# Tag Info

3

To check if your estimator is consistent, you want to, as you said, compute $\mathbb{E}[T]$. In your case that is $$\mathbb{E}[T] = \mathbb{E}[Y/n] = \frac{1}{n}\mathbb{E}[Y].$$ Now $Y$ is the is number of $i$ that $X_i=1$, or, equivalently, $Y=\sum_{i=1}^n\mathbb{1}_{X_i=1}$, where $\mathbb{1}_E$ denotes the indicator function of an event $E$, i.e. $$... 3 The distributions of numbers of drivers having one or more accidents in each month are Binomial in the case of both the low risk group of drivers and the high risk group of drivers. When N is large and p is small the Binomial can be approximated by a Poisson distribution with same expected value (Np), so total number of drivers having accidents in a month ... 2 In this case Since its not asking you in how many ways can it be ordered to be dealt 3 hearts and 4 other suites the question is asking about Combinations. There are 52 cards in a deck. There are 4 suites in a deck. so 52/4 = 13 cards of each suite. So we need to CHOOSE 3 cards with a heart suite out of the 13 cards with a heart suite availiable to us. ... 2 What you need is the double expectation formula:$$ \DeclareMathOperator{\E}{E} \E(X) = \E \E (X|Y) $$In the double expectation, the inner expectation is a function of Y, that is, a random variable, and the outer expectation then is then taken over the distribution of Y. Assume \E X = \E Y = 0 (we can do that without loss of geneality, since ... 2 If the question is what is the meaning of the function g defined by g(y)=\displaystyle\int f(y\mid x)\mathrm dx, the answer is: None, except in special cases such as X being uniform. When X is uniform on some set U with finite measure 1/c, then f=c\mathbf 1_U hence$$ \int_U f(y\mid x)\mathrm dx=c^{-1}\int_U f(y\mid x)f(x)\mathrm ...

2

Substract $68$ to all numbers in you interval.You get the interval from 1 to 17. therefore there are the same quantity as the numbers between 1 and 17. Clearly there are 17 of these. Assuming each number is equally likely there are 17 favourable cases out of 100 possible cases. So the odds are $\frac{17}{100}=0.17$

2

The Maclaurin expansion idea is good. Note that if we find the expansion of $\frac{1}{1-s+p+ps}$, multiplying this by $p(1+s)$ will not be difficult. Rewrite the bottom as $1+p-s(1-p)$, and then as $(1+p)\left(1-s\frac{1-p}{1+p}\right)$. To make things look nicer, let $a=\frac{1-p}{1+p}$. Then we have $$\frac{1}{1-s+p+ps}=\frac{1}{1+p}\cdot\frac{1}{1-as}.$$ ...

2

The principal reason for the widespread use of mean square deviation instead of mean absolute deviation is that if random variables $X_1,\ldots,X_n$ are independent, then $$\operatorname{var}(X_1+\cdots+X_n)=\operatorname{var}(X_1)+\cdots+\operatorname{var}(X_n).$$ That makes it possible to related the dispersion of the sum, and hence the average, of the ...

2

Notice that faces of higher codimension than $1$ are negligible (they have measure zero), so to generate your random point you just need to: Pick a random integer from $0$ to $k.$ That will tell you which (top dimensional) face of the surface of the simplex you are working with. The top dimensional faces are themselves simplices, so Use the algorithm ...

1

Maximum Likelihood Estimation should give you the parameter values, for the given data set. (of course you need to decide what distribution you're going to use before hand) http://research.microsoft.com/en-us/um/people/minka/papers/minka-gamma.pdf

1

A First Course in Probability by Sheldon Ross covers all of those, with the possible exception of Monte Carlo (it isn't in the 9 chapters i've gone through, and it isn't in the title of any subsequent section). The book is what i'd consider detailed, especially with regards to having many well-elaborated example problems. Additionally, if you ever get ...

1

I full-heartedly recommend Introduction to Mathematical Statistics by Hogg, Craig and McKean, 7th edition. My professors have led to believe that this is the mainstream text for a first course in mathematical statistics. It covers all your desired topics and has strong emphasis on monte carlo simulations via R. It teaches you how to write good R-code and ...

1

The solution basically boils down to: $$P(X=x) = \int_0^\infty \dfrac{e^{-\lambda} \lambda^x}{x!} e^{-\lambda} \, \mathrm{d}\lambda$$ which after a little manipulation evaluates to $$P(X=x) = \dfrac{2^{-(x+1)}\Gamma(x+1)}{x!}=\left(\dfrac{1}{2}\right)^{x+1}$$ since for discrete r.v. (i.e. Poisson), $\Gamma(x+1) = x!$

1

Cowboys wins at 3:0 WWW $p(3:0)=(0.59)^3$ Cowboys wins at 3:1 The first 3 games gives 2:1 (no matter what is the order), and the Cowboys wins the fourth. There are $C_3^1=3$ possible ways that first three games gives 2:1 (WWL, WLW, LWW) with the same probability $(0.59)^2*0.41$. So $p(3:1)=3*(0.59)^2*0.41*0.59$ Cowboys wins at 3:2 The first 4 games ...

1

As mentioned by @Greg Martin, you can use the binomial distribution here. Treat each time they give out a souvenir as a Bernoulli trial, where a success is Davis getting a souvenir (the probability being $p = \frac{1}{4936}$) and a failure is not getting a souvenir (the probability being $1-p = 1-\frac{1}{4936}$). The question asks what the probability of ...

1

Hint : Here is the edited solution: Ist part is correct 2nd Part x1 - time until failure happens with power supply x2 - time until failure happens with screen failure Define x1 be the R.V with pdf = f(x1) = $\lambda_1$*$e^{-\lambda_1*x1}$ Define x2 be the R.V with pdf = f(x2) = $\lambda_2$*$e^{-\lambda_2*x2}$ P(X1<=10000) = $1-e^{-\lambda_1*x1}$ ...

1

(a) correct (b) wrong, first of all here you should use the CDF but not the PDF. Also you cannot simply add them up. PG for power good, SG for screen good. $P(PG)=1-p(PS)$, $P(SG)=1-P(SF)$. BS1300 is not dead, equivalent to both PG and SG. Since PG and SG are independent. So $P(PG\cap SG)=P(PG)P(SG)=...$

1

The intended answer is C. Having the mean above the median is (usually) caused by some very large entries. In a case like this, think about 900 frogs at 4.1 oz plus 100 frogs at 18.1 oz. This distribution has the desired properties. It is far from the only one, but the mean gets pulled by things far from the median. One could argue that a distribution ...

1

For question 1: $A:=\{THH, HHT, HHH\}, B:=\{HTH, THH, HHT, TTT\}, P(A|B)=P(\{THH,HHT\}|\{HTH, THH, HHT, TTT\})=\frac{2}{4} \neq P(A)=3/8$ so (1) is not independent, as A becomes more likely if we know B is true. You just need to find one counterexample to the rule given by MPW above: P(A|B)=P(A) or $P(A\cap B)=P(A)P(B)$

1

Here are some hints: 1) What is the probability of rolling a $6$? Knowing this, what is the probability of not rolling a $6$? If you lettered the dice $A$ through $J$, what is the probability that $A$ through $C$ are $6$ and $D$ through $J$ aren't? How would you count all of the other ways that this can happen? (That is, $A, D, H$ are $6$ but the ...

1

"Window length" is used without comment on the Wikipedia page for moving averages and it's been there for quite a while without being edited out (despite the page getting other edits), so I think that suggests that "window length" is a fine way to refer to this idea.

1

Hint: Note that $X$ is a geometric random variable. $Y=7$ implies that rolls one through to 6 was not a $1$. So we can consider two cases: $X \le 6$ and $X\gt 7$ By definition \begin{align*} E(X \, | \, Y = 7) & = \sum_{k=1}^{\infty} \, k \,P(X = k \, | \, Y = 7)\\ & = E(X \, | \, Y = 7, X\lt 7) \cdot P(X \le 6 \, | \, Y = 7) \\ ... 1 Your answer to number 2 provides the future value of Rs 100,000 deposited semiannually after 25 years at 4% compounded semiannually. You needed to calculate a semi annual deposit amount so that after 25 years you would have Rs 100,000. \begin{align*} 100000 &=d\frac{\left(1+\frac{0.04}{2}\right)^{50}-1}{\frac{0.04}{2}}\\ ...

1

I think you need to revise your CDF from $P\left[ \theta< Y_n<\epsilon+\theta \right]$ to $P\left[ \theta-\epsilon< Y_n<\theta \right]$ as $\theta$ is the maximum value for any Y, which means your CDF, as written, will always equal 0. Now, if you are using the maximum, then the above probability is the same as the complement of the probability ...

1

Hint: The correlation is defined in terms of $\mathrm{Cov}(X,Y)$, $\mathrm{Var}(X)$ and $\mathrm{Var}(Y)$ which can be computed if we know the following quantities $${\rm E}[X],{\rm E}[X^2],{\rm E}[Y],{\rm E}[Y^2],{\rm E}[XY].$$ These should be straightforward to find. For instance, $${\rm E}[X^2]={\rm E}[X]=P(X=1)=a+b.$$

1

If the probability of at least one accident in a calendar month is $0.2$ for a high-risk driver, then the probability of no accident is $0.8$. Thus the expected number of accident-free months is $(0.8)(12)$. Add up over all high risk drivers. We get that the expected total bonuses to high risk drivers is $(5)(600)(0.8)(12)$. There is a similar expression ...

1

In statistics, if you want to make inference about characteristics of the population ($N=5000$) based on small samples ($n=455$), then two things play a role. Sample size, what you are worried about, and variance of the data. Hence, if there is very little variance in the data, then you only need a very small sample size to be confident that the sample ...

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