# Tag Info

5

Edit. Your reasoning is almost perfect for the single application case. I would say that $P(J|I) = \frac 12$ if there is equal chance that you do or don't get the job once you are in the interview, thus yielding $P(J) = P(J|I)P(I) + P(J|I^c)P(I^c) = \frac 12 \cdot \frac 12 + 0$ by the law of total probabilities. Anyway, I will stick with the $1/8$ for this ...

5

I think I find this line of reasoning rather too speculative, I'm afraid. :-) First, you assert that $P(I) = 1/2$. I don't see any symmetry that justifies this kind of application of the principle of indifference; it's not as though the only difference between being selected for an interview and not being selected is the opportunity to go to the interview. ...

4

Assuming that the numbers you use are correct ($1/2$ and $1/4$) everything if pretty fine, except the last sentence and the first sentence of your thoughts. Let $A$-is the getting job from the first company, $B$ - from the second (for now just with two jobs). Then you just do this: $$P\{A\cup B\}=P\{A\}+P\{B\}$$ And here the problem rises, you can't just ...

2

Your intuition is correct. That does not sound, and it isn't, right. Let $x$ be the conditional probability that a hair is left at the scene given that somebody else did it. By the law of total probability, the probability that a hair is left at the scene is $$0.99 = 0.99 \times 0.1 + 0.01 \times x$$ So $x=89.1>1$, which cannot be a probability. The ...

2

If $p_i$ is the discrete probability distribution and $x_1$ is the set of outcomes, then the mean $\mu$ and standard deviation $\sigma$ are given by $$\mu =\sum_{i=1}^Nx_ip_i \tag 1$$ and $$\sigma =\sqrt{\sum_{i=1}^N(x_i-\mu)^2p_i} \tag 2$$ Here, we find from $(1)$ that $\mu=4.552$. Then, with $x_i=i$, $p_i$ as given in the table of the OP, and ...

2

The 'E-4' is just notation for the exponent, in this case, to convert to a decimal, the decimal place needs to be moved 4 places to the left. Any blank spaces are to be filled with zeroes. For example: 8.45E-3 = 0.00845

2

It should be $$\int_0^1\int_0^{1-x}(x+y) \, dy \, dx$$ which evaluates to $\frac{1}{3}$. Because the density is $x+y$, regions with high $x,y$ are more likely. So intuition should tell you that $\{x+y\leq 1\}$, the lower triangle of the unit rectangle, should have probability less than that of the upper triangle of the same rectangle. That is, you should ...

2

You are on the right track. Look at the extreme cases for (A) and (B). (A) If $C=\varnothing$, then $P(A\cup B)=0.7$. If $A=B=\varnothing,P(C)=0.7$, then $P(A\cup B)=0$. So $$0\le P(A\cup B)\le0.7$$ (B) If $C=A\cup B$, then $(A \cup B) \cap C^c=\varnothing$ so $P((A \cup B) \cap C^c)=0$. If $C=\varnothing$ , then $P(A\cup B)=0.7$ and $(A \cup B) \cap ... 2 The first equality that you don't understand is indeed a MacLaurin series expansion of the characteristic function, using the link between the moments of a random variable and the derivatives of the characteristic function. See https://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29#Moments . So Shalop is right, the$\sigma$is a ... 2 The cdf of$Y$is $$F_Y(y)=P(Y<y)=P(X^3<y)=P(X<\sqrt[3]y)= \begin{cases} 0,&\text{ if } y<0\\ \int_0^{\sqrt[3]y}\frac{x^2}{9} \ dx&\text{ if } 0\le y \le 27 \\ 1&\text{ otherwise} \end{cases}.$$ Between$0$and$27$$$F_Y(y)=\int_0^{\sqrt[3]y}\frac{x^2}{9} \ dx=\frac1{27}\left[x^3\right]_0^{\sqrt[3]y}=\frac1{27}y.$$ It follows ... 2 $$E[\hat \sigma^2_x]=\frac{1}{n} E \left[ \sum_{i=1}^n (X_i^2-2X_i\mu_x+\mu_x^2) \right] = E[X_1^2]-2\mu_x^2+\mu_x^2=\sigma_x^2$$ 2 No and this is a very common misconception!!! First of all, let's write$H=1$if the alternative is true and$H=0$if the null is true. You are asking whether the p-value is equal to$\Pr[H=0]$. First observation: From a frequentist point of view, your question does not even make sense, because either your hypothesis is true or it is not (i.e.$\Pr[H=0]=0$... 2 Let$W$have exponential distribution with parameter$\lambda$. Let$X=W^{1/3}$. Then if$x\gt 0$, we have$\Pr(X\le x)=\Pr(W^{1/3}\le x)=\Pr(W\le x^3)=1-e^{-\lambda x^3}$. 2 If the true model is that you have homogeneity across the units, i.e. in the models $$y_{it}=X_{it}\beta_i+\alpha_i+u_{it}$$ we have $$\beta_i=\beta$$ then pool mean group estimator i.e. pooling everything and estimate$\beta$is up to a statistically vanishing (with growing$(N,T)$) error the same as estimating each equation separately and the averaging ... 2 We want$20-Y\gt 7$. So we want$20-2X\gt 7$, so we want$X\lt 6.5$. We also want$X\gt 7$. The two conditions are incompatible. 2 Consider a Poisson process of rate$\lambda$. Independently, each occurrence of the Poisson process is "special" with probability$q = 1/(p+1)$. Your$T$is the waiting time until the first special occurrence. You can also consider this from a different point of view: the special and the non-special occurrences form independent Poisson processes with ... 1 Hint: The given information in the problem is filled out in the table below. Try filling in the rest of the table.$\begin{array}{c|c|c||c}&\text{did bring backpack}&\text{did not bring backpack}&\text{total}\\ \hline \text{did bring hat}&\bullet&\bullet&50\\ \hline \text{did not bring hat}&\bullet&40&\bullet\\ \hline ...

1

The probability of rolling a 2 or 6 on a fair die is $1/3$ and the probability of rolling an odd number (so 1, 3 or 5) is $1/2$. Since these two events are independent, $P(A) = 1/2 \cdot 1/3 = 1/6$. In general with problems like these, you can sometimes split up the event $A$ into more manageable parts (divide and conquer) that are independent and then ...

1

It seems to me that you are in danger to misuse Cluster Analysis, which is a technique to discover structure in data and not to describe a structure that one just assumes to exist. To see what I mean, ask yourself the following questions for example: Why are you clustering into 3 groups? Not 4 or 9? When plotting the data (luckily there are 3 attributes ...

1

You can integrate it, it's just that the result is not expressible in elementary functions. You can express it using the error function, or using the standard normal CDF.

1

Not assume, but prove... By definition, $\bar{A}$ and $B$ are independent iff $P(\bar{A}\cap B)=P(\bar{A})\cdot P(B)$. We can express $B$ as $B=(A\cap B)\cup(\bar{A}\cap B)$. (If this is not clear, look at a two-set Venn diagram.) So we can write $P(B)=P(A\cap B)+P(\bar{A}\cap B)$. Then \begin{align*} P(A\cap B)&=P(B)-P(\bar{A}\cap B)\\ ...

1

On this website http://economictheoryblog.com/2015/02/19/ols_estimator/ you can find the proof that the beta estimator of OLS is indeed an unbiased estimator of the true beta in the population. HTH

1

You can say that $$E[X_{n}] = E[X_{n} \cdot I_{\{X_{n}> \varepsilon\}}] + E[X_{n} \cdot I_{\{X_{n} \leq \varepsilon\}}]$$ where $I_{\{X_{n}> \varepsilon\}}$ and $I_{\{X_{n} \leq \varepsilon\}}$ are indicator functions. Note then that $$E[X_{n}] \leq E[1 \cdot I_{\{X_{n} > \varepsilon\}}] + E[\varepsilon \cdot I_{\{X_{n} \leq \varepsilon\}}]$$ ...

1

Basically this equivalence holds true, because in this case convergence in probability (statement 1) is induced by a metric, as previously shown here. In particular this shows that (since in your case $1 \geq X_i \geq 0$): $$X_i \overset{p}{\to} 0 \Leftrightarrow E\left[\frac{X_i}{1+X_i}\right] \to 0$$ Now the left hand site is equivalent to your ...

1

The answer is yes. First, in many cases, the sample is not normally distributed but you can use the central limit theorem to make a normal approximation and obtain an asymptotical confidence interval. But you can also find exact confidence intervals even if the data is not normal. You need to find a pivotal quantity, i.e. a statistic whose distribution ...

1

$E(X)=pE(Y)+(1-p)E(Z)$ $E(X^2)=pE(Y^2)+(1-p)E(Z^2)$ $var(X)=E(X^2)-E(X)^2$ Simplifying we get $$\sigma_Z^2=p\sigma_Y^2+(1-p)\sigma_Z^2+p(1-p)(\mu_Y-\mu_Z)^2$$ As noted by Bernard below, the above represents the decomposition of the total variance into expected conditional variance and variance of conditional expectation (using Bernoulli variable $I$ that ...

1

The cumulative distribution function $$F_{X}(x) = 1-e^{-\lambda x^3}\text{, }x \geq 0$$ implies that $X$ follows a (two-parameter) Weibull distribution. Using the parametrization in the link, the density function is $f\left(x, \dfrac{1}{\sqrt[3]{\lambda}},3\right)$. As Andre has shown, if $Y$ is exponential with mean $\dfrac{1}{\lambda}$, $X = Y^{1/3}$ ...

1

You have used indicators to do: \begin{align} \mathsf E[X] & = \sum_{j=1}^k \mathsf P[X_j{=}1] \\ & = \sum_{j=1}^k (1-\mathsf P[X_j{=}0]) \\ & = k (1-(1-\frac{1}{k})^n)\end{align} So continue: \begin{align}\mathsf E[X^2] & = \sum_{j=1}^k \mathsf P[X_j{=}1] + \mathop{\sum\sum}_{\substack{j\in\{1;k\}\\i\in\{1;n\}\setminus\{j\}}}\mathsf ...

1

Outline: We are assuming that the null hypothesis holds, that is, that the median is $100$. We want to find out how unlikely is a result of $263$ or more test results $\gt 100$, under the hypothesis that the median is $100$. Under that hypothesis, $Y$ has binomial distribution with mean $(482)(1/2)=241$, and variance $(482)(1/4)$. (You are using the wrong ...

1

If $X_1,X_2\sim\operatorname{Exp}(\lambda)$ are independent with densities $f$ and $g$ then the density of $X_1+X_2$ can computed as \begin{align} h_2(t) &= (f\star g)(t)\\ &= \int_0^t f(s)g(t-s)\ \mathsf ds\\ &= \int_0^t \lambda e^{-\lambda s}\lambda e^{-\lambda(t-s)}\ \mathsf ds\\ &= \lambda^2 e^{-\lambda t} \int_0^t \ \mathsf dt\\ &= ...

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