New answers tagged

1

So, I will explain a solution that does not introduce a binomial random variable. Concretely, there is only one road that will lead him to the mall and three other roads that will not lead him to the mall, but back to the starting position. The problem is asking: what is the probability $A$ that in his first attempt, he chooses one of these three incorrect ...


0

Presumably the traveler won’t choose the same road again on his second try, so this is sampling without replacement. For the second choice to be correct, the first has to be incorrect, so the probability is, assuming that the choices are uniformly distributed, $\frac34\cdot\frac13=\frac14$.


1

Consistency is an asymptotic property, which roughly asks, as our sample becomes large, does our estimator become accurate. Bias on the other hand, is a not an asymptotic property. The bias tells us, given a sample, how off our estimator is in expectation. There are consistent estimators that are biased and unbiased estimators that are not consistent. Based ...


0

A multilinear map is a function which is linear in all arguments, i.e. of the form (for three vector variables in two dimensions) $$m(\mathbf p,\mathbf q,\mathbf r)=\\Ap_xq_xr_x+Bp_yq_xr_x+Cp_xq_yr_x+Dp_yq_yr_x+Ep_xq_xr_y+Fp_yq_xr_y+Gp_xq_yr_y+Hp_yq_yr_y.$$ It is possible to perform linear regression on such a model, but this is uncommon.


0

Assuming you mean Multivariate Linear Regression by multi-linear coefficient: In Multiple Linear Regression you have multiple predictors /independent variables ($x_1,\cdots,x_n$) and only one depentend variable $y$: $$y=\beta_0+\beta_1x_1+\cdots+\beta_nx_n$$ In Multivariate Linear Regression you have multiple predictors ($x_1,\cdots,x_n$) and multiple ...


0

When you calculate the p do not round up, instead it'll just be (55%*24)/100=13.33 which is the correct answer.


1

Your procedure is fine. Alternately, let $A$ be the event $0.9\le X\le 1$, and let $B$ be the event $0.9\le Y\le 1$. It is clear that $\Pr(A)\ne 0$ and $\Pr(B)\ne 0$, but $\Pr(A\cap B)=0$. So $\Pr(A\cap B)\ne \Pr(A)\Pr(B)$, and therefore $X$ and $Y$ are not independent.


1

a) each flavor must be different and the order of flavors is unimportant? $31! / 3!(28)!$ Yes. $^{31}C_3$ or $\binom{31}{3}$ counts the ways to select 3 unique items from 31. b)each flavor must be different and the order of flavors is important? $31! / (28)!$ Likewise, $^{31}P_3$ or $\binom{31}{3}3!$ is the ways to select ...


1

a)if each couple is to sit together? $$4!~2!^4 = 8\cdot 6\cdot 4\cdot 2$$ $\checkmark$ LHS counts ways to arrange 4 couples, then ways to arrange partners in each couple.   RHS counts ways to select a person, their partner, another person, their partner, and so on.   Either way is okay (but see part c). b)if all men sit together? ...


1

a) and b) are correct. For c), we will use Inclusion/Exclusion. There are $8!$ arrangements without restriction. From this we need to subtract the number of bad arrangements, where at least one couple are next to each other. First we count the number of arrangements where Couple A are together. Tie them together with rope. There are then $7$ objects to be ...


0

We have been given: $$f_{X,Y}(x,y) = y^{-1}e^{-y} ~[0\leq x\leq y]$$ So then we know: $$\begin{align}f_Y(y) ~ = ~& \int_0^y y^{-1}e^{-y}~\operatorname d x~~[0\leq y]\\[1ex] =~& e^{-y}~[ 0\leq y]\end{align}$$ Then, by change of variables (chain rule, Jaccobian, etc.): $$\begin{align}f_{XY\mid Y}(z\mid y) ~=~ & \lvert \dfrac{\mathrm d ...


0

If you have a term in a sum which is independent of the index then it's constant and you can simply factor it out of the sum. Do that with the denominator inside the square brackets: \begin{align} \sum_{i=1}^n\left[\frac{(x_i - \bar x)}{\sum_{j=1}^n (x_j - \bar x)^2}\right]^2 &= \sum_{i=1}^n \frac{(x_i - \bar x)^2}{\left(\sum_{j=1}^n (x_j - \bar ...


0

A much bettter estimate of the ratio of the range over the SD is given by the formula 1.897 + 0.6728*ln (N) where N is the sample size. This was derived from simulating 1000 data set in R. Stephen


1

My answer based on the second version of the original post: According to the central limit theorem, the standard deviation of the sample mean of $n$ data from a population is $\sigma_{\overline{X}}=\sigma_X/\sqrt{n}$, where $\sigma_X$ is the population standard deviation. In your case, $\sigma_{\overline{X}}=40/\sqrt{100}=4$. My answer based on the first ...


1

There are $\binom{7}{2}$ equally likely ways to choose $2$ people. There are $\binom{3}{2}$ ways to choose two females, and $\binom{3}{1}\binom{4}{1}$ ways to choose one female and one male, and $\binom{4}{2}$ ways to choose two males. Thus the (simplified) probability of two females is $\frac{1}{7}$, the probability of one of each is $\frac{4}{7}$, and the ...


0

An error in your calculation: you're using the standard error of the mean (which divides by $\sqrt n$) when you should use just the plain standard deviation. You can see that 42% is the wrong answer for the probability of 17 or younger, since only 1/6 of the population is below 18.


0

The question is not stupid at all. The R help files are not very useful on this topic IMHO. The answer is somewhat hidden in the following quote from the R help file for anova.lm(): Normally the F statistic is most appropriate, which compares the mean square for a row to the residual sum of squares for the largest model considered. If something is ...


1

(1) The statistic $T$ has Poisson($n\lambda$) distribution so the expectation of $\delta(T)$ must involve $n\lambda$: $$ E[\delta(T)] = \sum_{t=0}^\infty \delta(t)P(T=t)=\sum_{t=0}^\infty\delta(t)e^{-n\lambda}{(n\lambda)^t\over t!} $$ They then abbreviate $n\lambda$ as $\gamma$ for the sake of saving ink. (2) The equality comes from the line above it (which ...


2

Let $X_n=B(n,p)$ be a binomially distributed random variable. Also notice that $X_n=Y_1+Y_2+\cdots+ Y_n$ where $Y_i$ are i.i.d. Bernoulli with parameter $p$. Now observe that \begin{align} \sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}&= \operatorname{E}(X_n)\\ &= \operatorname{E}( Y_1+Y_2+\cdots Y_n)\\ &=\operatorname{E}( ...


12

We have this sum: $$ \sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} \tag 1 $$ First notice that when $k=0$, the term $k\dfrac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}$ is $0$, and next notice that when $k\ne0$ then $$ \frac k {k!} = \frac 1 {(k-1)!} $$ so that $$ k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} = \frac{n!}{(k-1)!(n-k)!} p^k (1-p)^{n-k}. $$ The two expressions ...


7

The trick is the using the identity $k { n \choose k} = n {n-1 \choose k-1}$. $$\begin{align*} &\sum_{k=1}^n k { n \choose k } p^k (1-p)^{n-k}\\ &= \sum_{k=1}^n n { n-1 \choose k-1} p^k (1-p)^{n-k}\\ &=np \sum_{k=1}^n {n-1 \choose k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}\\ &=np (p+(1-p))^n\\ &=np \end{align*}$$


2

The error is that you have have assumed that the distribution of $\frac{V_1}{V_1+V_2}$ is distributed as $F(5,14)$. This is not the case, as $V_1$ and $V_1+V_2$ are not independent (have a look at the Characterisation section in here) - on the other hand, $\frac{V_1/5}{V_2/9}$ would be distributed as $F(5,9)$. Going back to the problem at hand, we can ...


1

There are $10!$ possible arrangements for the first person, and any of these will do. For the second person to match the first person's order, they must arrange the numbers in the exact order that person 1 did, meaning there is a probability of $1$ in $10!$ that they are in the same order. So yes, you are correct. Mathematically, the most intuitive way to ...


2

For a), what about $x<0$ and $x>1$?. For c), use the rule \begin{align*} \mathbb{P}(A | B ) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}. \end{align*} And for d), you have for all $n \in \mathbb{N}$, so in particular for $n=1,2$ \begin{align*} \mathbb{E}(X^n) = \int_{-\infty}^\infty x^n f(x) \text{d}x = \int_0^1 x^n f(x) \text{d}x. \end{align*}


0

Use the formula $$ Pr({X=x| Y=0 }) = \frac{Pr({X=x \cap Y=0})}{Pr(Y=0)} $$ $$\begin{align} Pr(Y=0) = 20/50 \\ \end{align} $$ $Pr({X=x \cap Y=0})$ is the first row in the table. This results in the probabilities 15/20, 2/20 and 3/20 respectively.


1

The authors' claim can be proved using the Skorohod representation theorem. To prove one direction: if the ratio $R_n:=S_n/V_n$ converges in distribution to a limit $R$, then there exist variables $R_1^*, R_2^*, \ldots$ and $R^*$ with the same distributions as $R_1, R_2,\ldots$ and $R$ respectively such that $R_n^*\to R^*$ almost surely. The identity $$ T_n ...


1

The authors reference a 1969 paper by Efron. The relevant section in Efron appears to be a reference to what at that time was an unpublished paper by Logan, Mallows, Rice and Shepp (see p.16, last paragraph of Efron). However, the article you are reading give the actual paper, published in 1973 :Limit distributions of self-normalized sums. They reference ...


0

Suppose $U_1 = n$. That event is more probable if the $Y$s are smaller than they are in a typical sample. That the $Y$s are smaller than they are in a a typical sample makes it more probable than it would otherwise be that $U_2= n$. Therefore $U_1, U_2$ are not independent.


0

Let event A=The player is using the drug B=The player is not using the drug P=The test result is positive. N=The test result is negative. We have $Pr(A)=0.03, Pr(B)=0.97$ and $Pr(P)+Pr(N)=1$ We also know $Pr(P|A)=0.93$ and $Pr(N|B)=0.98$. (a) By Law of Total Probability, we have ...


0

1) What do you know? The first step is to summarise the information provided using the usual probability symbols. Let $T$ be the event of a positive test; $D$ be the event of using; and $T^\complement,D^\complement$ their complements.   Then we have been told: $$\begin{align}\mathsf P(T\mid D)=0.93 \\ \mathsf P(T^\complement\mid D^\complement)=0.98 ...


0

For a), just use the basic formula $P(A\cup B)+P(A\cap B)=P(A)+P(B)$. For b), just write what the conditional probability $P(B|A)$ is (by formula), use a) and you're done.


2

We see that by inclusion-exclusion, $$P(B \cap A) = P(B) + P(A) - P(A \cup B)$$ $$= .36 + .45 - .55 = .26,$$ So you ar correct about that. Since $P(A) \cdot P(B) = .36 \cdot .45 = .162 \neq .26.$, It is apparent that $P(A) \cdot P(B) \neq P(A \cap B),$ and so $A$ and $B$ are not independent.


1

There is an error in your calculations involving $\hat\mu_j$ and $\hat\mu_i$ (i.e., the last three terms in your expansion). Using a non-clashing index of summation, we have $$ \hat\mu_j=\frac1n\sum_kx_{kj}, $$ so that $${\rm Cov}(x_{li},\hat\mu_j)=\frac1n\sum_k {\rm Cov}(x_{li},x_{kj}).\tag1$$ The terms in the sum (1) are zero when $k\ne l$ (by ...


0

For a, you draw four cards the second time (so $Y_2=4$) unless there is at least one $2$ in the first draw. What is the chance of no $2$'s in the first draw? What is the chance of one $2$ in the first draw? In that case you have $Y_2=3$ and so on. Your result should just be four numbers. As the problem is stated, if you draw four twos on the first draw, ...


0

Here is a simulation using R statistical ssoftware of a million performances of this experiment, where $P(Heads) = .3$ for the biased coin. Results should be accurate to a couple of decimal places. You can use them as a 'reality check' for your work. m = 10^6; x = y = numeric(m) for(i in 1:m) { x[i] = rbinom(1, 2, .5) y[i] = rbinom(1, x[i], .3) } ...


0

First calculate out Cov(X,Y) using $Cov(X,Y)=\sum(X-\mu_X)(Y-\mu_Y)f(X,Y)$ where f(X,Y) is the correspondig pdf. Then use the formula of correlation coefficient: $cor(X,Y)=\frac{Cov(X,Y)}{\mu_X\mu_Y}$You can take a look at this example: https://onlinecourses.science.psu.edu/stat414/book/export/html/94 However, I think you made a few mistake in part a) and ...


1

Let me address your confusion about the argument. The density of the product of two independent random variables is not their convolution of their densities. It is more complicated. And it does not, therefore, correspond to the additive framework after you take the Fourier transform. This is possible, but it requires the Mellin transform.


0

Indeed, one must need the joint distribution. You can make further assumptions by introducing a copula https://en.wikipedia.org/wiki/Copula_(probability_theory)


-1

for analog continous signals, we have time average.Time average is averaged quantity of a single system over a time inetrval directly related to a real experiment. or discrete signals, we have ensemble average. Ensemble average is averaged quantity of a many identical systems at a certain time.


0

The CLT should work very well for the sum of 10 uniform distributions. As a check, here is a brief simulation in R which should approximate the answer to a couple of decimal places--without direct appeal to the CLT. m = 10^6; n = 10; x = runif(m*n) DTA = matrix(x, nrow=m) # each row a sample of 10 uniforms s = rowSums(DTA) mean(s > 7) ## 0.013626 ...


-1

have you tried this? any way I don't know if it works just hope it helps \begin{equation} P(Z-Y< t)= \int P(Z-Y<t \lvert Y=y) P(Y=y) dy = \int P(Z<y+t) P(Y=y) dy \end{equation} If you could find the distribution function of $Z-Y$ you would be able to determine it as a specific random variable. I mean you may want to work with distribution function ...


2

Here is the current edition of the question: If $X$ is a symmetric $n$-dimensional random vector with mean $0$ then is it true that: \begin{align*} & X \text{ follows a multivariate normal law} \\ & \text{iff} \\ & \|X\| \text{is a chi random variable with $n$ degrees of freedom?} \end{align*} Let $X=(X_1,\ldots,X_n)$. As phrased above, ...


1

In general it depends on the string. $P_1$ does not depend on your chosen string. It can be expressed as $1-\left(1-2^{-N}\right)^M$. $2^{-N}$ is the probability of matching each element of $B_1$, so $\left(1-2^{N}\right)^M$ is the probability of not matching any of the $M$ elements of the array. Consider for simplicity the case where $M=N=2$. In this ...


1

It seems likely that your question will be closed. But I think I see some of the issues that are causing you trouble. So I will try to give you some detailed help in case 'Answers' get shut down. Misprint rate for one page. In a Poisson problem, you need to make sure the rate $\lambda$ matches the random variable of the problem. @Henry is right that the ...


1

$\Phi(.)$'s are not uniform on $[0,1]$ if $\mu \neq \beta$. This idea comes from the fact that: $Y=F(X) \sim Unif[0,1]$ if $F$ is a CDF of $X$. In your case, $\Phi(X-\mu)$ is CDF of $Z \sim N(\beta-\mu,1)$. So at least the drift $\mu$ matters in this expectation, that can be interpreted as expectation $E[g(x)]$ of the function $g(x) = ...


2

The error is that the first term should have $z-x$ while the second should have $x-z$, so that the integrand is always positive. Then you get that result using integration by parts. For the first term: $$\int_{-\infty}^z (z-x) \,dF(x) = (z-z) F(z) - \lim_{y \to -\infty} (z-y) F(y) + \int_{-\infty}^z F(x) \,dx.$$ The first term is trivially zero, the second ...


1

HINT: You made a mistake in this step: $$P((y-2)^2>a^2z)=0.9 \implies P\left(3\frac{(y-2)}{3}>a\sqrt{\frac{z}{7}}\sqrt{7}\right)=0.9$$ What you should have found was: $$P((y-2)^2>a^2z)=0.9 \implies P\left(3\left|\frac{(y-2)}{3}\right|>\left|a\right|\sqrt{\frac{z}{7}}\sqrt{7}\right)=0.9$$ If we simplify this and assume $a\geq 0$, then we get: ...


0

The observed data implies $r^2\ge3^2+1^2=10$, so $r\ge\sqrt{10}$. Thus the hypothesis $r\le2$ has already been refuted. The likelihood function for $r$ is now $f(r)=\frac{\sqrt{10}}{r^2}$ for $r\ge\sqrt{10}$ and $0$ otherwise.


0

Since a sum of independent poisson variables (which is going to be my assumption) is poisson distributed, the total number of accidents in a year is poisson distributed with mean $12\times 10$. Fromt the CDF of this distribution, you should find your answer immediately.



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