}{c}\text{Var}(2X)=4{\sigma }^2\text{Var}(X+Y)={\sigma }^2+{\sigma }^2=2{\sigma }^2\\ \\ 4{\sigma }^2=2{\sigma }^2\gg \gg \sigma =0\end{array}$$

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(4) Thus, $\mu \in ℝ$ and $\sigma =0$ (i.e., $X,Y$ are constants) are the only values that satisfy the condition.

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(b)

(1) Define $W=X-Y-2$:

$$\begin{array}{c}{\mu }_W=\mu -\mu -2=-2\\ \\ {\sigma }_W^2={\sigma }^2+{\sigma }^2=2\end{array}$$

Thus $W\sim 𝒩(-2,2)$, so $W=\sqrt{2}Z-2$.

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(2) Compute $P[X>Y+2]=P[W>0]$:

$$\begin{array}{c}P[W>0]\gg \gg P[\sqrt{2}Z-2>0]\gg \gg P[Z>\frac{2}{\sqrt{2}}]\\ \\ \gg \gg 1-\mathrm{\Phi }(1.4142)\gg \gg 1-0.9213\gg \gg 0.0787\end{array}$$

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(c)

(1) Let $W=X-3Y$:

$$\begin{array}{c}{\mu }_W=0\\ \\ {\sigma }_W^2={\sigma }_X^2+9{\sigma }_Y^2\end{array}$$

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(2) Compute $P[X-3Y>0]=P[W>0]$:

Since ${\mu }_W=0$, we have P[W > 0] = P[Z > 0] \quad \gg\gg \quad \colorbox{cyan}{ ParseError: Unexpected end of input in a macro argument, expected '}' at end of input: …colorbox{cyan}{0.5} ParseError: Expected 'EOF', got '}' at position 1: }̲