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4

Just do some in-your-head approximations, no table-lookups: $95\,\%$ is about $2\sigma$, here $\sigma=\sqrt{npq}=5$, so anything between $40$ and $60$ heads will not raise your suspicion at this confidence level


4

The simplest way is probably using the CDF, which is known in this case. Let $X \sim Exp(1)$ and $Y = \sqrt{X}.$ The CDF of $X$ is $F_X(x) = 1 - e^{-x}$, for $x > 0.$ Then $$F_Y(y) = P(Y \le y) = P(\sqrt{X} \le y) = P(X \le y^2) = 1 - e^{-y^2},$$ for $y > 0.$ Then take the derivative of the CDF of $Y$ to get $F_Y^\prime(y) = f_Y(y) = 2ye^{-y^2},$ ...


3

I have no trouble understanding $E(X)=\int xf(x)dx $ and $E(Y)=\int y f(y)dy$ Actually, your formulas should read \begin{align} E[X] &= \int_{-\infty}^\infty x f_X(x)\,\mathrm dx = \int_{-\infty}^\infty t f_X(t)\,\mathrm dt \tag{1}\\ E[Y] &= \int_{-\infty}^\infty y f_Y(y)\,\mathrm dy = \int_{-\infty}^\infty u f_Y(u)\,\mathrm du\tag{2} ...


3

Want to find $$ P(X \leq x \mid U \leq \mathrm e^{-X}) = \frac{P(X \leq x \cap U \leq \mathrm e^{-X})}{P(U \leq \mathrm e^{-X})}. $$ Note the joint density factors by independence: $f_{X,U}(x,u) = f_X(x)f_U(u) = \lambda \mathrm e^{-\lambda x} \cdot 1 = \mathrm e^{-x}$ with $\lambda = 1$. Numerator: $$ P(X \leq x \cap U \leq \mathrm e^{-X}) = \int_0^x ...


3

The median for a random variable $X$ is $m$ such that $P(X \le m) \ge 1/2$ and $P(X \ge m) \ge 1/2$. In the first example the correct answer is $0$: $P(X \le 0) = P(X = 0) = 0.728303$ and $P(X \ge 0) = 1$. In the second example it is $2$: $P(X \le 2) = 0.10 + 0.20 + 0.30 = 0.6$, $P(X \ge 2) = 0.30 + 0.25 + 0.15 = 0.7$. Your method is completely wrong.


2

As $X$ and $Y$ are independant, $X-Y\sim\mathcal{N}(0,1)$, so $V(|X-Y|)=V(|Z|)$ where $Z\sim\mathcal{N}(0,1)$. $V(|Z|) = E(|Z|^2)-E(|Z|)^2 = E(Z^2)-E(|Z|)^2$. $E(Z^2)=V(Z)=1$. Now all you have to do is finding $E(|Z|)$ where $Z\sim\mathcal{N}(0,1)$ (calculate the corresponding integral for example).


2

$$\text{Boy, High income} =4\\ \text{girl, High income} = 6\\ \text{Boy, low income} = 6\\ \text{Girl, low income} = x$$ $$ P(\text{Male - M}) = \frac{10}{16+ x}\\ P(\text{High income - H}) = \frac{10}{16+ x}\\ P(\text{Male High income}- MH) =\frac{4}{16+x} $$ Independence means that $P(M \text{and} H) = P(M) P(H) $ $$\frac{10}{16+x}\frac{10}{16+x} = ...


2

The formula for the variance of 'a random sum of random variables' is given in many probability texts. It is derived by conditioning as suggested by @Augustin. If $X_i, \dots, X_N$ are iid and $N$ is independent of the $X$'s, then the sum $S$ has variance $$V(S) = E(N)V(X) + V(N)[E(X)]^2.$$ Roughly speaking, the second term expresses the additional ...


2

Please do look at the discrete example in Wikipedia; here is a little more in direct answer to your question. Consider tossing two fair dice to get realizations of independent random variables $X$ and $Y$. Most games use the sum of the numbers $X + Y$. It is easy to see that $E(X) = E(Y) = 3.5$ and $$E(X+Y) = E(X) + E(Y) = 3.5 + 3.5 = 7.$$ However, it ...


2

The marginal (unconditional) distribution of $X$ under this hierarchical model is $$\begin{align*} \Pr[X = k] &= \sum_{n = k}^\infty \Pr[X = k \mid N = n]\Pr[N = n] \\ &= \sum_{n=k}^\infty \binom{n}{k} p^k (1-p)^{n-k} e^{-\lambda} \frac{\lambda^n}{n!} \\ &= \frac{(p\lambda)^k e^{-\lambda}}{k!} \sum_{n=k}^\infty \frac{((1-p)\lambda)^{n-k}}{(n-k)!} ...


2

This can be realized with a compound Poisson distribution. Each occurrence, independently, is a "success" with probability $p$ and otherwise a "failure"; the number of occurrences is a Poisson$(\lambda)$ random variable, and $X$ is the total number of successes. The successes and failures can also be considered as two separate independent Poisson random ...


1

Your question is not well posed. When you specify $X \sim \text{Binomial}(n,p)$ you are telling us what $n$ is. It is a fixed constant, not a random variable. In that sense, $E[n] = E[n|\text{anything}] = n$.


1

The crucial point is the level of significance of the test, if you consider $\epsilon = 2^{-1}$ then you are rejecting the null conjecture (It is a random array) in many cases. For some reason you are bound to think that many $0$ may occur. So you will reject randomness at the event the first $m$ digits are zero. You could also think that it is not very ...


1

Without seeing your data, it seems to me you might think they are lognormal (that means they would be normally distributed if you took their logs). You might try taking taking logs of a few of your samples and making histograms to see if you get something like a normal shape. A sample of several hundred would be better than several dozen because it takes a ...


1

I guess "intuitively" means without using the term measure, etc? Firstly, $X$ is a function on a set of event $\Omega$, and E(X) means $\int_{\Omega}X(\omega)\,{\rm d}P(\omega)$ by definition. $P$ is the probability measure (sorry, just here) we consider. The important thing is that you integrate over $\Omega$. Now, you wrote $E(X)=\int xf(x)\,{\rm d}x$. ...


1

A $\chi^2$ test would be appropriate since we are dealing with observed frequencies from mutually exclusive groups


1

If the $\chi^2$ test David suggested fails $H_0$ , i.e., that all means are equal, you can then use, either Fisher's LSD test or, for non-parametric, you can use Kruskal-Wallis , to decide which two ( or more ) means are not equal ( at the given choice of significance).


1

Let $X$ be a random variable with density function $f_X(x)$, $g(x)$ be a monotone increasing function of $x$, and $Y = g(X)$. We seek the density function $f_Y(y)$ of $Y.$ In this instance it can be shown that $$f_Y(y) = f_X(g^{-1}(y)) \frac{dg^{-1}(y)}{dy},$$ where $g^{-1}$ is the inverse function of $g$. In your Question, $X \sim Exp(1),\;$$g(x) = ...


1

To find $(Z < 1.246),$ possible entry points into the table are 1.24 and 1.25. The distance between them is .01. The distance from 1.24 to 1.246 is .006. Then $.006/.01 = 6/10.$ Here are exact values from R statistical software: pnorm(.024); pnorm(.0246); pnorm(.025) ## 0.5095737 # P(Z < .024) ## 0.509813 # P(Z < .0246) ## 0.5099725 # ...


1

The probability that any question is answered correctly is $p=0.2$. If the number of questions is $N=180$, the number you get right is $n$ and the maximum score $S_{max}=2$ your score is $s=\frac{n}{N}\times S_{max}$. Now $n\sim B(N,p)$, that is the number of correct answers has a binomial distribution with the given parameters. Then using the Normal ...


1

I don't see how to work the problem unless $X$ and $Y$ are independent. Let $\hat \alpha_1 = (2X + Y)/2.5$ as stated. Then $E(\hat \alpha_1) = \alpha,$ so that $\hat \alpha_1$ is an unbiased estimator of $\theta.$ Now let $\hat \alpha_2 = (X + 2Y)/2$ be another estimator of $\alpha.$ It also has $E(\hat \alpha_2) = \alpha,$ so it is also unbiased. ...


1

Consider the random variable $$X = \begin{cases}1 & \text{with probability } 1/2\\ -1 & \text{with probability } 1/2 \end{cases}$$ this random variable has variance $1$. Now consider $$Y = X + 3$$ this random variable has also variance $1$. The reason we make such an hypothesis is that this make explicit the properties used in the proof ...


1

No $m$ need not be equal to $g(n)$. In fact $g(n) = m + f(n)$. The construction is concerned with the sequences of lenght $n$ that follow $x_1 \ldots x_m 1\ldots 1$ we are not appending $n - g(n)$ digits (as far as I understand). For example $m = 2$ $n = 4$ $g(n)=2$ $2^{n-g(n)}-1=3$. The sequences we have are $$x_1 x_2 1 1\\ x_1x_2 10\\ x_1x_2 01\\ $$ note ...


1

First we can assume without loss of generality that $n = 2$ since a sum independent Poisson r.v.'s is itself Poisson. For $0 \leq k \leq t, k \in \mathbb{N}$ we have \begin{align} P(X_1 = k \mid X_1 + X_2 = t) &= \frac{P(X_1 = k \cap X_1 + X_2 = t)}{P(X_1 + X_2 = t)} \\ &= \frac{P(X_1 = k) P(X_2 = t - k)}{P(X_1 + X_2 = t)} \\ &= ...


1

Hint: $T=\sum_{k=1}^nX_k \sim Poisson(\lambda)$ $T_1=\sum_{k=2}^nX_k\sim Poisson(\lambda-\lambda_1)$ and $P(X_1=x_1,T=t)=P(X_1=x_1, \sum_{k=2}^nX_k=t-k)=P(X_1=x_1)P( \sum_{k=2}^nX_k=t-k)$ now just use Poisson formula and you will get the result.


1

$Z_i$'s i.i.d $\sim N(0,1)$ then $X=\sum Z_i^2 \sim \chi^2_n$. Now $E[Tr\{(ZZ')^2\}]=E[\{\sum Z_i^2\}^2]=E(X^2)$. Now As $X\sim \chi^2_n$ we have $E(X^2)=Var(X)+E^2(X)=2n+n^2$ (i.e. the answer is neither $3n$ nor $3n^2$)


1

I would suggest looking at "On the Identification of Variances and Adaptive Kalman Filtering" by R. Mehra as a starting point. This paper is highly cited and the methods are not difficult to implement.


1

Pst: The density function for the $M$ largest ordered statistics is: $$f_{{X}_{(L)}, {X}_{(L-1)},\ldots,{X}_{(L-M+1)}} ({x}_1,{x}_2,\ldots,{x}_M)$$ This is the probability that: $M$ of the results have the given values $\langle x_1,x_2,\ldots,x_M\rangle$ $${f}_X({x}_1)\cdot {f}_X({x}_2)\cdots {f}_X({x}_M)$$ There's no particular placement of these ...


1

If $X_1,X_2,\ldots,X_{19}$ are i.i.d random variables, what it the probability that $$ \min(X_i)\geq 120 $$ ? Obviously, it is given by: $$ \mathbb{P}[X_1 \geq 120]^{19}.$$ Since $e^{-x^2/2}$ is a fixed point of the Fourier transform, the arithmetic mean of $n$ i.i.d normally distributed $N(\mu,\sigma^2)$ random variables is still a normal variable with the ...


1

I don't think there's an elegant solution but there is a correct solution. Note that $(N_i,N_j,N-N_i-N_j)$ follows a multinomial distribution with parameters $(N,p_i,p_j,1-p_i-p_j)$. Therefore the probability that $N_i > N_j$ is $$P(N_i > N_j) = {\sum_{k=1}^N {\sum_{h=0}^{k-1}{ N! \over {k! h! (N-k-h)!}} p_i^{k} p_j^{h} (1-p_i-p_j)^{N-k-h}}}$$



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