# Tag Info

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?

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