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You are given a piecewise probability density function: $$f_X(x)=\begin{cases} 125 x^2/18 &:& ~~~0\leq x\lt 0.6\\ 9/(10x^2)&:& 0.6\leq x\leq 0.9\\0&:&\textrm{elsewhere}\end{cases}$$ To find an expect …

Take $\mathsf{Var}(\alpha X + \beta Y+\gamma) = \alpha^2\cdot \mathsf{Var}(X) + \beta^2\cdot \mathsf{Var}(Y) + 2(\alpha\cdot \beta)\cdot\mathsf{Cov}(X, Y)$ Substitute $\alpha\gets 0,\\ \beta\gets b^2 …

$$\mathsf E(\bar X)=\sum_x x\, \mathsf P(\bar X{=}x)$$ The variance of the sample mean is $((2500-3000)^2+(3500-3000)^2)/2$ which is $250\,000$. …

$\mathsf E(X)~{= 1\cdot p_{\small X}(1)+ 0\cdot p_{\small X}(0)\\=q}$ I guess this is trivial but how do I find what would be the probability of Z at z=(1−q)2 and q2? By definition $Z=(X-q)^2$ so.. …

My Attempt (I think I have the right answer. I just want some verification. Thank you) Verified. You have the right answer. All your work checks out, and in short, since the variables are independ …

Variance is not a linear operator. Rather, Covariance is a Bilinear operator: $$\mathsf{Cov}(S+T,U+V)=\mathsf{Cov}(S,U)+\mathsf{Cov}(S,V)+\mathsf{Cov}(T,U)+\mathsf{Cov}(T,V)$$ ...and so... … The independence means that the covariance between two distinct members of the sequence is zero, while identical distribution means that the variance of any member of the sequence equals the variance of …

They are for different cases. When $n$ is some constant and $X$ is the same variable. $\mathsf {Var}(\sum_{k=1}^n X)~{=\mathsf {Var}(nX)\\= n^2\mathsf{Var}(X)}$ When $N$ is a random variable an …

Hint: Assume that it is, and calculate the correlation coefficient for any two samples, say $X_s,X_t$.

For the variance, $\mathsf{Var}(Z)= \mathsf E(X^2Y^2)-\mathsf E(XY)^2$ and, again, because of independence... …

As you know, $\mathsf {Var}(Z)=\mathsf E(Z^2)-\mathsf E(Z)^2$. Substituting $Z\gets X+Y$ , then expanding and rearranging using the Linearity of Expectation, and such, we call the "cross terms" menti …

d~~}{\mathrm d p}\left(1-p^{-1}\right)&&\text{algebra} \\[1ex]\tag 9 &=p~\cdot~p^{-2}&&\text{derivation} \\[1ex]\tag {10} &=\dfrac 1{p}&&\text{algebra} \end{align}$$ Also, this is the mean, not the variance …

$$\begin{align}&\mathsf E(X) \\[2ex]=~&\mathsf E(\mathsf{Var}(X\mid L))+\mathsf{Var}(\mathsf E(X\mid L))&&\text{Law of Total Variance} \\[2ex] =~&\mathsf E(\mathsf {Var}(\sum_{t=1}^LD_t\mid L))+\mathsf … mathsf E(L~\mathsf{Var}(D_1))+\mathsf{Var}(L~\mathsf E(D_1))&&\text{identical distribution of all $D_t$} \\[2ex]=~&\mathsf E(L)~\mathsf{Var}(D_1)+\mathsf{Var}(L)~\mathsf E(D_1)^2&&{\text{expectation, and variance …

$$\begin{align}\mathsf E(X)&=\sum_{i=1}^{25}\mathsf E(X_i)\\[1ex]&=\dfrac{25\cdot{^{50}C_3}}{^{75}C_3}\end{align}$$ Regarding the variance, we recognize this is just a binomial distribution, so the variance … $$W\mid X\sim\mathcal{Bin}(3X, 0.4)\\\mathsf E(W\mid X)= (3X)(0.4)\\\mathsf {Var}(W\mid X) = (3X)(0.4)(0.6)$$ So use the Laws of Total Expectation, and Variance:$${\mathsf E(W)=\mathsf E(\mathsf E(W\mid …

Then just use bilinearity of (co)variance:$$\mathsf{Var}(X)~{=\mathsf {Var}(Y)+\mathsf {Var}(Z)+2\mathsf{Cov}(Y,Z)\\=\mathsf {Var}(Y)+\mathsf {Var}(Z)\qquad+0}$$ Now, you may use Indicator random variables … to find the variance of these counts seperately from first principles. …

(The variance of which is classically obtained by the indicator random variable method anyway.) …