5

Disclaimer: I'm not an expert, so please correct me if there are serious flaws in my argument. Hand-waving/guessing First observe that if $X^2=t$ then $X$ is either $\sqrt{t}$ or $-\sqrt{t}$, so this is a discrete distribution, and to specify the distribution you only need to compute $P(X=\sqrt{t} \mid X^2=t)$ and $P(X=-\sqrt{t} \mid X^2=t)$. In analogy with ...


3

It seems the task is to find $P(X > 2) = 1-P(X \le 2)$ when $X \sim \mathsf{Binom}(n = 100, p=0.02).$ So the computation reduces to finding $P(X \le 2) = 0.6767,$ to four places. Exact computation in R is as below: first using the R binomial CDF function pbinom; second, using R as a calculator to compute the necessary three terms of the binomial PDF ...


2

I'm wondering if this conditional distribution is even well$-$defined. Let $\epsilon_1,\epsilon_2$ be very small positive numbers and $t>0$ be arbitrary. Define $I_1=(-\sqrt{t}-\epsilon_1,-\sqrt{t}+\epsilon_1)$ and $I_2=(\sqrt{t}-\epsilon_2,\sqrt{t}+\epsilon_2)$. Assume $\epsilon_1, \epsilon_2$ are small enough to make $I_1,I_2$ disjoint. Notice how $$P\...


2

MAJOR EDIT Better check @angryavian's answer. $\Pr\{X=x|X^2=t\}$ $\;\;\;=\Pr\{X=x|X>0, X^2=t\}\Pr\{X>0|X^2=t\}$ $\;\;\;\;\;\; +\Pr\{X=x|X<0, X^2=t\}\Pr\{X<0|X^2=t\}$ Now: $\Pr\{X=x|X>0, X^2=t\}=1$, if $x=\sqrt{t}$, and zero otherwise, also $\Pr\{X=x|X<0, X^2=t\}=1$, if $x=-\sqrt{t}$, and zero otherwise. And: $\Pr\{X>0|X^2=t\} = \dfrac{\...


2

Determinant of $\Sigma$ may be zero. Let $k$ be the dimension of the space to which the sample vectors $x_n$ belong. If $N < k$ then $\Sigma$ is not full rank and therefore $\det \Sigma = 0$. Moreover, even when $N \ge k$ it is still possible that $\det \Sigma = 0$. For a simple example, suppose that $k=2$ and that the sample vectors $x_n \in\mathbb{R}^2$ ...


2

Let's start form $$\frac{X}{\sigma}\sim N(0;1)$$ It is well known, and anyway easy to prove that $$W=\Bigg(\frac{X}{\sigma}\Bigg)^2\sim \chi_{(1)}^2=Gamma(1/2;1/2)$$ Thus $$Y=\sigma^2\cdot W=Gamma(1/2;1/(2\sigma^2))$$


1

Hint: Write your function as a difference of CDF evaluating at two different points. Then differentiate. Also, you should know that $F'(z)=f(z)$ where $F$ is the CDF and $f$ the PDF. Notes. $$ P(-\sqrt{z}\le X\le \sqrt{z})=P(X\le \sqrt{z})-P(X\le -\sqrt{z})=F(\sqrt{z})-F(-\sqrt{z}) $$ Now use the chain rule.


1

Could you please elaborate last step? How did $\sigma^2$ fit in distribution? Let's have $$X\sim Gamma(n;\theta)$$ (I mean $\theta$ is the rate parameter) $$f_X(x)=\frac{\theta^n}{\Gamma(n)}x^{n-1}e^{-\theta x}$$ Let's set $$y=\sigma^2 x$$ $$x=\frac{y}{\sigma^2}$$ $$x'=\frac{1}{\sigma^2}$$ thus $$f_Y(y)=f_X[g^{-1}(y)]\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|=\frac{\...


1

In normal distribution we use the notation $P(X\le a)=\phi(a)$ $P\bigg(\dfrac{x-\mu}{\sigma} < \dfrac {25-\mu}{\sigma}\bigg)=P\bigg(Z< \dfrac {25-\mu}{\sigma}\bigg)=\phi\bigg(\dfrac{25-\mu}{\sigma}\bigg)=0.1082$ Now look at normal table and find $a$ such that $\phi(a)=0.1082$ then, $a=\dfrac{25-\mu}{\sigma}$ similarly, you have $P(X>42)=.1303$ So ...


1

Let $X = y + \sigma Y$, where $Y \sim N(0,1)$. We know that $k = \mathbb P(X<x) = \mathbb P(Y < \frac{x-y}\sigma)$, therefore $$\mathbb P(X < 2y-x) = \mathbb P(Y < \frac{y-x}\sigma) = 1- \mathbb P(Y \geq \frac{y-x}\sigma) = 1 - \mathbb P(Y \leq \frac{x-y}\sigma) = 1 - k$$


1

The point is that you calcluated the probability that Ali makes exactly one jump over $6$m. It is possible that Ali cleared $6$m on two, or even all three, jumps. So you need to take those possibilities into account too. The easiest way to do this: Ali fails to make the team if and only if he jumps less than $6$m on all three attempts. This occurs with ...


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