# Tag Info

## Hot answers tagged normal-distribution

3

Mean: You integrate $y=e^{-x^2}$ times a Gaussian pdf. Hence, you get an exponential with exponent $$-x^2-\frac{(x-\mu)^2}{2\sigma^2}=-\frac{2\sigma^2x^2+x^2-2\mu x+\mu^2}{2\sigma^2}.$$ By completing the square, $$2\sigma^2x^2+x^2-2\mu x+\mu^2=(2\sigma^2+1)(x-\frac\mu{2\sigma^2+1})^2-\frac{\mu^2}{2\sigma^2+1}.$$ This yields the pdf of a Gaussian of ...

2

To get the pdf of $X$, let us first get the cdf of $X$: $$F_X(x) = P(X \le x) = P(2Z + 1 \le x) = P(Z \le \frac{x-1}{2})$$ Now $P(Z \le \frac{x-1}{2}) = F_Z(\frac{x-1}{2})$ so: $$F_X(x) = F_Z(\frac{x-1}{2})$$ $$\to f_X(x) = f_Z(\frac{x-1}{2}) (\frac{x-1}{2})'$$ $$\to f_X(x) = f_Z(\frac{x-1}{2}) (\frac{1}{2})$$

2

$$\Phi(x)=\frac{1}{2}+\frac{1}{2}\mathrm{erf}(x/\sqrt{2})$$

2

$$\mathbb{E}[X^2+X]=\mathbb{E}[X^2]+\mathbb{E}[X]=\mathbb{E}[(X-\mathbb{E}[X])^2]+\mathbb{E}[X]=\sigma^2 +0=1$$

2

I decided to move from a comment to an answer because some times I ended up using some facts without digging too much in the whys. So, I decided to dig a little more and hopefully answer the questions of the OP. This answer is mainly about why the RMS of the noise is equal to its standard deviation. As a side note, I will also do some comments about the ...

1

Since $p>\frac{1}{2}$, we can write $p= \frac{1}{2}+q$ for some $q>0$. Now $$\frac{(n-1)^p}{\sqrt{n}} = \sqrt{\frac{n-1}{n}} (n-1)^q \geq \frac{1}{2} (n-1)^q, \qquad n \in \mathbb{N},$$ implies $$\mathbb{P} \left(B_1 > C \frac{(n-1)^p}{\sqrt{n}} \right) \leq \mathbb{P} \left(B_1 > \frac{C}{2} (n-1)^q\right).$$ Choose $k \in \mathbb{N}$ ...

1

You appear to be considering something like: $$P\left ( x-\frac{\Delta x}{2} \leq X \leq x+\frac{\Delta x}{2} \right )$$ where $X$ is Gaussian with mean $\mu$ and standard deviation $\sigma$. This is exactly equal to $$\int_{x-\Delta x/2}^{x+\Delta x/2} \frac{1}{\sqrt{2 \pi} \sigma} e^{-(y-\mu)^2/(2\sigma^2)} dy$$ which can be straightforwardly ...

1

The univariate normal p.d.f. is $$x \mapsto \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{1}{2}\left( \frac{x-\mu} \sigma \right)^2 \right). \tag 1$$ The expression $\left(\dfrac{x-\mu} \sigma\right)^2$ is the same as $(x-\mu) (\sigma^2)^{-1} (x-\mu)$. The $1\times1$ matrix $x-\mu$ is its own transpose, and the inverse of the $1\times1$ matrix whose ...

1

Outline: Let $a$ be the probability that a randomly chosen female Smurf is between $1$ and $1.3$, and let $b$ be the corresponding probability for male Smurfs (Smurves?). Then our required probability is $(0.6)a+(0.4)b$.

1

Let $X \sim N(\mu,\sigma^2)$. Take $g := 1$. Then Stein's lemma gives $$\mathbb{E}(1 \cdot (X-\mu)) = 0, \tag{1}$$ i.e. $\mathbb{E}(X) = \mu$. For $g(x):= (x-\mu)^{n-1}$, $n \geq 2$, we have, by Stein's lemma, $$\mathbb{E}((X-\mu)^n) = \sigma^2 (n-1) \mathbb{E}((X-\mu)^{n-2}).$$ If $n$ is odd, i.e. $n=2k+1$, then we get by iteration \begin{align*} ... 1 I will assume upvotes validate the accuracy of the following answer: The first value is not a probability, but a density value, D_p given byD_p=\frac1{\sigma\sqrt{2\pi}}e^\frac{-(x-\mu)^2}{2\sigma^2}.$$The density for a value x=a should be used to compare whether a value x_i is more likely to be chosen from the population if it is near (within a ... 1 By the properties of a continuous density,$$ \mathbb P(0 < Y \le b) = \int_0^b f_Y(y) \, dy. $$Therefore, for Y = 1/Z,$$ \mathbb P(0 < Y \le y) = \mathbb P(Z \ge 1/y) = \int_{1/y}^{\infty} f_Z(z) \, dz = 1/2 - \int_0^{1/y} f_Z(z) \, dz. $$Differentiating with respect to y to recover Y's density,$$ f_Y(y) = \frac d{dy} \left( 1/2 - ...

1

When you add two independent variables, you can calculate the distribution with the convolution of the two pdfs. If Z = X + Y, $f_Z(z) = \int_{-\infty}^{\infty} f_X(z-x)f_Y(x) dx$ This question explains the density of two uniform random variables: density of sum of two uniform random variables $[0,1]$ For subtraction, use the pdf of -Y, which is ...

1

I don't know an answer in the general case but concerning the specific case $b=2$ we may start by defining : \begin{align} \tag{1}f(a,c)&:=\int_0^{\infty} \frac{x^{-1}}{a+x^{-2}} e^{-c x^2} dx\\ &=\int_0^{\infty} \frac{x}{1+ax^2} e^{-c x^2} dx\\ &=e^{\,\large{\frac ca}}\int_0^{\infty} \frac{c\;x}{c\;(1+ax^2)} e^{\,\large{-\frac ca (1+ax^2)}} dx\\ ...

1

You don't want $P(X-\mu < \sigma)$ if $X>\mu$ and $P(\mu-X < \sigma)$ if $X < \sigma$, you want $P(|X-\mu|<\sigma) = P(|z|<1) = P(z<1) - P(z<-1)$.

1

What "nullUser" said is correct. Looking at the table you find P(z<-1) = .16 and p(z<1) = .84 roughly. You can solve then by just subtracting .84-.16 to get the probability in-between. If you just calculate p(z<1) like it seems you did, it overlaps with z<-1 so there is no way to discern the correct answer unless you "delete" the overlapping ...

1

I think your first statement is incorrect, leading to confusion right from the start. In terms of $Y \sim Norm(\mu=430, \sigma=100),$ you want to find $b$ such that that $P(Y < b) = .75.$ Then standardize to express the question in terms of standard normal $Z \sim Norm(0,1).$ $$P\left\{Z = \frac{Y-\mu}{\sigma} < \frac{b - 430}{100} \right\} = .75.$$ ...

1

The standard normal distribution is symmetric. And therefore $P(X\geq 40)=1-\Phi \left( \frac{40-\mu}{0.8} \right)=\Phi \left( \frac{\mu-40}{0.8} \right)=0.96$ $\frac{\mu-40}{0.8}=\Phi^{-1}(0.96)$ $\frac{\mu-40}{0.8}=1.75$ To get your solution you multiply both sides by $(-1)$: $\frac{40-\mu}{0.8}=-1.75$

1

Your question posits the existence of a normally distributed random variable $X$ with standard deviation $\sigma = 0.8$, for which $\Pr[X > 40] = 0.96$; that is, to say, the chance that $X$ is above $40$ is $96\%$. That is what those two conditions mean. From this, we are supposed to determine the mean $\mu$ of $X$. To do this, we can standardize $X$: ...

1

Suppose $X$ is normal with mean $\mu$ and standard deviation $\sigma$. Then $Z=\frac{X-\mu}{\sigma}$ is normal with mean $0$ and standard deviation $1$, and $X=\sigma Z + \mu$. Then $$E[X \mid X \in [a,b]]=E[\sigma Z + \mu \mid X \in [a,b]] \\ = \mu + \sigma E[Z \mid X \in [a,b]] \\ = \mu + \sigma E \left [Z \left | Z \in \left [ ... 1 Hint. You are looking for a number a such that$$ P(L>a)=0.8 $$or equivalently, by the change of variable L \to Z,$$ P(Z>\frac{a-50000}{5000})=0.8 $$or$$ P(Z\leq\frac{50000-a}{5000})=0.8 $$Using a table your Z-table you can find that$$ P(Z\leq \color{red}{0.84}) \approx 0.8. $$Then$$ \frac{50000-a}{5000}=\color{red}{0.84} $$and ... 1 The easier way to proceed is to note that 2Z+1 \leq x if and only if Z \leq \frac{x-1}{2}. This gives you the CDF, which you can differentiate as necessary. 1 Let me write P_k=P(X=x_k). The plan is to take the trivial estimate 1\ge \sum\limits_{k=m}^{m+d} P_k with some d\approx \sqrt{n}, and replace P_k by c\cdot P_m. So, we need a suitable upper bound on \frac{P_m}{P_k}. For 1\le j\le d\le\sqrt{n},$$ \frac{P_m}{P_{m+j}} = \frac{\binom{n}{m}p^{m}q^{n-m}}{\binom{n}{m+j}p^{m+j}q^{n-m-j}} = ...

1

The mean of $e^{-X^2}$ is the integral of $e^{-x^2}$ weighted by the density function of $X$ (which is another exponential function). For an analytic expression, notice that the integral now contains a product of two exponentials. Combine the product of exponentials into an exponential of a sum and re-write the resulting exponent as a single negative square ...

1

No, in general, if $X$ follows a normal distribution, then in shorthand it is written as $$X\sim N(\mu,\sigma^2).$$ Thus, for example, the variance of $X_1$ is 3 not $3^2$. Or as you have written it $$\sigma_1^2 = 3$$ not $\sigma_1 = 3$.

1

If $X_1$ and $X_2$ are iid random variables such that $X_1\sim\mathcal N(0,\sigma^2)$ and $X_2\sim\mathcal N(0,\sigma^2)$, then $$X_1-X_2\sim\mathcal N(0,2\sigma^2).$$ If $X\sim\mathcal N(0,\sigma^2)$, then $Y=|X|$ has the half-normal distribution and $$\operatorname EY=\frac{\sqrt2\sigma}{\sqrt\pi}.$$ Hence, we have that $$\frac12\operatorname ... 1 This question seems to belong to StackOverflow's R-tag In R we do not generally care about whether a variable is integer or floating if they are used for calculations as an operation such as 2L/3.0 will return a floating variable anyways. For example an integer is also a floating variable: is.numeric(2L) returns TRUE while is.integer(2) returns FALSE. For ... 1 You need to complete the top-right of the covariance matrix by reflecting the bottom-left to give$$ \Sigma = \left( \begin{array}{cccc} 1037.21 & -80.02 & 1430.7 & 271.44 \\ -80.02 & 219.84 & 92.1 & -91.58 \\ 1430.7 & 92.1 & 2624 & 210.3 \\ 271.44 & -91.58 & 210.3 & 177.36\end{array} \right)$$... 1 We will transform |Y−μ|<1.96\sigma into the form a(Y)<\mu<b(Y):$$|Y−μ|<1.96\sigma\Leftrightarrow1.96\sigma<Y−μ<1.96\sigma\Leftrightarrow1.96\sigma-Y<−μ<1.96\sigma-Y\Leftrightarrow Y-1.96\sigma<μ<Y+1.96\sigma.$$From P(\left | Y-\mu \right | < 1.96\sigma) \approx0.95 and transforming done above we obtain ... 1 General forecasting rule Suppose X_1 has mean \mu_1, X_2 has mean \mu_2, and that they are Gaussian with covariance matrix$$ \left[\begin{matrix} \Omega_{11},\Omega_{12} \\ \Omega_{21},\Omega_{22} \end{matrix} \right]$$then$$ X_2∣X_1 \sim N(\mu_2+(\Omega_{21}\Omega_{11}^{-1}(X_1-\mu_1),\Omega_{22}-\Omega_{21}\Omega_{11}^{-1}\Omega_{12})  ...

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