Recent content by Francobati

Yes and I obtain $E(X^2)=var(X)+(E(X))^2=\frac{(b-a)^2}{12}+(\frac{a+b}{2})^2=\frac{1}{12}+(\frac{1}{2})^2=\frac{1}{12}+\frac{1}{4}=\frac{1}{3}$ This result equal to this $E(X^2)=\frac{1^3-0^3}{3(1)}= \frac{1}{3}$, how I can translate this $E(X^2)=\frac{1^3-0^3}{3(1)}$ in a general formula?

$ E(X^2)=\frac{1^3-0^3}{3*1} $ $ E(X^4)=\frac{1^5}{5}$ $ E(Y^2)=1+\frac{1}{5}-2*\frac{1}{3}=1+\frac{1}{5}-\frac{2}{3}=\frac{15+3-10}{15}= \frac{8}{15}\ne2\ne\frac{1}{2} $, so first and second are false. $ var(Y)=var(1-X^2)=var(1)-var(X^2)=0-var(X^2) $ $...

Hello. Let $ Y=1-X^2 $, where $ X~ U(0,1) $. What statement is TRUE? -$ E(Y^2)=2 $ - $ E(Y^2)=1/2 $ - $ var(Y)=1/12 $ - $ E(Y)=E(Y^2) $ -None of the remaining statements. Solution: I compute: $ E(Y^2)=E(1-X^2)^2=E(1+X^4-2X^2)=1+E(X^4)-2E(X^2) $, then?

Hello. Can you help me figure out how to pass, integrating, by the marginal cost: $MC_{i}(q)_{i}=q_{i}+10$ to the total cost: $TC=\frac{1} {2}q_i^2+10q_{i}$? $i=1,2$, are the two companies. $q_{i}$ is the quantity. What are the calculations?

Hello. Many thanks. You are absolutely right. I am studying probability, I am trying to read the theory, but unfortunately for the exercises and the applications I need somebody routes, I addresses, because the practice is very different from the theory and is at the same time useful to better...

Let $X\sim U(0,1)$ and define $Y=1-X$. What statement is TRUE? (1): $F_{X}(u)\neq F_{Y}(u)$, for every $u\epsilon \left [ 0,1 \right ]$; (2): $Y$ is not a rv; (3): $E(X+Y)=2$; (4): $Y\sim U(0,1)$; (5): none of the remaining statements.

Ok. Many thanks.

Yes, the answer is 3. But as I explain in an exhaustive way?

$E(X)=:x_{1}P(A_{1})+...+x_{k}P(A_{k})\geqslant 0$ $F(x)=P(X\leqslant x)$

I have no idea. What information do I need to resolve it? Help me, please.

I have to find the true answer among these five.

$E(tX)=tE(X)$ $E(\frac{1}{\varepsilon }X)=\frac{1}{\varepsilon }E(X)$ $E(X)=7P(X\leqslant 7)+7\varepsilon P(X>7)$ $E(\frac{1}{\varepsilon }X)=\frac{1}{\varepsilon }(7P(X\leqslant 7)+7\varepsilon P(X>7))$ $E(\frac{1}{\varepsilon }X)=\frac{7}{\varepsilon }P(X\leqslant 7)+7P(X>7)$ Correct?

Ok, thanks. Perfect! Now I have to go by $E(X)=7P(X\leqslant 7)+7\varepsilon P(X>7)$ to $E(\frac{1}{\varepsilon}X)=\frac{7}{\varepsilon}P(X\leqslant 7)+7P(X>7)$. What calculations do I need to do? There is perhaps some recollection to common factor to do?

And after?

Unfortunately I do not know to calculate it. Is $E(I_{X>7})=P(X>7)$ and $E(I_{X\leqslant 7})=P(X\leqslant 7)$?