New answers tagged probability-distributions
0
I'm not sure I fully understand the question.
You have the CDF $F(\alpha) = P \{ \omega | X(\omega) \in (-\infty,\alpha] \}$, and since $(-\infty, a] \cup (a,b] = (-\infty, b]$, we have $ P \{ \omega | X(\omega) \in (-\infty,a] \} + P \{ \omega | X(\omega) \in (a,b] \} = P \{ \omega | X(\omega) \in (-\infty,b] \}$, from which we get $P \{ \omega | X(\omega) ...
2
Both are approximations, though the second is a very good one. The actual percentage of the area under the normal curve that is within two standard deviations of the mean is about $95.44\%$; the cutoff within which you get $95\%$ of the area is very close to $1.96$ standard deviations on each side of the mean.
The rule of thumb that says that about $68\%$ ...
2
1.96 is equal to 2 for many practical purposes.
1
Note that $E[X\mid\Lambda]=\Lambda$ hence $E[X]=E[E[X\mid\Lambda]]=E[\Lambda]$. If $\alpha$ denotes the mean of the exponential distribution, then $E[\Lambda]=\alpha$ hence $E[X]=\alpha$. If $\alpha$ denotes the rate of the exponential distribution, then $E[\Lambda]=1/\alpha$ hence $E[X]=1/\alpha$.
1
Let random variable $X$ be the distance the ambulance has to travel. The time $T$ is then $\frac{X}{60}$. We will find the distribution of $X$. From that it is not hard to find the distribution of $T$. We do $X$ instead of $T$ for two reasons: (i) It is closer to the intuition and (ii) There might as well be something left for you to do.
We will find the ...
0
Yes, you should check that
$$\int_0^1\int_0^1(x+y)\,dydx=1\,.$$
By definition, $F_X(t)=P(X<t)$, and this can be expressed by the integral:
$$F_X(t)=\int_0^t\int_0^1 (x+y)\,dydx\,.$$
1
This is just an application of the definition of the third central moment. What's the mean of a Bernoulli variable? Subtract that from the two possible values, take the third powers, multiply by the probabilities, add, factor out $p(1-p)$, and you're done.
1
By the definition, any covariance matrix $\Sigma$ consists of $\Sigma_{i,j} = Cov(X_i, X_j) = E[(X_i - \mu_i)(X_j - \mu_j)]$, where randoms $X_i$ and $X_j$ have expectations $E[X_i] = \mu_i$ and $E[X_j] = \mu_j$ accordingly. So your problem is to prove $Cov(m_i + \mathbf{b}_i \mathbf{X}, m_j + \mathbf{b}_j \mathbf{X}) = (B \cdot \Sigma \cdot B^\top)_{i,j}$, ...
1
First we calculate the probability of : $P(X_i=k | X=m)=\binom mk (\frac 1n)^k(1-\frac 1n)^{m-k}$ . This is true because we pick k out of m balls to put in the $i$'th bin, so k balls fall into that bin with probability $(\frac 1n)$ and the rest of the balls are are not in it with probability $(1-\frac 1n)^{m-k}$.
Now using the law of total probability:
$ ...
0
There are some general methods, but at that stage in the book I think you're supposed to work it out manually.
Let $p_i$ be the probability of reaching square nine before square one given that I'm on square $i$.
I need to come up with an equation involving $p_5$ that I can solve.
From square five I can reach only squares four five and seven before I reach ...
0
Hint: For $0\leq n\leq N$ : $P(X_{N}=n)$ is the probability that
exactly $n$ people got their hat back.
In how many ways this can be done ?
If want to fix some $n$ hats
in their place (in how many ways can we choose them ?) and dearrange
the other $N-n$ hats (in how many ways can you do that ?)
0
You need to try and reverse the steps in your previous answer.
The important point is that each ball is goes independently and uniformly (equally as likely to go in any of the buckets)
So for a sequence $X_1,\dots, X_n$, with $\sum_{i=1}^n X_n = X$ you can write down the probability that I get $X$ from a Poisson distribution and the probability that when ...
0
MGF's are easy and by far the most convenient way to solve your problem.
$$
\begin{align}
\psi_X(u) &= \mathbb{E} e^{uX}\\
&=\sum_{k=0}^{\infty} e^{uk} \frac{\lambda^k}{k !} e^{-\lambda}\\
&=e^{-\lambda} \sum_{k=0}^{\infty} \frac{(\lambda e^{u})^k}{k !}\\
&=\exp\{ \lambda(e^u -1) \}
\end{align}
$$
Now, since you know that $X_i$'s are ...
0
By using the following (which can be computed using standard integrals, integration by parts and this limit)
$$\int^\infty_{a+h}\frac{1}{(x-a)^{\alpha+1}}dx=\left.-\frac{1}{\alpha(x-a)^\alpha}\right|^\infty_{a+h}=\frac{1}{\alpha h^\alpha},$$
...
0
It does indeed depend on the distribution and number of hits.
For example, suppose everyone always logged on at the same moment. Then the peak doubles.
Now suppose that the distribution is completely random throughout the day, and that you go from 1 user to 2 users. The chance that they log on at the same time is 1/86400. Therefore the peak will almost ...
1
I think that you're attacking this from the wrong angle.
First, your idea is fraught with technical difficulties. You didn't mention what kind of space you envision this space of PDFs to be. You could start with just the set, and simply define a $\sigma$-algebra on its powerset. But there's a myriad of ways to do that, which one do you pick?
Or you could ...
2
Yes. One possible definition of multivariate normal vector is that every linear combination of its components is (univariate) normally distributed. Since $X$ is multivariate, every linear combination of $(X_1, ..., X_n)$ is normally distributed. But every linear combination of components in $Z$ is also a linear combination of $(X_1, ..., X_n)$ because ...
3
What you are describing is called a Poisson Binomial Distribution (PBD). In what sense do you want to estimate it?
If you are looking to some approximate learning of the distribution, based on samples drawn from it, you can have a look at this paper. It gives a (randomized) algorithm which, given access to i.i.d. draws from a PBD $p$ and a parameter ...
1
I'm not sure exactly what you're trying to ask here. You'll probably be able to find this answer in a textbook somewhere. If you were asking something else it would be helpful if you gave a little more detail in your question.
Suppose $p_i< 2^{-k}$. If I toss a coin $k$ times I have $2^k$ possible events each of probability $2^{-k}$. So I cannot ...
0
You can use Borel's Zero-One Law (prop. 2.2 here), using $$\mathbb{E} X_1 \leqslant \sum_{n=0}^\infty \mathbb{P}\{X_1 > n\} = \sum_{n=0}^\infty \mathbb{P}\{X_n > n\}$$ with $A_n\stackrel{\rm{}def}{=}\{X_n > n\}$.
0
Joint density is $f(x,y) = \frac{1}{2} (x+y) e^{-(x+y)},\, x,y \geq 0$ and the distribution function is
$F_Z(z) = P(Z < z) = P (X+Y < z) = P ( X < z, Y < X -z)$
then we have for the density of Z
\begin{array}[rl]{l}
f_{Z}(z) &= \frac{\partial}{\partial x} F_z(z) = \frac{\partial}{\partial x} \int\limits_{0}^{z}\int\limits_{0}^{x-z} ...
0
Sampling $m$ elements simply means obtaining $m$ values from $X_i$, call them $x_i$, according to the distribution $D$, exactly as for the discrete case. If you prefer, think about it as choosing a number which has the correct probability of being in any interval.
If your observation $x_i$ of $X_i$ satisfies $x_i>z$, then with probability one it ...
1
Your answer $\left(\frac{1}{20}\right)^{17}$ makes sense to me, given your description of the problem.
I recognize the pattern of the teacher's answer, and that is an answer to a different question. There are $\binom{20+17-1}{17}$ ways to distribute $17$ indistinguishable balls into 20 urns. If all such distributions were equally likely, then that would ...
1
I think you may have some confusion (as do we) about what you mean by the first basket.
If you mean basket number 1 (or 7 or 12 or any other particular number) then your formula will give the correct result $\left(\frac{1}{20}\right)^{17}$.
If you mean the basket that the first ball drops into then you are interested only that the remaining 16 balls fall ...
1
The key is that the balls are indistinguishable.
Let us consider an example, where we can work out explicitly. Consider $2$ balls and $3$ baskets.
If the balls are distinguishable, say colored red and blue, then the first ball can go into any of the three baskets, while the second ball can also go into any of the three baskets. This thereby gives us a ...
0
This is essentially a reworking of Did's answer. Fix $i=1$, $q = P\{X > 1\}$
and note that
$$P\{X = s+1\mid X > 1\} = \frac{P\{X = s+1, X > 1\}}{P\{X > 1\}}
= \frac{P\{X = s+1\}}{q} = P\{X=s\}$$
which gives
$$\begin{align}
P\{X=2\} &= qP\{X=1\}\\
P\{X=3\} &=q P\{X=2\} = q^2P\{X=1\}\\
P\{X=4\} &=q P\{X=3\} = q^3P\{X=1\}\\
...
1
First, observe that
$$
\mathbb{P}\{X>t\}=\sum_{n > t}\mathbb{P}\{X=n\}
$$
so one can show that the memorylessness property implies ($s,t>0$)
$$
\mathbb{P}[X>s+t| X > t]=\mathbb{P}\{X> s\}
$$
Note also that since $s>0$
$$
\mathbb{P}[X>s+t| X > t\} = \frac{\mathbb{P}\{X>s+t\}}{\mathbb{P}\{X > t\}}
$$
Define $f\colon ...
2
Let $p=P[X\geqslant2]$, then $P[X\geqslant i+1\mid X\geqslant i]=p$ for every $i\geqslant1$ hence $P[X\geqslant i]=p^{i-1}$ for every $i\geqslant1$. Surely you can deduce the distribution of $X$ from this observation.
0
Consider the binomial random variable:
$P(X=i)=\frac{n!}{(n-i)!i!}p^i(1-p)^{n-i}$
If you let $n\rightarrow \infty$ and $p \rightarrow 0$ for fix $np$ and consider $p=\frac{\mu}{n}$
Use the following identities:
$\lim_{n \to \infty} (1-\frac{\mu}{n})^n= e^{-\mu}$ and $\frac{n!}{(n-i)!} \approx n^i$ and you arrive at the Poisson random variable:
...
2
If $p$ is uniform on the simplex $\Delta_n=\{(p_i)_{1\leqslant i\leqslant n}\mid p_i\geqslant0,p_1+\cdots+p_n=1\}$, then each (continuous, for every $n\geqslant2$) random variable $p_i$ has density $(n-1)(1-x)^{n-2}\mathbf 1_{0\leqslant x\leqslant 1}$ with respect to the Lebesgue measure $\mathrm dx$. Hence
$$
\int_{\Delta_n}\sum_{i=1}^np_if(p_i)\mathrm ...
0
I assume the OP is referring to this http://en.wikipedia.org/wiki/Poisson_limit_theorem for $n, p$.
$p$ is probability that a person makes a phone call during a particular 10-second interval, which is very small. $n$ is the number of people in the city. The large city assumption guarantees $n$ is large, so this can be modeled by $Poisson(np)$.
0
This is not quite a definition, just an illustration of a quantity, well modeled by a Poisson random variable. I am not sure what $n,p$ in your question are, but I think that you need a large city because if you have a small population sample, they don't call each other according to the same frequency distribution...
0
A Poisson r.v. $N$ follows a Poisson distribution, which has only one parameter $\lambda$; and takes values in $\mathbb{N}$ — what do you refer to with $n$ and $p$? (are you thinking instead of a binomial distribution $\operatorname{Bin}(n,p)$?)
I assume the "large city" assumption is just so that you don't have to worry about the number of calls being too ...
1
Since $E\left[\sum{\dfrac{x}{a}}\right] = \dfrac{\sum{E[x]}}{a}$, you have that
$$
E\left[\frac{X_1+X_2 + ...+X_n}{n^3}\right] = \dfrac{n*E[X]}{n^3} = \dfrac{n*(-1/2*\alpha + 1/2*(1-\alpha))}{n^3}=\dfrac{n*(1/2-\alpha)}{n^3}=\dfrac{1/2-\alpha}{n^2}
$$
The rest is simple.
1
I'd say
$\displaystyle\left|\sum_{k=0}^n X_i\right|<n/2$, hence $\dfrac{E[\sum_{k=0}^n X_i]}{3^n}\to 0$ quite fast.
2
I am not sure I can give a reference for this particular question, but it can be done easily enough.
If the two Brownian motions are independent you can find the probability distribution for $Y_t$ conditioned on a path of $X_t$. You can write
$$ \log Y_t = \log Y_0 - \int_0^t X_t dW^Y_t - \frac12 \int_0^t X_t^2\,dt. $$
The point is that $\int_0^t X_t^2\,dt ...
1
This is a magic property of Poisson processes.
Let $\mu = 2$ and $\lambda = 3$ be the rates of your two machines
Now I want to know the probability that I get nothing out of either machine in the first $t$ minutes.
I get nothing out of machine $A$ with probaiblity $e^{-\mu t}$ and nothing out of machine $B$ with probability $e^{-\lambda t}$.
So as the ...
0
As explained in comments, neither $f_X$ nor $f_Y$ is a PDF in general. For example, using the change of variable $t=\mathrm e^{-x/\alpha}$ in $f_X$ yields $\mathrm dt=-t\mathrm dx/\alpha$, hence
$$
\int_0^{+\infty}f_X(x)\mathrm dx=\int_0^1(1-t)^{\tilde{r}-1} t^{K-\tilde{r}}\mathrm dt=\mathrm{Beta}(K-\tilde{r}+1,\tilde{r}).
$$
For each $\bar r\gt0$, there is ...
0
Actually, the weighted exponential random variable is still an exponential random variable. Only the mean is scaled with the same weight. Therefore, the sum $Z$ is an Hypo-exponential random variable and it is summarized in wiki's webpage Hypoexponential.
1
$P\left( n,\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)T \right)$ Disaggregating Tail of Poisson
If $X$ is Poisson with mean $a+b$, a way to exhibit $X_a$ and $X_b$ independent and Poisson with respective means $a$ and $b$ such that $X=X_a+X_b$ is to consider an i.i.d. sequence $(U_n)_{n\geqslant1}$ of Bernoulli random variables with $P[U_n=1]=a/(a+b)$ and $P[U_n=0]=b/(a+b)$ for every $n\geqslant1$, independent of $X$, and to define
$$
...
0
$P\left( n,\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)T \right)$ Disaggregating Tail of Poisson
If I understand your question, yes you can. It is a well known result that the sum of two independent Poisson random variables is another Poisson random variable whose parameter is the sum of the constituent parameters. One can show this in a straight forward way, but it is easier to do so via mgfs:
If $X_1 \sim $Poisson$(\lambda_1)$ and $X_2 \sim ...
0
We prove the first implication.
Let's take $a>0$ such that $\sum_{n \in N } P(X>an) < \infty$.
First of all, $P(X>an)=\sum_{k=n}^{\infty}P(X\in(an,a(n+1)])$. Hence, we write
$\infty>\sum_{n \in N } P(X>an) = \sum_{n \in N } \sum_{k=n}^{\infty} P(X\in(ak,a(k+1)])$. By Tonelli's theorem we can change the order of summation. We arrive at
...
1
Not a stupid question at all!
For a Poisson process, if one and only one event occurs in the interval between 0 and $t$, then the timing of when the event occurs is uniform between 0 and t. The total number of occurrences $N(t)$ is a Poisson random variable. It's when we "zoom in" and look at a single occurrence that we observe a uniform distribution (or an ...
1
Try to integrate the pointwise inequality $X\leqslant a\sum\limits_{n=0}^{+\infty}\mathbf 1_{X\gt an}$ to get one implication. For the other implication, try to find a similar pointwise inequality, only reversed.
Edit: Consider $a\sum\limits_{n=1}^{+\infty}\mathbf 1_{X\gt an}$ (note that the sum is starting at $n=1$, not $n=0$), can you compare it to $X$?
0
Here's a hint: for $X>0$,
$$E[X] = \int_0^\infty x f(x) \mathrm d x = \sum_{n=0}^\infty \int_{na}^{(n+1)a} x f(x) \mathrm d x$$
Can you bound these integrals in terms of integrals of $f$? What can you do with the factors?
2
Let $t_i$ denote the expected number of children to be born for 3 of the same sex to be born consecutively, after $i$ children of the same sex were born. Thus, $t_3=0$ and one is looking for $t_0$. Conditioning on the sex of the next child, one gets $t_0=1+t_1$, $t_1=1+\frac12t_1+\frac12t_2$ and $t_2=1+\frac12t_1$. Hence $t_0=7$.
For example, assume the ...
3
They have a kid, cost $1$. We will add this at the end.
After the first kid, the parents are in one of two states: State A if the two preceding children are of the same sex (so they are almost there); State B is there is a previous child, but they are not in State A.
Let $a$ be the expected number of additional children if they are in State A, and $b$ the ...
1
Hint: Try
$$
Z=UX+(1-U)Y
$$
where $X\sim F$, $Y\sim H$ and $U\sim \mathrm{bin}(1,p)$ is a binomial distributed random variable such that $X,Y$ and $U$ are independent
0
If you want to show that two random variables follow the same distribution, there's not just one procedure to do that. This is because there are several quantities that uniquely determine the distribution of a random variable.
For instance, if $X$ is a random variable, then the cumulative distribution function $$F(x)=P(X\leq x),\quad x\in\mathbb{R},$$ ...
1
I think Alice can always assure herself at least $1 \over 4$ cdf, in the following way.
First, in each of the 4 corners, mark a square that contains $1 \over 4$ cdf. Since the pdf is finite, it is always possible to construct such a square, by starting from the corner and increasing the square gradually, until it contains exactly $1 \over 4$ cdf.
There is ...
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