New answers tagged

0

if $Z=f(X)$ and $f$ is injective function so $$\sigma(Z)=\sigma(X)$$ since $Z=f(X)$ so $$\sigma(Z) \subset \sigma(X)$$ and because $f$ is injective so $X=f^{-1}(Z)=g(Z)$ so $$\sigma(X) \subset \sigma(Z)$$ so $\sigma(Z)=\sigma(X)$ or $\sigma(f(X))=\sigma(X)$ now $$E(Y|X)=E(Y|\sigma(X))=E(Y|\sigma(f(X)))=E(Y|f(X))$$


1

If given $X$; $Y$ and $Z$ are conditionally independent, we'll have $$E[Y|X=x,Z=z]=E[Y|X=x]$$ which will make $\gamma=0$ always.


3

Hint: if $X\sim \Gamma (a,\lambda)$ so $E(X^{k})=\frac{\Gamma(a+k)}{\Gamma(a)}\lambda^k$ so $X\sim \chi^{(2)}_{(n)}=\Gamma (\frac{n}{2},2)$ so $E(X^{k})=\frac{\Gamma(\frac{n}{2}+k)}{\Gamma(\frac{n}{2})}2^k$ $$F_{(n-1,m-1)}=\frac{\frac{\chi^{(2)}_{(n-1)}}{n-1}}{\frac{\chi^{(2)}_{(m-1)}}{m-1}}$$ $$E(F_{(n-1,m-1)})=E\left( \frac{\frac{\chi^{(2)}_{(n-1)}}{n-...


1

$$E(XY)=E(E(XY|X))=E(XE(Y|X))$$ $$E(X_i \mu_i)=E \left( E(X_i \mu_i|X_i) \right)=E \left(X_i E( \mu_i|X_i) \right)=E(X_i 0)=0$$


0

This may be easier to see by expanding the matrix notation (which, through its inconsistent use of transposes, might be a little confusing): $$f(\mathbf x) - y = x^\prime\alpha - (\beta^\prime \mathbf{x} + \varepsilon) = \sum_{i=1}^p (\alpha_i-\beta_i)x_i - \varepsilon.$$ Its square expands into three terms according to the power of $\varepsilon,$ $$(f(\...


0

For $M_n^{(1)}$ we must first check that the process is integrable. Note that $\theta\mapsto |e^\theta+e^{-\theta}|$ is an even function which is increasing on $(-\infty,0)$ and decreasing on $(0,\infty)$ and hence attains its maximum $2^{-n}$ at $\theta=0$. Now, \begin{align} \mathbb E[|M_n^{(1)}|] &= \mathbb E\left[\left|\frac{2e^{\theta S_n}}{(e^\...


1

There is an interesting paper proposing to maximize not the observed likelihood, but the expected likelihood Expected Maximum Log Likelihood Estimation. In many examples this gives the same results as MLE, but in some examples where it is different, it as arguably better, or at least different in an interesting way. One example: Take the usual multiple ...


1

Since it must hold for all $n$ (dimension of $X$ and $T$), consider $n=1$. The expectation reduces to $$\mathbb{E}\left[\frac{XT}{T^2}\right]=\mathbb{E}\left[\frac{X}{T}\right]$$ Some intuition: Consider zero-mean $X$, then this expression is equal to $\operatorname{cov}(X,1/T)$. Intuitively, if the correlation/covariance between $X,T$ is positive; e.g. ...


13

Ok. It's a bit long to include the whole proof here, so I will just sketch: Apply a first-order Taylor expansion around some, initially arbitrary point, $x_0$, $$y = m(x_0) + [x-x_0]'\nabla m(x_0,\theta) + R_1 + \epsilon.$$ where $R_1$ is the Taylor remainder. Set $$b_0 = m(x_0),\; b = \nabla m(x_0,\theta),\;\beta = (b_o, b)' $$ $$\tilde x = x-x_0,\; u =...


1

$X_n$ is a Gaussian RV for each $n$; since iid, they all have the same mean, variance, moments etc. i.e. simply we have $$E[X_1^2]=E[X_2^2]=\dots=E[X_n^2]$$ Therefore, $\frac{1}{N}\sum_{n=1}^N E[X_n^2]=\frac{1}{N}NE[X_1^2]=E[X_2^2]=E[X_n^2]$ for any $n$. This is why we can remove the summation. The proof for scaling property of variance can be: $$\...


1

Note that $x_n$'s are random variables, not specific numbers. Assuming that $x_n$ is gaussian distributed with density $p$, it's expectation is per definition $$ E[x_n] = \int_\mathbb{R} x \cdot p(x) dx. $$ If we apply a function to $x_n$, it's expectation is $$E[f(x_n)] = \int_\mathbb{R} f(x) \cdot p(x) dx, $$ in your case $f(x) = x^2$. The calculation ...


2

They're correct. For the first one, you could also use $E[X_i^2]=\theta$ and obtain the result w/o going into gamma distribution. For the second one, an alternate solution would be using Chi-squared distribution, by defining $Z_i=X_i/\sqrt\theta$, yielding the same result since: $$\operatorname{var}\left(\frac{1}{n}\sum X_i^2\right)=\frac{\theta^2}{n^2}\sum_{...


1

Your PDF doesn't integrate to $1$, so you need a suitable scalar in front of it, e.g. $3/16$ instead of $1/16$. But, let's assume it's valid. You don't need to transform boundaries because the integration is in terms of $x$. The last term is also correct because both your boundaries and the integral $x^2f(x)$ is calculated correctly.


1

Whether the arithmetic mean is a good measure for a strongly skewed distribution depends on what you are using it for -- what does "good" mean? One variable that is usually very right skew is income and, usually, you see median income rather than mean income. But why? Because, usually, when we look at some measure of central tendency, we are looking for ...


0

It always take $7$ min to succeed, which happens once. It takes an average of $10$ min to fail. The probability of failing is $2/3$ on each try, or $(2/3)^n$ for n tries. So the total time is ($7 + 10\cdot \sum_n (2/3)^n$) min. Which is $7$ min + $2 \cdot 10$ min $ = 27$.


5

Typical way to solve this: The ant will either take path a and finish or take path b or c and be back in it's starting position. Let $k$ be the number of times that the ant already has taken path b or c. Let $T_k$ be the expectation value for the time to finish for an ant that took already $k$ times the path. Then the expectation value for an ant with $k$ ...


2

Commenting on @Igor F.'s post (not enough reputation in this subforum to simply comment): Both terms are geometric series: $\sum_{i=0}^\infty i\cdot (\frac{2}{3})^i \overset{q=2/3}{=}\sum_{i=0}^\infty i\cdot q^i=q \frac{d}{dq}\sum_{i=0}^\infty q^i=\frac{q}{(1-q)^2}, \text{ for } |q|<1$, so for $q=\frac{2}{3}$ this equals $6$. $\sum_{i=0}^\infty (\frac{...


24

$$T=7/3+(8+T)/3+(12+T)/3=9+2T/3$$ $$T/3=9$$ $$T=27$$ From start point, each of 3 paths are equally possible. Two paths lead you back to start point.


5

Let me join with an explanation which, to me at least, seems even easier: First, observe that the paths $B$ and $C$ can be joined into a single path $X$, with the passage time equal to the mean times of $B$ and $C$, i.e. $10$ minutes. This is due to the fact that these paths have the same probability. If they didn't, we'd need to take a weighted average. ...


8

Hello donkordr and welcome to CV. No, the mean does not answer the question unfortunately. You can read your problem as a Markov Chain with two states: woods, and ant house. Now, your transition probabilities are: From the woods: (does not really matter as it's the goal) $p=1$ to stay in the woods From the ant house: $p_1=\frac{2}{3}$ to stay at the ant ...


10

To answer this question, you have to sum over all possible paths the ant can take, and get the duration of that path, multiplied by the probability of taking that path. That is, $$ E[T] = \sum_{\text{path} \in \text{possible paths}} p(\text{path}) T(\text{path}) $$ Every possible path takes Passage $A$ only once, but can take passages $B$ and $C$ any number ...


1

There is an even simpler answer. If we know that there are $6$ heads in total, we'd expect $3$ heads in the first five tosses and $3$ heads in the last five tosses. However, according to your answer, expected number of heads in the last 5 tosses is $6-{29\over9}={25\over 9}$. So, you would expect different number of heads in the first and last five tosses. ...


2

Yes, that equation holds: The law of iterated expectation works for functions of random vectors, as well as random variables. Thus, you can let $\mathbf{X}_* = (X_2,...,X_n)$ and you then have: $$\begin{equation} \begin{aligned} \mathbb{E}(g(X_1,...,X_n)) &= \mathbb{E}(g(X_1,\mathbf{X}_*)) \\[6pt] &= \mathbb{E}(\mathbb{E}(g(X_1,\mathbf{X}_*)|\...


0

Continuing from @whuber's comment, $-Y$ has normal distribution with mean $-3$ and variance $1$. So $Z = X - Y = X + (-Y)$ has normal distribution with mean $12-3=9$ and variance $4+1=5$. The moment generating function of a normal distribution with mean $\mu$ and variance $\sigma^2$ is $e^{\mu t + \sigma^2 t^2/2}$, and so the moment generating function of $Z$...


2

The probability generating function (pgf) of $X_i$ is $$p_{b}(x) = (1 + x^1 + x^2 + \cdots + x^{b-1})/b = b^{-1} \frac{1-x^b}{1-x}$$ with $b=10.$ (For related examples and explanations see, inter alia, 1, 2, 3, and 4; or consult Wikipedia.) Because the $X_i$ are independent, the pgf of the sum of $n$ of them is $$p_b(x)^n = b^{-n} \frac{(1-x^b)^n}{(1-x)^...


4

You can calculate this recursively, using a (backwards) dynamic programming-like iteration. Let us construct a state variable $s \in \{0, \dots, 10\}$. $s$ represents the current value of the sum, except for state $10$, which represents any sum $>=10$. You start in state $s=0$. If you are in state $10$, obviously you are done, and the expected ...


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