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2

If you choose $b$ blue balls, then you pick $7-b$ red balls. So $r=7-b$. So you have perfect negative correlation.


2

Here are some corrections to your derivatons: $EY^2=\frac{1}{2}$, $EXY=\frac{1}{4}$, $Cov(X,Y)=\frac{1}{36}$, $Var(y)=\frac{1}{18}$, hence $\rho=\frac{1}{2}$ is correct. The other values you calculated correctly.


1

No. Counterexample: Let $\xi$ and $\eta$ be iid with $\mathbb{P}\left(\xi=1\right)=\mathbb{P}\left(\xi=-1\right)=\frac{1}{2}$ and let $\zeta:=\xi\eta$. Then $\xi$, $\eta$ and $\zeta$ are pairwise uncorrelated. However $\mathbb{E}\xi\zeta\eta=\mathbb{E}\xi^2\eta^2=1\neq0=\mathbb{E}\xi\mathbb{E}\zeta\mathbb{E}\eta$.


1

If the function $f$ is even then the function $x\mapsto xf(x)$ is odd hence its integral is zero, that is, $E(X)=0$. If the functions $f$ and $h$ are both even then the function $fh$ is even hence the function $x\mapsto xf(x)h(x)$ is odd hence its integral is zero, that is, $E(Xh(X))=0$. Thus, if the functions $f$ and $h$ are both even, then ...


1

Only if the travel times of AB and BC are uncorrelated. For general random variables: $Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y)$.


1

Let $Q(x)$ be the Q-function, i.e., $Q(x)=P(N\geq x)$, where $N$ is a Gaussian random variable with mean $0$ and variance $1$. Then $$ \begin{align*} P(Y \geq y) & =P(XZ \geq y) \\ &= P(Z=1)P(XZ \geq y|Z=1) + P(Z=-1)P(XZ \geq y|Z=-1) \\ &= \frac{1}{2}P(X \geq y|Z=1) + \frac{1}{2}P(-X \geq y|Z=-1) \\ &= \frac{1}{2}P(X \geq y) + \frac{1}{2}P(X ...


1

The distribution is discrete. Since the given probabilities add up to $1$, the "missing" entries are all $0$, so are not missing at all. To compute the correlation coefficient, you will need the covariance and the two variances. Let us start. From the table, $E(XY)=(1)(1)(0.25)+(1)(2)(0.25)+(2)(1)(0.25)+(0)(0)(0.25)$. Now let us compute $E(X)$. This can be ...


1

We know that the values of $a^*$ and $b^*$ are simply the regression coefficients (by definition) $ a^* = \frac{cov(X,Y)}{Var(X)}$ $ b^* = \bar{Y} - a*\bar{X}$ It should be straight forward to plug these values into $E[(a^{*}X + b^{*} - Y)]$ and solve for the given identity



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