Tag Info

Hot answers tagged

11

This sketch might help. You want the red area as a proportion of the red and blue areas.


9

It would generally not be true even if they were independent. For example if $X,Y,Z$ were identically and independently continuously distributed then they can come in any order with equal probability so $ P(X \leq Y \leq Z) = \frac16$ but similarly $P(X \leq Y)P(Y \leq Z) = \frac12 \times \frac12 = \frac14$.


7

Let $X=Z$, and $Y$ such that $P(X \le Y) = \frac{1}{2}$ and $P(X=Y)=0$, then in order for the equality to hold we must have \begin{eqnarray} 0 &=& P(X=Y) \\ &=&P(X \le Y \le Z) \\ &=& P(X \le Y) P(Y \le X)\\ &=& P(X \le Y) \left( 1- P(X<Y) \right) \\ &=& \frac{1}{4} \end{eqnarray} So the equality does not hold in ...


5

Assuming at least that $\Bbb E|X_j|<\infty$: Clearly yes if $\sum|a_j|<\infty$; then Chebyschev says that $\sum P(|a_jX_j|>\epsilon)<\infty$ for every $\epsilon>0$. No in general. Assuming $X_j$ is not essentially bounded there exist $a_j\to0$ such that $\sum P(|a_jX_j|>1)=\infty$, so the less trivial half of Borel-Cantelli says that ...


5

Let $A_n$ be independent events with $\mathbb{P}(A_n)=1/n$, and define $X_n=1_{A_n}$. Then $X_n\to0$ in probability, but $X_n$ does not converge almost everywhere. Apply the second Borel-Cantelli lemma twice; once to the sequence $A_n$ and also to the sequence $A_n^c$, to conclude that $$P([X_n=1\mbox{ infinitely often}] \cap [X_n=0\mbox{ infinitely ...


4

The simplest way is probably using the CDF, which is known in this case. Let $X \sim Exp(1)$ and $Y = \sqrt{X}.$ The CDF of $X$ is $F_X(x) = 1 - e^{-x}$, for $x > 0.$ Then $$F_Y(y) = P(Y \le y) = P(\sqrt{X} \le y) = P(X \le y^2) = 1 - e^{-y^2},$$ for $y > 0.$ Then take the derivative of the CDF of $Y$ to get $F_Y^\prime(y) = f_Y(y) = 2ye^{-y^2},$ ...


3

It involves covariance between the various $x_i$. $xx^T$ is a $n\times n$ matrix. Let $C=E\{xx^T\}$. The answer is the trace of $AC$. If $E\{x_ix_y\}=E\{x_i\}E\{x_j\}$, so the $x_i$ are independent of each other, then the answer is $E\{x\}^TAE\{x\}$


2

$$\text{Boy, High income} =4\\ \text{girl, High income} = 6\\ \text{Boy, low income} = 6\\ \text{Girl, low income} = x$$ $$ P(\text{Male - M}) = \frac{10}{16+ x}\\ P(\text{High income - H}) = \frac{10}{16+ x}\\ P(\text{Male High income}- MH) =\frac{4}{16+x} $$ Independence means that $P(M \text{and} H) = P(M) P(H) $ $$\frac{10}{16+x}\frac{10}{16+x} = ...


2

I do not know if this is what you are after ... there is another answer given too, but if you know the mean and the covariance of $X$ then: using your comment ($A=B^TB$) and assuming $EX=\mu$ and $cov(X)=\Sigma$ we may write \begin{align} EX^TAX&=E(X-\mu)^TA(X-\mu)+\mu^TA\mu\\ &=E[B(X-\mu)]^TB(X-\mu)+\mu^TA\mu\\ &=\mbox{tr} B\Sigma ...


2

$$X\leq Y\implies \mathbb E[X]\leq \mathbb E[Y],$$ and thus $$X\geq 0\implies \mathbb E[X]\geq \mathbb E[0]=0.$$


2

$$\begin{align} \mathsf P(X < Y) &=\int_1^3 \left\{\int_0^y f_X(x) \operatorname dx\right\}g_Y(y) \operatorname dy\\ &\neq \tfrac{1}{4} \int_1^3 \int_0^y \operatorname dx \operatorname dy \end{align}$$ Here's the problem.   The inner integral's upper bound should be $\min(2, y)$ because the support for $X$ is $(0;2)$.   Watch out for ...


2

By the definition of $T$ we have $Z_T>2\alpha$ whenever $T<\infty$. If, in addition, $|Z_n-Z_T|\leq \alpha$ we get $$|Z_n|-\alpha=\alpha + |Z_n|-2\alpha > \alpha + |Z_n|-|Z_T| \geq \alpha - |Z_n-Z_T|\geq 0.$$


2

$$M(t)=\frac{1}{100}e^{-2t}+\frac{3}{20}e^{-t}+\frac{237}{400}+\frac{9}{40}e^{t}+\frac{9}{400}e^{-2t}\\=\frac{e^{-2t}}{400}\left(4+60e^{t}+237e^{2t}+90e^{3t}+9e^{4t}\right)\\=\frac{e^{-2t}}{400}\left(2+15e^{t}+3e^{2t}\right)^{2}$$ So $$M_{X}(t)=\frac{1}{20}(2e^{-t}+15+3e^{t})$$ And by definition, $$M_{X}(t)=\mathbb{E}(e^{tX})=\sum p_{i}e^{x_{i}t}$$ So ...


1

The characteristic polynomial of $A$ is $$ p_A = t^2-4X t + (5X^2-X) $$ The eigenvalues of $A$ are $$ \lambda_{1,2} = 2X \pm \sqrt{ 4X^2 - 5X^2+X} = 2X\pm \sqrt{X(1-X)} $$ There expected values are $$ \mathbb E[\lambda_{1,2}] = 1 \pm \mathbb E[\sqrt{X(1-X)}]. $$ This is equal to the eigenvalues you found only if $E[\sqrt{X(1-X)}] = \frac12$. If $X$ is ...


1

No, you have to find the random eigenvalues of $A$ and then find its mean. For example here $A$ has the characteristic equation $$\lambda^2-4\lambda X+5X^2-X=0\implies \lambda=2X\pm \sqrt{X-X^2}$$ To find mean eigenvalue of $\lambda$ you need to find expectation of this quantity, which is obviously not going to be the answer you found, in general.


1

Split the inteval and use Baye's Theorem to get $$P(X<Y) = P(X<Y | X<1) P(X<1) + P(X<Y|X\geq1)P(X\geq 1))$$ sind $X$ is uniform on $(0,2)$, we know that $P(X<1)=P(X>1) = \frac12$. Furthermore, $P(X<Y|X<1) = 1$ since $Y$ is uniform on $(1,3)$. This yields $$P(X<Y) = 1 \cdot \frac12 + P(X<Y|X\geq1)\frac12$$ Now $P(X<Y| ...


1

If you want to solve the problem using integrals then you should notice that you have wrong upper limit in the inner integral. It should be min(y,2).


1

We .can draw the rectangle and it's interior $ 0 \leq x \leq 2$ and $1 \leq y \leq 3$. Then we can draw the line $y=x$. Let's look at our event. So we should draw the line $y=x$. Now, the region delimited is given by the triangle whose vertices are $(1,1)$, $(2,2)$ and $(2,1)$.The probanility is $\int\limits_{1}^{2}\int\limits_{1}^{x} \frac{1}{4}dydx$.


1

Let $X$ be a random variable with density function $f_X(x)$, $g(x)$ be a monotone increasing function of $x$, and $Y = g(X)$. We seek the density function $f_Y(y)$ of $Y.$ In this instance it can be shown that $$f_Y(y) = f_X(g^{-1}(y)) \frac{dg^{-1}(y)}{dy},$$ where $g^{-1}$ is the inverse function of $g$. In your Question, $X \sim Exp(1),\;$$g(x) = ...


1

You have $$ (1)=\mathbb E[1_{\{\tau_a\leq t\}}1_{\{X_{t-\tau_a}+W_{\tau_a}\leq a\}}] = \mathbb E[\mathbb E[1_{\{\tau_a\leq t\}}1_{\{X_{t-\tau_a}+W_{\tau_a}\leq a\}}|F_{\tau_a}]] = \mathbb E[1_{\{\tau_a\leq t\}}\Pr(X_{t-\tau_a}\leq 0]|F_{\tau_a})],$$ where you use that $W_{\tau_a}=a$ and measurability of $\tau_a$ wrt. $F_{\tau_a}$. Moreover, ...


1

$$D = b^2-4a = (4k)^2 - 4 \cdot 4 \cdot (k+2) = 16(k^2 - k - 2) = 16(k-2)(k+1)$$ We need $D \ge 0$ for real roots, which happens when $k \ge 2$ or $k \le -1$. Can you take it from here?



Only top voted, non community-wiki answers of a minimum length are eligible