Question

6. Let X and Y be two r.v.s. with E(X) = E(Y ) = 0, Cov(X, Y ) = 1, V (X) = 1 and V (Y ) = 2.
What is the expectation of (X + 1)(X − 3Y )?

Need help I know the method but answer is wrong help...

Answer 1

Well, as a educated guess I know what you're talking about

Answer 2

this is kind of statistics or probability or something in between.

Answer 3

\[\begin{align*}E\left[(X+1)(X-3Y)\right]&=E\left[X^2+X-3XY-3Y\right]\\
&=E\left[X^2\right]+E\left[X\right]-3E\left[XY\right]-3E\left[Y\right]\\
&=E\left[X^2\right]+0-3E\left[XY\right]-3(0)
\end{align*}\]
Recall that for a random variable \(X\), its variance is \(V(X)=E\left(X^2\right)-\left(E(X)\right)^2\). Since \(E(X)=0\), you have \(V(X)=E\left(X^2\right)-0=1\).
\[\begin{align*}E\left[(X+1)(X-3Y)\right]&=E\left[X^2\right]+0-3E\left[XY\right]-3(0)\\
&=1-3E\left[XY\right]
\end{align*}\]
Now let's use what you know about covariance.
\[\begin{align*}\text{Covar}(X,Y)&=E\bigg[(X-E(X))~(Y-E(Y))\bigg]\\
&=E\bigg[XY-E(X)Y-E(Y)X+E(X)E(Y)\bigg]\\
&=E(XY)-E\bigg[E(X)Y\bigg]-E\bigg[E(Y)X\bigg]+E\bigg[E(X)E(Y)\bigg]\\
&=E(XY)-E(X)E(Y)-E(X)E(Y)+E(X)E(Y)\\
&=E(XY)-E(X)E(Y)
\end{align*}\]
Use the given information. The covariance is 1, and the two expected values are both 0:
\[1=E(XY)-0\cdot0~~\Rightarrow~~E(XY)=1\]
\[E\left[(X+1)(X-3Y)\right]=1-3(1)=-2\]

Answer 4

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