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0

You could attempt to control it as follows: Each time you choose a box to put a ball in you sample from the the distribution over boxes $P(B)$ Let $P(B)$ initialise as a uniform probability distribution such that $P(B=b_i) = \frac{M}{KT}$, start T=M Make a sample from the distribution, hence picking a box (lets say $b_x$) Now before repeating step 3, ...


0

First, $$\binom{20}1(0.9)^1(1-0.9)^{19}$$ isn’t what you want: you got the exponents back to front. You want the probability that exactly one ball is flat, which is $$\binom{20}1(0.9)^{19}(1-0.9)^1\approx0.27017\;.$$ Secondly, your arithmetic went astray somewhere, because $$\binom{20}1(0.9)^1(1-0.9)^{19}=1.8\times10^{-18}$$ exactly.


1

Hint: $$\begin{align*} \operatorname{Var}[Y] &= \operatorname{Var}[X_1 + 2X_2 - X_3] \\ &= \operatorname{Var}[X_1] + 4\operatorname{Var}[X_2] + \operatorname{Var}[X_3] \\ &\quad + 4 \operatorname{Cov}[X_1, X_2] - 2 \operatorname{Cov}[X_1, X_3] - 4 \operatorname{Cov}[X_2, X_3] \end{align*}$$


2

The difference between the terms "probability measure" and "probability distribution" is in some ways more of a difference between terms rather than a difference between the things that the terms refer to. It's more about the way the terms are used. A probability distribution or a probability measure is a function assigning probabilities to measurable ...


0

I've tried something like this: ${x_1,x_2,...,x_N}$ is sequence of length N and $x_i$ are uniformly distributed random numbers from interval (a,b) . Let's say that increasing sub-sequence of length L appears at the beginning, i.e. it's a sub-sequence ${x_1,x_2,...,x_L}$. Probability that we will draw a number $x_1$ is 1. Actually, I was thinking about this ...


4

The claim is false, as shown by the following counterexample: let $A$ and $C$ be independent events with $P(A) = P(C) = 1/3$, and let $B$ be the event "$A$ or $C$", which has probability $P(B) = 5/9$. Then $P(B \mid A) = 1 > P(B)$ and $P(C \mid B) = P(C)/P(B) > P(C)$, but $P(C \mid A) = 1/3 = P(C)$. EDIT: An even more concrete example (in the lines of ...


1

For any measure space $(X,\mathcal M,\mu)$, $\mu(\varnothing)=0$ by definition. Since a probability measure is just a special case where $\mu(X)=1$, we still have $\mu(\varnothing)=0$.


2

Hint: Note that $ A\cup \emptyset = A$, thus: $P(A\cup \emptyset) = P(A)+P(\emptyset)+P(A\cap\emptyset) = P(A)$, now $P(A)>0 \implies ?$


1

Hint: $$P(\emptyset)=P(\emptyset\cup\emptyset)=P(\emptyset)+P(\emptyset)$$ This because $\emptyset\cap\emptyset=\emptyset$, i.e. the sets are disjoint.


1

There are several paper for this problem. See for example: On the Distribution of the Product of Independent Beta Random Variables – Applications Products of Beta distributed random variables BETA PRODUCTS WITH COMPLEX PARAMETERS


1

You can have a look at the following paper http://www.dm.fct.unl.pt/sites/www.dm.fct.unl.pt/files/preprints/2012/7_12.pdf which tackles your problem


0

Hint: For $r\in(0,1)$ $$P\left(R<r\right)=\int_{0}^{\infty}P\left(R<r\mid X_{1}=x\right)f_{X_{1}}\left(x\right)dx$$


1

$F^+(t)=P(T\leq t|t>0)$ and $F^-(t)=P(T\leq t|t<0)$ Now, let $X_i=B_iX_i^+-(1-B_i)X_i^-$ where $B_i\sim Ber(0.5)$. Thus, the distribution of $X_i$ is just a mixture of $X^+$ and $X^-$, where the distribution of the two signed random variables are the conditional poisson distributions given $X\geq 0$ and $X<0$. Since each realization is either ...


1

Expected values can be misleading. We have $$ \mathbb E[V_{n+1}|V_{n}] = \dfrac{1}{2} \cdot \dfrac{3}{2} V_n + \dfrac{1}{2}\dfrac{3}{4} V_n = \dfrac{9}{8} V_n $$ so that $\mathbb E[V_n] = (9/8)^n V_0 \to \infty$. However, the Strong Law of Large Numbers is about sums, not products. To make a product into a sum, we take logarithms. $$ V_n = V_0 ...


0

One way to do this is to rescale: let $Y = t X_1$, so $X_0/X_1 > t$ is equivalent to $X_0 > Y$. Now $X_0$ and $Y$ are independent exponential random variables with rate parameters $\alpha$ and $\alpha/t$ respectively. Think of two independent Poisson processes with these rates. One way to realize this is to have a combined Poisson process with rate ...


0

As the comments suggest, $1$ test is equivalent to $x=0$, since none of the plugs would succeed. Likewise, $4$ tests means $x=3$ and $6$ tests means $x=5$. Since you know the probability distribution function, you just want to sum this over values from $x=3$ to $x=5$ (inclusive). This gives the answer that the answer key has.


0

The CLT does hold here: $\frac{S_n-\lambda}{\sqrt{\lambda}}\to \mathcal{N}(0,1) \implies S_n\dot{\sim} \mathcal{N}(\lambda,\lambda)\implies \frac{S_n}{\lambda} \dot{\sim} \mathcal{N}(1,\lambda^{-1})\implies \frac{S_n}{\lambda}\xrightarrow{p} 1$ So, your ratio converges to $1$ in probability; but we can prove more: Using your construction: $Var(S_n)=n ...


1

Each round, you have $1/2$ chance of multiplying your money by $3/2$ and $1/2$ chance of multiplying it by $3/4$. You are correct that the expectation after one round is $9/8$. For any finite number of rounds, your expectation is then $(9/8)^n$. After $n$ rounds you have a distribution of results. If you have won $m$ times, you have $\frac {3^n}{2^n ...


0

In Bayesian inference, you've calculated the posterior distribution of $X$ using a uniform prior, which may or may not lead to a proper posterior distribution...


0

The probability you want is twice the probability where the left pocket empties first, by symmetry. The left pocket empties when there are $r$ matches left over in the right pocket precisely when: The first $2n - r - 1$ times the left pocket is chosen $n-1$ times and the right pocket $n-r$ times. The $2n-r$-th time the left pocket is chosen. So the total ...


0

Some things to note. $\operatorname{Pr}[X_1 = x_1] = q^{x_1 - 1}p$. You had forgotten the "$- 1$". Likewise for $\operatorname{Pr}[X_2 = n - x_2]$. But, more importantly, the sum of geometric random variables gives negative binomial. $$ \operatorname{Pr}[X_1 + X_2 = n] = \sum\limits_{\substack{i + j = n \\ i, j \in \{1, 2, \ldots\}}} \overbrace{q^{i - ...


1

This is a balls and urns problem solvable using a combination of inclusion-exclusion and stars&bars $|a|\leq 10 \Rightarrow -10\leq a \leq 10$. Make a change of variable $x_1 = a+10\Rightarrow 0\leq x_1 \leq 20$ Do similarly to the other letters, so we are at $x_1 + x_2 + x_3 + x_4 = (a+10) + (b+10) + (c+10) + (d+10) = (a+b+c+d)+40 = 18+40 = 58$ ...


1

If you multiplied the integrand by $f_Y(y)$ then you would get the overall probability/density $f_X(x)$. However without that I'm not sure if there's any interesting meaning to the integral. If you integrate with respect to $y$ without taking the distribution of $y$ into account then in some sense you could probably get anything.


10

Assuming that the misprints are independently distributed across the book, and equally likely to be on any page, the number of misprints $X$ on a given page is binomially distributed with parameters $n = 1000$ ($=$ number of misprints) and $p = 1/1000$ ($=$ probability of a misprint ending up on the given page). Thus, the probability of there being at least ...


4

The number of ways 1000 mistakes can happen in 1000 pages is $1000^{1000}$. Assuming all of them are equally likely, we calculate the required probability as follows. The number of ways given page can have at most 2 mistakes is $999^{1000} + 1000(999)^{999}+\frac{1000*999}{2}999^{998}$. Therefore the required probability is $$1-\frac{999^{1000} + ...


7

No, your expression has a number of problems: First of all, even if you assume a binomial distribution applies to your situation, your expression only calculates the probability of exactly three errors, not at least three. Second, the binomial coefficient is incorrect: it should be $\binom{1000}{3}$ if you were to apply such a formula. Third--and this ...


0

Maybe you are looking for a statement like this: The higher the probability of high values of $X$, the more likely it is that the firm defaults. That is, you are comparing probability distributions of $X$. Mathematically, you could use the notion of first order stochastic dominance (FOSD). Suppose you have two different distributions for $X$. Denote the ...


3

The usual model here is a Poisson model: the number $X$ of misprints in a randomly chosen page has roughly Poisson distrobution, mean (and therefore parameter) $\lambda=\frac{1000}{1000}$. We have $\Pr(X\ge 3)=1-\Pr(X\le 2)$. And $\Pr(X\le 2)\approx e^{-1}\frac{1^0}{0!}+e^{-1}\frac{1^1}{1!}+e^{-1}\frac{1^2}{2!}$.


1

$A_n$ is increasing in $n$ because the event $$ \bigg\{ \delta \sum_{i=1}^{n-1} X_i > b \bigg\} $$ is a subset of the event $$ \bigg\{ \delta \sum_{i=1}^n X_i > b \bigg\}, $$ simply because $X_n$ is nonnegative.


1

The expression for the probability is gotten in this way: What is the probability that some specific $j$ elements are all less than or equal to $t$, but that the rest are greater than $t$? That is obviously $$t^j(1-t)^{n-j}$$ Now how many ways can I choose a that specific set of $j$ elements? That answer is $\binom{n}{j}$. So the probaility that exactly ...


0

I was able to solve this explicitly: Given a set of moment constraints $\mathcal{G}_I(a,b,m_I)$, the maximally uniform density function, $S^*(t)$, will take the form: $$S^*_I(t) = \frac{\sum\limits_{i \in I} \lambda_i t^i}{\sqrt{1-\left[\sum\limits_{i \in I} \lambda_i t^i\right]^2}}\mathbf{1}_{a\leq t\leq b}(t)$$ Where the Lagrange multipliers ...


2

Your answer is correct. The probability of the test passing is the probability that all oranges will pass the test, and since the oranges are independent, it is equal to the probability that the first orange will pass the test times the probability that the second orange will pass the test times etc, meaning it is equal to $$(1-\frac15)^{10}\approx 0.1074$$ ...


1

We have $E(X_1+\cdots +X_n)=E(X_1)+\cdots +E(X_n)$, and, by independence, $\text{Var}(X_1+\cdots+X_n)=\text{Var}(X_1)+\cdots +\text{Var}(X_n)$. The $X_i$ have exponential distribution parameter $\lambda=\frac{1}{2}$. It is a standard fact that the mean of such an exponentially distributed random variable is $\frac{1}{\lambda}$, and the variance is ...


2

The following, quoted from the question, is wrong: $$ \Pr[N=n] = \Pr[X_1,\ldots,X_{n-1} \le X_0] = \Pr[X_1 \le X_0] \cdots \Pr[X_{n-1} \le X_0] = F(X_0)^{n-1}. $$ This would be correct if it said $$ \Pr[N=n\mid X_0] = \Pr[X_1,\ldots,X_{n-1} \le X_0\mid X_0] = \Pr[X_1 \le X_0\mid X_0] \cdots \Pr[X_{n-1} \le X_0\mid X_0] = G(X_0)^{n-1} $$ where $G$ is the ...


1

The distribution is discrete. Since the given probabilities add up to $1$, the "missing" entries are all $0$, so are not missing at all. To compute the correlation coefficient, you will need the covariance and the two variances. Let us start. From the table, $E(XY)=(1)(1)(0.25)+(1)(2)(0.25)+(2)(1)(0.25)+(0)(0)(0.25)$. Now let us compute $E(X)$. This can be ...


0

Since $F(x)$ is continuous, $P(X_n=X_m)=0$ for all $m,n$, since otherwise $F(x)$ would need to have a point mass, which means it cannot be continuous. Then since $\{X_n \}$ is i.i.d., $P(X_n>X_0)=\frac{1}{2}$ for all $n$, by symmetry. Therefore $N$ is simply a geometric random variable with parameter $\frac{1}{2}$, and $\mathbb{E}[N]=2$. Note: This ...


1

Let $ X_1 \sim N(23.95,7.44), X_2 \sim N(16.29,7.79) $, then, as mentioned by Dilip Sarwate in the comments, $ X_1 - X_2 \sim N(7.66,{\sqrt{7.44^2 + 7.79^2}}) $ Thus the problem becomes $$\mathbb{P}(X_1<X_2) = \mathbb{P}(X<0),\ \ where \ \ X = X_1 - X_2 $$ This can be calculated directly or reverting to the standard $ N(0,1) $ distribution as ...


1

A more convenient way is to use moment generating functions. MGF of Poisson$:= E[e^S]=e^{\lambda (e^t-1)}$. Define a sequence of random variables $S_n$, where $S_n\sim Poi(n)$, then $S_i=\sum\limits_{j=1}^{i}X_j\;\;X_j\sim Poi(1)$ Also, $S_i-i = \sum\limits_{j=1}^{i}(X_j-1)$ We can now use a convenient property of moment generating functions ...


1

It seems -looking at point 1 and the comments- to me that you didn't quite understand the answer (and comments) to the previous question, even informally. Quick example: we are told that two variables $(X,Y)$ have some joint density $f_{X,Y}$ with support (that is, taking strictly positive values) over a unit circle centered in the origin: $f(x,y) > 0 ...


2

Integral: $$\int_{\mathbb{R}}f(0)e^{px^2}dx=f(0)\int_{\mathbb{R}}e^{px^2}dx$$ is Gaussian integral, see here how to calculate this integral, so: $$\int_{\mathbb{R}}e^{px^2}dx=\sqrt{\frac{\pi}{-p}}$$ So $p=\sqrt{\frac{-p}{\pi}}$.


1

One approach is to observe that $$e^{-x^2}$$ is the "core" (i.e. the part where $x$ appears) of the normal distribution, which has density $$f(x)=\frac{1}{σ\sqrt{2\pi}}\mathrm e^{-(x-μ)^2/(2σ^2)}$$ where $μ$ and $σ$ are the parameters of the distribution. In your case $μ=0$ and $f(0)$ and $p$ should be such that $$\begin{cases}f(0)=\dfrac{1}{σ\sqrt{2\pi}} ...


2

First off, clearly it doesn't matter if we assume the boys and girls are all different or all the same, so we assume all boys are identical, and all girls are identical. We now create a pictorial representation for each arrangement: In this arrangement we place a bar for each boy and a star for each girl. The following is a valid configuration: ...


1

For every $i$, the density $f_X(x_1,\ldots,x_n)$ is an even function of $x_i$ hence, for every $i\ne j$, $(-X_i,X_j)$ is distributed like $(X_i,X_j)$, in particular, $E(-X_i)=E(X_i)$ and $E(-X_iX_j)=E(X_iX_j)$. Thus, $E(X_i)=E(X_iX_j)=0$, which is enough to conclude that $E(X)=0$ and that $\mathrm{Cov}(X,X)$ is diagonal. To go further, note that the ...


1

(a) Distribution of $X_i^m$. Let $I=\{1,2,\ldots,100\}$. For any $x\in I$, \begin{eqnarray*} P(X_i^m=x) &=& P \left(\left[(p_m=x) \cap (p_m^{'} \leq x)\right] \cup \left[(p_m\leq x) \cap (p_m^{'} = x)\right]\right) \\ && \\ &=& P((p_m=x) \cap (p_m^{'} \leq x)) + P((p_m\leq x) \cap (p_m^{'} = x)) - P((p_m=x) \cap (p_m^{'} = x)) \\ ...


2

The variance of this random vector is $\begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix}$. Level sets of the density are ellipses, which are circles only when the correlation $\rho$ is $0$. With a positive-definite symmetric matrix, one can always rotate the two coordinate axes to diagonalize the matrix; then one has two linear combinations of the ...


1

Note the convolution of density functions is the density of the sum: $$g^+ = f \ast f$$ Depending on the given $g^+$ and knowing about the symmetry and support of $f$ you can try to solve the equation $$g^+(s) = \int_{a}^{b} f(t) f(s-t) \ \mathrm dt = (f \ast f)(s)$$


0

\begin{align*} \int_{-\infty}^\infty \int_{-\infty}^\infty g(x,y)dxdy &= \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{f(x+y)}{x+y}dxdy \\ &= \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{f(z)}{z}dzdu \\ &= \int_{-\infty}^\infty du \int_{-\infty}^\infty \frac{f(z)}{z}dz \\ &= \infty \end{align*} as long as $$ ...


0

For computing the density of the sum of more than two i.i.d random variables, it is sometimes easier to work with characteristic functions than with convolutions. The characteristic function of the exponential distribution is given by: \begin{align} \frac{\lambda}{(\lambda - it)} \end{align} For the sum of 3 exponentials, the Characteristic function of ...


0

HINT 2 If $X_i \sim \text{Exp}(λ)$ then the moment generating function of $X_i$ is $M_i(t)=\frac{\lambda}{t-\lambda}$ for $t<\lambda$. Then, by indipendence, the m.g.f of $X_1 + \cdots + X_k $ is $M(t)=M_1(t)\cdots M_k(t)= \left(\frac{\lambda}{t-\lambda}\right)^k$ which is just the m.g.f of a $\text{Gamma}(k, λ)$ with an integer shape parameter $k$. ...


2

If we replace uniforms on $(0,1)$ by uniforms on $(0,w)$, the resulting random variable $Y$ has the same distribution as $wX$, where $X$ has Irwin-Hall distribution. In particular, $\Pr(Y\le y)=\Pr(wX\le y)=\Pr(X\le \frac{y}{w})$. It follows that if $f_X$ is the density function of the Irwin-Hall, then $Y$ has density $f_Y(y)=\frac{1}{w}f_X(y/w)$. In a ...



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