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The best approximation in the $L^2$ sense is given by the value of $\alpha\in\mathbb{R}^+$ for which: $$\frac{d}{d\alpha}\left(\frac{1}{2\pi}\int_{|x|\geq \alpha}e^{-x^2}\,dx + \int_{-\alpha}^{\alpha}\left(\frac{1-|x/\alpha|}{\alpha}-\frac{e^{-x^2/2}}{\sqrt{2\pi}}\right)^2\,dx\right)=0,$$ i.e. by minimizing the $L^2$ norm of the difference between the pdf ...

5

It depends in what sense you want your triangular distribution to "approximate" the normal distribution. The normal distribution is symmetric about $0$ and unimodal, so you probably want your triangular distribution to be symmetric about $0$ and unimodal as well. In order for your triangular distribution to be a probability distribution, the area under the ...

3

The sum of independent normal random variables is normal. That assumption of independence is very important: don't leave it out! The mean of the sum is the sum of the means (expected value is always additive), the variance is the sum of the variances (variance is additive for independent random variables).

3

The transformations in $\frac {X -\mu}{\sigma}$ are just translation and scaling, i.e., this is a change of coordinate from $X$ to $\frac {X-\mu}{\sigma}$ given by the function $X'=\phi(X)=\frac {X-\mu}{\sigma}$. More formally, you can use the Jacobian to study the effect of the change of variables in the density function: ...

2

Just write out the explicit formula for the gaussian (https://en.wikipedia.org/wiki/Normal_distribution) and then perform the transformation. The algebra shouldn't be difficult.

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

On the line you don't understand, you are using the standardization of your random variable $X$ to find your standard deviation value $\sigma$. Recall that in order to standardize a Normally distributed random variable, you subtract its mean, and divide by its standard deviation. Essentially your formula is $$Z = \dfrac{x-\mu}{\sigma}$$ where $x$ denotes ...

2

Set $$f_n(s) := \sum_{j=1}^n s_j 1_{[t_{j-1},t_j]}(s).$$ Then your calculation shows $$\Phi_{X^m}(s) = \exp \left(- \frac{1}{2} \sum_{k=1}^m \langle f_n, e_k \rangle^2 \right).$$ Letting $m \to \infty$, we obtain $$\Phi_X(s) = \exp \left(- \frac{1}{2} \sum_{k \geq 1} \langle f_n, e_k \rangle^2 \right). \tag{1}$$ Since $(e_k)_{k \geq 1}$ is an ONB, we ...

2

If $X$ and $Y$ are independent and $Y$ takes values in $\{-1,1\}$ while the distribution of $X$ is symmetric about $0$, then $XY$ has the same distribution as $X$. Hint: condition on $Y$.

2

I think this is the "continuity correction". You observe single units occurring, so you are in the middle of a value. Hence add or subtract 0.5 to actually switch to the next block and make the result more appropriate.

2

First observe that the required probability $<\ \approx 0.68^2=0.4624$ Then observe that you might as well assume $\sigma_1=\sigma_2=1$, then your question becomes what is the probability that a point lies inside the unit circle. Then proceed by converting to polars, when: $$p=\int_{r=0}^1 f_R(r)\;dr=\int_{\theta=0}^{2\pi} ... 1 The terminology is a weighted mixture of normal distributions or, in short, a mixture distribution. For a mixture of two normals (N(\mu_1,\sigma_1^2) and N(\mu_2,\sigma_2^2)) the density is$$f(x)=\frac{w \exp \left(-\frac{\left(x-\mu _1\right){}^2}{2 \sigma _1^2}\right)}{\sqrt{2 \pi } \sigma _1}+\frac{(1-w) \exp \left(-\frac{\left(x-\mu ...

1

In your book $\Phi$ is the cumulative distribution of $Y \sim \mathcal{N}(0,1)$ And $X$ doesn't follow (approximatively) a $\mathcal{N}(0,1)$, but follow (approximatively) a $\mathcal{N}(5000,50^2)$ So $$P( 4900 \leq X \leq 5100 ) =P( -2 \leq \frac{X-5000}{50} \leq 2 )$$ And now $\frac{X-5000}{50} \sim \mathcal{N}(0,1)$, so $$P( 4900 \leq X \leq ... 1 If X and Y are not independent, I doubt that there's much that can be said. There are just too many ways to obtain Z as the sum of X and Y. For example, here are some ways to get an arbitrary distribution as Z = X + Y with X and Y identically distributed. Let B be Bernoulli(1/2) and independent of Z. Let f be an arbitrary Borel ... 1 It is immediately obvious that there is insufficient information to obtain the desired probability: you have, for a given constant, three parameters that may vary: the common marginal mean \mu, the common marginal variance \sigma^2, and the correlation \rho; yet you have only one condition which is that when the constant is zero,$$\Pr[X < 0 \cap ...

1

Since $X$ has a symmetic distribution about $0$, you have $P(X-0 \le x) = P(0-X \le x)$, i.e. $P(X \le x) = P(-X \le x)$. Now consider the cumulative distribution function $P(Z \le x)$. This is $P(XY \le x)=P(XY \le x|Y=1)P(Y=1)+P(XY \le x|Y=-1)P(Y=-1)$ which is $P(X \le x|Y=1)P(Y=1)+P(-X \le x|Y=-1)P(Y=-1)$ which by the symmetry and independence is ...

1

If $\sigma = 1.5$, then $P\{X < c\} = P\{\frac{X-\mu}{\sigma} < \frac{c-\mu}{\sigma}\} = P\{Z < \frac{c-4}{1.5}\} = 0.35,$ where $Z$ is standard normal. Then from normal tables $(c - 4)/1.5 \approx -0.3853.$ Solve for $c.$ If $\sigma^2 = 1.5$ as you say, then $\sigma = 1.224745$ and adjust the denominator accordingly before solving. If $\sigma = ... 1 In part (a) you calculated$P(T > 36 \mid Y^\complement),$where$Y^\complement$is the event that the computer does not have a hardware problem and$T$is the time taken to assemble the computer. In part (b) you calculated$P(T > 36 \mid Y),$where$Y$is the event that the computer has a hardware problem and$T$(as before) is the time taken to ... 1 The highlighted "at least one credit card" corresponds to the "at least one credit card" on lines 1-2. And the$X$you have been using is the number of persons, not number of cards. 1 Note that$-Y$is also$N\left(0,\frac12\right)$. Thus,$X-Y$is the sum of two$N\left(0,\frac12\right)$variables, which is a$N(0,1)$variable. The absolute value of a$N(0,1)$variable has the PDF $$\sqrt{\frac2\pi}e^{-x^2/2}[x\ge0]$$ where$[\cdot]$are Iverson Brackets. Using the substitution$t=x^2/2, we get \begin{align} ... 1 I agree with the comment by @eigenchris for 'well-known' distributions encountered early on in a probability course. However, one does not have to venture too far into the study of probability distributions to find examples in which knowing the population mean and variance does not easily specify the distribution. It is useful to make a distinction between ... 1 For X\sim N(\mu,\sigma^2) the first quartile isq_{0.25}=\sigma\cdot \Phi^{-1}(0.25)+\mu$$where \Phi is the CDF of the standard normal r.v. because q_{0.25} is the solution of$$P\{X\le q_{0.25}\}=P\left\{\frac{X-\mu}{\sigma}\le \frac{q_{0.25}-\mu}{\sigma}\right\}=\Phi\left(\frac{q_{0.25}-\mu}{\sigma}\right)=0.25$$1 For a standard normal distribution, you should use \Phi for the cumulative distribution function, and \phi for the density. I also suspect you want -\frac{x-\mu}{\sigma} trather than \frac{-x-\mu}{\sigma}. Making both changes will give you the answer in the book, or close enough. Since a normal distribution N(\mu,\sigma^2) is symmetric about the ... 1 Since the normal distribution is symmetric, what the question is essentially asking is: Can you find a z-score such that the area between 0 and z is 0.6 / 2 = 0.3? Since a z-score is for a standardized normal distribution, you'll have to then convert this back to unstandardized form by setting z equal to one of the given inputs for the cdf; note ... 1 Yes, you've done it correctly. However, there is some inaccuracy in your answer due to the floating-point roundoff error which happens because you are subtracting P(Z \leq 8), which is close to 1, from 1: > 1-pnorm(8) [1] 6.661338e-16 Instead of this, a better method is to use the option `lower.tail=FALSE' to give the upper tail directly: ... 1 This is easy if you look for the probability distribution of a-b. As a \sim \mathcal{N}(\mu_a, \sigma_a^2) and b\sim \mathcal{N}(\mu_b, \sigma_b^2) are independant, you have that$$a-b = Z \sim \mathcal{N}(\mu_a-\mu_b, \sigma_a^2+\sigma_b^2)$Then you can calculate$P(Z<0)$as usual 1 See my comments under the question. What I write below will assume my guesses there are right and that$a=1$. Your symmetric matrix$C$will have one row corresponding to each unordered pair$\{i,j\}$. The entry in row$\{i,j\}$and column$\{k,\ell\}$will be$2$if$\{i,j\}=\{k,\ell\}$. Those are just the diagonal entries in the matrix. In row ... 1 Different books have different formats for normal tables. It will go something like this. The normal table has cut-off points in the margins and probabilities in the body. Look in the vertical margin for 1.6. Under the column headed .00 you will see .0548. That is the probability corresponding to cutoff 1.60. But you want the probabilities corresponding to ... 1 Using a standard normal table, which you can find a good example of one here: https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf, for$P(x < 31.5) = 0.05$, that means the area to the left of the z-score is$0.05$. Because we know that the distribution is normal, we would go into the middle of the table and find the closest value to$0.05$... 1 If I have understood the question correctly, you have a list of n elements {$x_i\$} some of which are "special" (exceed some level you set) and the rest of which are non-special. You want to define two probabilities p and q where p is the probability that a given special element is chosen and q is the probability that a given non-special element is chosen. ...

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