Tag Info

New answers tagged

0

Let $X_1, X_2, \dots$ denote successive random draws from a Uniform(0,1) parent, where the sample is terminated as soon as $X_1 + X_2 + \dots + X_n > a$ is attained. [To avoid notational confusion, I am using $a$ (rather than $x$) for the target sum.] Importantly, it is given that $0< a< 1$. Then: $P(N>1) = P(X_1 < a) = a$ $P(N > 2) = ...


0

$f(x | \theta)$ is not a proabbility density function; Let's take for example a random variable $X$ with exponential distribution; for example $\lambda = 3$ $f(x) = 3e^{-3 x}$. If you observe a value $x_1$ which you know has been generated according to the density function $f(x)$,you would like to know what was the probability a priori of that particular ...


0

Let Bag 1 contain (5R 20Y) and Bag 2 containn (15R, 10Y) $P(Bag1)*P(Y/Bag1) = 0.6\times\frac{20}{25} = 0.6\times 0.8 = .48$ $P(Bag2)*P(Y/Bag2) = 0.4\times\frac{10}{25} = 0.4\times 0.4 = .16$ Part I : $P(Y) = 0.48+.16 = 0.64$ $P(Bag1/Y) = \dfrac{P(Bag1)*P(Y/Bag1)}{(P(Bag1)*P(Y/Bag1)+P(Bag2)*P(Y/Bag2)}$ Part II: $P(Bag1/Y) = \frac{0.48}{0.64} = 0.75$


1

The skewed normal distribution is defined as follows: Let $\phi(x) := \frac{1}{\sqrt{2\pi}} e^{-\frac 1 2 x^2}$ be the PDF of the standard normal distribution. Let $\Phi(x) := \int_{-\infty}^x \phi(t) dt$ the corressponding CDF. Then we can define the PDF of the skewed normal distribution with a skewnessparameter $a$ as $$f_a(x) = ...


1

This is mechanized in Maple: with(Statistics): X := RandomVariable(Uniform(0, 1)): Y := RandomVariable(Uniform(0, 1)): Z := abs(X-Y)^a: Mean(Z) assuming a > 0; $$2\, \left( {a}^{2}+3\,a+2 \right) ^{-1}. $$


1

You should have something like: $$ E(|X-Y|^a)=\iint_{x>y}(x-y)^a\;dxdy+\iint_{y>x}(y-x)^a\;dxdy=2\iint_{x>y} (x-y)^a\;dxdy $$ Now: $$ \iint_{x>y}(x-y)^a\;dxdy=\int_{y=0}^1\int_{x=y}^1(x-y)^a\;dxdy=\frac{1}{(a+1)(a+2)} $$ So: $$ E(|X-Y|^a)=2\iint_{x>y}(x-y)^a\;dxdy=\frac{2}{(a+1)(a+2)} $$


2

There is no conditional density, since with probability $1/n$, $Y_1 = z$. With probability $1-1/n$, $Y_1$ is uniform on $[0,z]$. So the conditional law is a mixture of an atom at $z$ and a continuous distribution supported uniformly on $[0,z]$.


0

From the onset, due to the $X_n$ being i.i.d. continuous random variables, $X_n{}={}X_{n'}$ for $n \ne n^{'}$ are zero probability events. Consequently, we concentrate on orderings of these random variables. For a given $n$, there are $n!$ exhaustive ways of ordering the random variables $X_1,\ldots,X_n$. By symmetry (from the random variables being ...


0

The pmf of $min(X,Y)$ Given: $X \sim Geometric(p)$ for $x \in \{0,1,2, \dots\}$ with pmf $f(x)$: There are a number of ways to find the pmf of $min(X,Y)$ where $X$ and $Y$ are iid. Since $X$ and $Y$ are iid, the method I prefer is ... think of the problem as generating a sample of size 2 from a single random variable $X$, and then picking the smaller ...


1

You are mixing up indices. $N$ is defined to be the least number such that $X_1+...+X_N>x$, (notice it's $N$ not $n$). The problem is just recycling $n$. In other words the $n$ in the definition for $N$ has nothing to do with the $n$ in $P(N>n)$.


1

Let $X$ be the random variable that takes the values $0,1,...35$ and represents the number of defective circuits. Compute $\mathbb P(X=0)$ given that $\mathbb P(X=x)=\binom{35}{x}p^x(1-p)^{35}$. Why are you taking $1$ as the chance of being defective? The probability that the product operates is the probability that none of the circuits are defective.


4

The $X_{n}$ are independent and absolutely continuous. This gives us the liberty to start with the assumption that $X_{n}\neq X_{m}$ for each pair $\left(n,m\right)$ where $n\neq m$ without any loss of generality. Note that the event $E_n$ can be interpreted as: $X_n$ is the "greatest" of $X_1,\dots,X_n$. The $X_i$ (for $i\in\{1,\dots,n\}$) all have equal ...


1

You've got the right answer. Easier way to solve this is Note that three line segments will be $l/2, x, l-x$ For these to form a triangle, sum of two sides must be greater than the third. So $l/2 + x > l - x$ Rearranging gives $x > l/4~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$ and $l/2 + l -x > x$ Rearranging gives $x < ...


4

Your answer is correct as explained and written. Completing the square gives $$0 = x^2 - 2ax + b = x^2 - 2ax + a^2 + (b-a^2) = (x-a)^2 + (b-a^2),$$ and since no real square is negative, then the equation has no solution if $b-a^2 > 0$; i.e., $b > a^2$. The desired probability is therefore $$\Pr[b > a^2] = 1 - \Pr[b < a^2] = 1 - \int_{u=0}^h ...


0

Don't have time for a full answer, but here's an approach that seems to use the hint: $f_X(x;\theta)= e^{\theta - x}I_{(\theta, \infty)}(x)\implies F_X(x;\theta)=\left(1-e^{\theta-x}\right)I_{(\theta, \infty)}(x):=p_{\theta}\implies e^{\theta-x}I_{(\theta, \infty)}=1-p_{\theta}$ Now take log of both sides: $$\theta-x=\ln(1-p_{\theta})\implies ...


1

The moment generating function of the random variable $\,\,\dfrac{1}{n}\sum\limits_{j=1}^{n}X_j\,\,$ is (for $\lambda<n$) $$ \mathbb{E}\left[e^{\lambda\frac{1}{n}\sum\limits_{j=1}^{n}X_j}\right]{}={}\left(1-\lambda/n\right)^{-n}e^{\theta\lambda}. $$ One way to justify that, as $n\to\infty$, this MGF tends to the MGF for the required degenerate ...


3

If we fix $Z=z$, this implies the linear relationship: $y=zx$, so we can see $Z$ as a random slope of a line through the origin. In fact, each $z$ simply specifies a rotated diameter of the unit circle. Thus, we can re-parametrize this in terms of a random angle: $$\theta \sim U\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$ However, the slope will certainly ...


0

Let $Y=X_1+X_2$. $Y$ has the density: $$g(x)=\int_0^t f(t-x)f(x)dx = \frac{e^{-t}t^3}{6}$$ for $t>0$. Now we want to calculate the final probability which is equal to: $$ \int_{ 0<y<4,x_3+y>4} f(x_3)g(y)d(y,x_3)= \frac{1}{6}\int_{0}^{4} e^{-x}x^3 ( \int_{4-x}^{\infty} e^{-y}y\,dy)$$ which is equal to $\frac{96}{5e^4}$. Could somebody check my ...


3

1) Draw the square $[-1,1]\times [-1,1]$. in the cartesian plane. The value of $X$ is on the $x$ axis, $Y$ on the $Y$ axis. 2) Asking the probability of $Z\leq s$ is equivalent to finding the area under the the line $x-y=s$, bounded by the square. 3) Find this area using simple geometry.


0

Assume that the $\Phi$ function is for some other independent standard normal random variable Y, and simply rewrite the problem like so: $$E\left(\Phi\left( \frac{a-bX}{c}\right)\right)=P(Y<\frac{a-bX}{c})=P\left( Y+\frac{bX}{c}<\frac{a}{c}\right)$$ Now, since $X$ ~ $N(0,1), \frac{bX}{c}$ ~ $N(0,\frac{b^2}{c^2})$, so $Y+\frac{bX}{c}$ ~ ...


0

Thanks for that! I've had a little play and it seems the same as I have produced in the modelling process (if I've misinterpreted please point it out). In the code below you'll see that I am comparing the distributions and performing a likelihood test at the end. The answer for hyp fit comes back as cannot accept null. Which leads to my conclusion that a ...


0

You can simply fit your data using one of the fit function in the ghpy package, demonstrated using random data, here: a_hyp_model<-fit.ghypuv(1/(1+abs(rnorm(100,0,1)))) And then you can use this to generate random observations following your "fitted" distribution (and plot it with hist): hist(rghyp(500,a_hyp_model)) For the other standard distribution ...


1

We always have $X<Y$. So if $x > y$ then $$F(x,y)-F(y,y)=\int_y^x \int_{-\infty}^y f(u,v) du dv = \int_y^x \int_{-\infty}^y 0 du dv = 0.$$ In other words, if $x>y$ then $F(x,y)=F(y,y)$. So that reduces the problem to case 3. In case 3 you just have to calculate the integral: $$\int_{-\infty}^y \int_{-\infty}^{\min \{ v,x \}} 2 e^{-u-v} du dv.$$ ...


1

In $h(p)$ it is defined as a sum that runs from 0 to x. $k{n \choose k } = k\frac{n!}{k! (n-k)!} = \frac{n!}{(k-1)! (n-k)!} = (n-(k-1))\frac{n!}{(k-1)! (n-(k-1))!} = (n-k+1){n \choose k-1}$ Note $(n-k+1) = (n-(k-1))$, as k runs from 1 to n in the sum, k-1 runs from 0 to n-1. So we can replace the (k-1) term by k and change the range our summand runs ...


1

That's a bit too many question, this is discouraged on MSE. Nevertheless: This is CDF, i.e. the expression is $P( Y \leq x)$, which, by the way, doesn't exist in closed form (partial sum of rows of Pascal's triangle), You take the derivative of the product of two functions of $p: p^k$ and $(1-p)^{n-k}$ Do the algebra. You may try logging the expression, ...


1

I will give a very thorough and practical approach to prove the mean, variance and covariance of a simple Wiener process. If we start at some value $W_0$ we can think of the process for $W_t$ as it being completely stochastic, completely deterministic or a mix. When it is stochastic we see each addition in time from $W_{t-1}$ to $W_{t}$ as being caused by an ...


-1

Denote the mean and the variance of $X_i$ by $m_i$ and $\sigma_i$, respectively. Then, if $n$ is large enough then $$X=\sum_{i=1}^{n}\frac{X_i-m_i}{\sigma_i}$$ can be considered as a random variable of standard gaussian distribution, whose cdf be denoted by $F$. Then $$ P(X<X_0)=E[P(X<X_0|X_0)]=\frac{1}{b_0-a_0}\int_{a_0}^{b_0}F(x)dx.$$ Unfortunately ...


0

\begin{align*} &P(X+Y<Z)\\ &=\int_0^4 P(X+Y<z)\cdot \left( \frac{1}{4} \right)dz \\ &=\frac{1}{4} \left[\int_0^1 P(X+Y<z)dz+\int_1^2 P(X+Y<z)dz+\int_2^3 P(X+Y<z)dz+\int_3^4 P(X+Y<z)dz \right]\\ &=\frac{1}{4} \left[\int_0^1 \left(\frac{z^2}{2}\right) dz +\int_1^2 \left(z-1+\frac{1}{2}\right)dz +\int_2^3 ...


0

For a simple method that'll still improve your method you might watch the following video about a generalization of the method of moments (GMM): Generalised Method of Moments Estimation Briefly, you can define the following error function: $$ g_i = b_i(\theta_1,\theta_2) - a_i$$ $$e = \sum_{i=1}^{K} w_i|g_i|^2$$ Where $w_i$ are positive weights. For ...


1

We need the volume of $\{(X,Y,Z):\ Z<X+Y\}$ inside $[0,1]\times[0,2]\times[0,4]$. This is equal to $$\int_{0}^{2}\int_{0}^{1}(x+y)dxdy=\int_{0}^{2}\left(\frac{1}{2}+y\right)dy=3$$ Then divide by $1\times 2\times 4=8$ to get $$P(Z<X+Y)=\frac{3}{8}$$ The reason why this works is that the definition of uniform distribution in an interval is the usual ...


0

S ince doors are opened between 07.00 and 09.00 I'm assuming the arrival time $T$ of a particular owner is uniformly distributed in $[0,115]$. The probability $p_t$ that this owner is present at time $t\in[0,120]$ is then given by $$p_t=\left\{\eqalign{ {\displaystyle{t\over115}}\qquad&(0\leq t\leq 5) \cr {\displaystyle{1\over23}}\qquad&(5\leq ...


0

Let $p$ denote the probability that any given person is present in the car park. The probability that $0$ out of $22$ people are present in the car park is: $$\binom{22}{0}\cdot(p^{0})\cdot(1-p)^{22-0}$$ The probability that $1$ out of $22$ people are present in the car park is: $$\binom{22}{1}\cdot(p^{1})\cdot(1-p)^{22-1}$$ The probability that $2$ ...


0

Assume that the drivers arrive at times uniformly distributed over $[0,120-5]$ minutes. At time $t\in [5,120]$ (no driver can leave before $t=5$) the probability that a given driver has left is $p(t)=\frac{t-5}{115}$. Now the number who have left before $t>5$ is $n(t) \sim B(p(t),22)$, and the number remaining is $r(t)=22-n(t)$


1

What we want to do is obtain a general expression for the (conditional) probability distribution of the random variable $X$ representing the minimum number drawn among $N = n$ numbers chosen from $\{1, 2, \ldots, 10\}$ without replacement. To this end, we make the following critical observation: If the minimum number is $X = x$, then the remaining $n-1$ ...


0

Let us consider what $f(x,y) = P(X=x, Y=y)$ is. Note that for $y>x$ you get $f(x,y)=0$ since it is impossible to have drawn a larger card than exists in the pile. It should follow quickly that among the remaining values for $y$ given a specific $x$ that the distribution would be uniform. $$f(x,y) = \begin{cases} 0\text{ for }y> x\\ \frac{1}{N}\cdot ...


1

Use Conditional Probability and the Law of Total Probability to determine that: $\begin{align} \mathsf P(X=x\mid Y=y) & = \frac{\mathsf P(X=x)\,\mathsf P(Y=y\mid X=x)}{\sum_{k=1}^{N} \mathsf P(Y=y\mid X=k)\,\mathsf P(X=k)}\; \mathbf I_{y\in \{1;x\}, x\in\{y;N\}} \\ & = \frac{\frac 1 N \cdot \frac 1 x}{\frac 1 N \sum_{k=y}^N \frac 1 {k}}\; \mathbf ...


1

A number of observations: (i) As functions of independent random variables, $\dfrac{X_n}{\sqrt{n}}\,\,$ is independent of $\,\,\displaystyle \sum\limits_{m=1}^{n-1}{X_m^2}\,$. (ii) $\,\,\dfrac{X_n}{\sqrt{n}}\sim\mathcal{N}\left(0, 1/n\right)\,\,$ and $\,\,\displaystyle \sum\limits_{m=1}^{n-1}{X_m^2}\sim\chi^2(n-1)\,$, since they are linear combinations of ...


1

$\newcommand{\var}{\operatorname{var}}\newcommand{\cov}{\operatorname{cov}}$You have not specified that the pair $\begin{bmatrix} X_1 \\ X_2 \end{bmatrix}$ is normally distributed, and that does not follow from the fact that the two components separately are normally distributed, nor have you said what the covariance between $X_1$ and $X_2$ is. You say $X_1 ...


0

If $0<y<x$ then \begin{align} F(x,y) = \Pr(X\le x\ \&\ Y\le y) & = \Pr\left(\Big(X\le y\ \&\ Y\le y\Big)\text{ or }\Big(y<X\le x\ \&\ Y\le y\Big)\right) \\[10pt] & = \Pr(X\le y\ \&\ Y\le y)+\underbrace{\Pr(y<X\le x\ \&\ Y\le y)}_{\text{This part is $0$.}} \end{align} The c.d.f. is perfectly well defined in the case ...


0

Where you can start : You have that : $Y = f(X)$ and you want the pdf of $X+Y$. Let $\phi : \mathbb{R}^2 \to \mathbb{R}$ be a smooth bounded function. Compute $\mathbb{E}[\phi(X+Y,X)]$. That should give you the joint pdf of $(X+Y,X)$. Then you should know how to get the pdf of $X+Y$.


1

Let $Y=2X$. Then, \begin{align*} G(t)&=E[t^{Y}]=\sum_{y=0}^{\infty}t^{y}P(Y=y) =\sum_{y=0}^{\infty}t^{y}P(2X=y) =\sum_{y=0}^{\infty}t^{y}P(X=y/2)\\ &=P(X=0)+t^{1}P(X=1/2)+t^{2}P(X=1)+t^{3}P(X=3/2)+t^{4}P(X=2)+\ldots \\ &=P(X=0)+0+t^{2}P(X=1)+0+t^{4}P(X=2)+\ldots \\ &=\sum_{x=0}^{\infty} t^{2x}P(X=x)+\ldots \\ ...


1

It isn't . See Renewal Theory. If you are talking about the time of the next event $T_{N+1}$ given the time of the last event $T_N$, then the Markov Property is that $P(T_{N+1}|T_N,T_{N-1}...)=P(T_{N+1}|T_{N})$. Nothing precludes the interarrival times from being any distribution, provided that inter-arriveal times are iid. Note also that the markov ...


0

First, you need to decide where you most care about normality...is it the tails of your distribution or the body? If your answer is "both/don't know", then a simple measure of fit is the Kolmorogov-Smirnov Statistic. It gives the worst-case bound on how wrong you could be if you used a normal distribution for probability calculations. If this statistic is ...


0

There is a 1-1 correspondence between the values of $X$ and the values of $2X$. You are correct in saying that $2X$ only assumes values that are twice the values that $X$ assumes. So,if the possible values of $X$ are $1,2,3,\ldots$, then the possible values of $2X$ are $2,4,6,\ldots$. The PDF (probability density function) of $2X$ looks exactly like that of ...


1

For a discrete distribution, $$P(2X=n)=P(X=n/2),$$ so it will be $0$ for odd $n$ and for $n=2k$ you get $P(2X=n)=P(X=k)$. For a continuous distribution, $$P(|2X-x|<\epsilon)=P(|X-x/2|<\epsilon/2),$$ so if the probability distribution function of $X$ is $f_X$, the probability distribution function of $2X$ is $$f_{2X}(x)=\frac12f_X\left(\frac ...


3

Correlation may have many meanings, but from the question, you are using the specific definition of the Pearson product-moment correlation coefficient. You are calling variables "correlated" when $\rho \neq 0$. That is solely when $\textrm{Cov}(X, Y) \neq 0$. In the case of $Y = aX$, regardless of how $X$ is distributed, we can state the following: $$ ...


1

One approach to intuition is, since Y is always positive, the positiveness of X has no impact on the positiveness of Y. Therefore, X being more positive has as much impact on Y being larger than X being more negative. This is the antithesis of high correlation. I think that is all there is to it.


1

I can't help you with the first case. What is your problem here? But to the second one: If two random variables are uncorrelated (i.e. covariance is zero), they are not necessarily independent. The example you have is the standard example to demonstrate this. Only if $X$ and $Y$ have a joint bivariate normal distribution, from $cov(X,Y)=0$ follows that ...


2

$Var(X)=E[X^2]$ is the variance (measures spread around mean), which for a standard normal distribution is $1$. $E[X^3]$ measures the skewness of the density and since a normal distribution is symmetric the skewness is $0$.



Top 50 recent answers are included