Can I say that  $(X^2-Y^2)/ \sqrt{X^2+Y^2}$ is normal since $X,Y\sim N(0,1)$ and $X,Y$ are normal and independent?  I am trying to do those problem using polar coordinates with $X=r \cos \theta$ and $Y=r \sin \theta$, but I got stuck when I was trying to simplify $(X^2-Y^2)/ \sqrt{X^2+Y^2}$. Can anybody please help me? Thanks!
53169
It is given that $X,Y \overset{\text{i.i.d.}}{\sim} N(0,1)$  Show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}} \overset{\text{i.i.d.}}{\sim} N(0,\frac{1}{4})$  I was thinking of making polar transformations $X=r \cos \theta, Y=r \sin \theta$  Then I am getting stuck at ranges of $\theta$
83455
Assume we have $n$ distinct points $x_1,x_2,...,x_n\in \Bbb{R}$, and observations which are independent random variables $y_1,y_2,...,y_n$ depending on formula: $$ y_i=\theta_0+\theta_1x_i^2+\theta_3e^{x_i}+\varepsilon_i $$ where $\varepsilon_i$ are distributed as $N(0,\sigma^2)$. Derive the maximum likelihood estimator for $\theta_{0},\theta_{1},\theta_{2}$ .  I know how to find the MLE for a known distribution like normal or Poisson, but for this question, I don't how I can find the likelihood function, should I consider it as normal distribution? And I don't know what the mean and variance would be since I can't figure out the distribution and likelihood function. Also, should I consider $\varepsilon_i$ as a useful variable, or just treat it as an error in estimation? Thanks for any help.
105744
Let $X, Y$ be independent standard normal random variables.  We already know that $X+Y$ is normal with mean 0 and variance 2. However, I am trying to prove this result in a slightly different way than usual, i.e., using convolutions.  We can address the problem in the following way: the sum of the two variables has a density $h(x)$ given by the convolution product of the densities of the two variables, say $f$ and $g$, so that   $$ h(x) = \int_{-\infty}^{\infty} f(x-y)g(y)dy. $$  However,  $$ f(y) = g(y) = \frac{1}{\sqrt {2\pi}}\exp(-y^2/2). $$  Hence now the integral becomes:  $$ h(x) =\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp(-(x-y)^2/2)\exp(-y^2/2)dy. $$  However, I am puzzled here. How should we go on from here? I cannot see how to transform this integral into the integral of a normal r.v of type $N(0,2)$. Thanks in advance for your help.
24322
If X and Y are independent random variables both with the same mean (0) and variance, how about $X^2$ and $Y^2$? I tried calculating E($X^2Y^2$)-E($X^2$)E($Y^2$) but haven't been able to get anywhere.
9154
I'm having a problem converting $\int\limits_1^2  \int\limits_0^ \sqrt{2x-x^2}   \frac{1}{(x^2+y^2)^2}  dy dx $ to polar coordinates.  I drew the graph using my calculator, which looked like half a circle on the x axis.  I know that $\frac{1}{(x^2+y^2)^2}  dydx$ turns to $ \frac{r}{(r^2)^2}drd\theta$, which would be $ \frac{1}{r^3}$  The region of integration in  $\theta $ is I guess $0 \leq  \theta  \leq  \frac{1}{4}  \pi $, but I'm stuck for r. The lower boundary is $x=1=rcos\theta \rightarrow r=1/cos\theta=sec\theta$, but what is the upper boundary?? Is it  $\sqrt{2rcos\theta-r^2cos^2\theta}$? I tried this but I couldn't integrate my equation...  Could someone please help me out?  Also, how can I draw the graph of $y= \sqrt{2x-x^2} $? I could only know how it looked like by using my calculator...
73606
$\int_{-2}^0\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \sqrt{x^2+y^2} dydx$  y describes the circle of radius 2, and is positive and negative, so doesnt limit any quadrants in our integral, however x can only be negative, which means the left half of our circle is what we are interested in.  This means, we need the integral:  $\int_{\pi/2}^{3\pi/2}\int_{0}^{2} rdrd\theta$  Giving us a value of $2\pi$.  Is this a correct assessment? I wasnt sure if making  $0$ and $2$ the limits of $r$ was entirely correct, since it describes the entire circle of radius $2$, and not just one half. Then again, a circle of radius $2$ does have an area of $4\pi$ which makes me feel I am along the right lines.   Any clarification would be fantastic. Thank you!
123137
Simplify: $x^2 &gt; 1$.  My solution: Taking square root on both sides:  $±x &gt; ±1$  So my results are:   $x &gt; 1$ $x &gt; -1$ $-x &gt; 1$ $\implies$ $(-1 &gt; x)$ $-x &gt; -1$ $\implies$ $(1 &gt; x) $   But I strongly feel this is wrong. What is wrong here?  A step-by-step explanation will help me.
115076
Please help me solve this problem:  Let $X$ and $Y$ be independent standard normal random variables, let $Z$ have an arbitrary density function, and form $Q = \frac{X + YZ}{\sqrt{1+ Z^2}}$. Prove that $Q$ also has a standard normal density function.   Thanh you
98400
Are $\cos^2 \theta$ and $\cos \theta^2$ the same? I mean be it $\sin,\cos, \tan ,\cot ,\sec,\csc$.  Are they same? Please help a maths noob here.
60383
If $X \sim \mathrm{Normal}(\mu,\sigma^2)$ and $Y \sim \mathrm{Normal}(\mu,\sigma^2)$ are independent random variables, how do I prove that $X+Y$ and $X-Y$ are also independent?  What happens with the independence between $X+Y$ and $X-Y$ when $X \sim \mathrm{Normal}(\mu_x,\sigma_x^2)$ and $Y \sim \mathrm{Normal}(\mu_y,\sigma_y^2) $  Thank you
62306
What would the normalising  constant be of the following, or atleast an approximation? I would like to avoid sampling.  $$f(\theta)=\exp(-k_1e^{-k_2\theta^2}-\theta^2)\qquad\theta\in(-\infty,\infty), \quad k_1,k_2&gt;0$$  In the event that its not possible, what would be a valid polynomial lower bound on $\log(f(\theta))$, that would take into account $k_1$  For anyone, who is interested, I end up with this expresssion because I place a Gaussian prior over (a squared exponential) kernel parameters, and trying to use variational bayes (hence, a lower bound).
94468
Consider the following 2-variable function:     $$f(x,y) = \frac{x^2y}{x^4+y^2}$$   I would like to find the limit of this function as $(x,y) \rightarrow (0,0)$.   I used polar coordinates instead of solving explicitly in $\mathbb R^2 $, and it went as the following:  $$ x = r \cos \theta, \qquad y = r\sin\theta $$   Hence,  $$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^4 + y^2} = \lim_{r \to 0}\frac{r^2\cos^2\theta(r\sin\theta)}{r^4\cos^4\theta + r^2\sin^2\theta}$$  This simplifies to,  $$ \lim_{r \to 0} \frac{r^3 \cos^2\theta\sin\theta}{r^2(r^2\cos^4\theta + \sin^2\theta)}$$  Simplifying $r^3/r^2$, we finally get;  $$\lim_{r \to 0} \frac{r (\cos^2\theta\sin\theta)}{r^2\cos^4\theta + \sin^2\theta}$$  Now from the above, we find that as $r \to 0$ the limit is $0$.   I wanted to verify this answer so I checked on Wolfram Alpha. Explicitly without changing to polar coordinates, it said that the limit does not exist at $(0,0)$ and rightly so. Then how is it that with polar coordinates, the limit exists and is $0$? Am I doing something wrong in this method?  Also, what should I do in this situation, and when should I NOT use polar coordinates to find limits of multi-variable functions?
103650
In polar coordinate how we can get $dx\;dy=r\;dr\;d\theta$?  with these parameters:    $r=\sqrt{x^2+y^2}$ $x=r\cos\theta$ $y=r\sin\theta$  Tanks.
1348
I am trying to prove the statement:  If $X\sim\mathcal{N}(0,\sigma_1^2)$ and $Y\sim\mathcal{N}(0,\sigma_2^2)$ are independent random variables, then $\frac{XY}{\sqrt{X^2+Y^2}}$ is also a Normal random variable.  For the special case $\sigma_1=\sigma_2=\sigma$ (say), we have the well-known result that $\frac{XY}{\sqrt{X^2+Y^2}}\sim\mathcal{N}\left(0,\frac{\sigma^2}{4}\right)$ whenever $X$ and $Y$ are independent $\mathcal{N}(0,\sigma^2)$ variables. In fact, it is more generally known that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $\mathcal{N}\left(0,\frac{\sigma^2}{4}\right)$ variables. A proof of the last result follows by using the transformation $(X,Y)\to(R,\Theta)\to(U,V)$ where $x=r\cos\theta,y=r\sin\theta$ and $u=\frac{r}{2}\sin(2\theta),v=\frac{r}{2}\cos(2\theta)$. Indeed, here $U=\frac{XY}{\sqrt{X^2+Y^2}}$ and $V=\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$. I tried to imitate this proof for the problem at hand but it appears to get messy. If I haven't made any error, then for $(u,v)\in\mathbb{R}^2$ I end up with the joint density of $(U,V)$ as $$f_{U,V}(u,v)=\frac{2}{\sigma_1\sigma_2\pi}\exp\left[-\sqrt{u^2+v^2}\left(\frac{\sqrt{u^2+v^2}+v}{\sigma_1^2}+\frac{\sqrt{u^2+v^2}-v}{\sigma_2^2}\right)\right]$$ I have the multiplier $2$ above as the transformation is not one-to-one. So density of $U$ would be given by $\displaystyle \int_{\mathbb{R}}f_{U,V}(u,v)\,\mathrm{d}v$, which isn't readily evaluated. Now I am interested to know if there is a proof where I can only work with $U$ and don't have to consider some $V$ to show that $U$ is Normal. Finding the CDF of $U$ doesn't look so promising to me at the moment. I would also like to do the same for the case $\sigma_1=\sigma_2=\sigma$. That is, if $X$ and $Y$ are independent $\mathcal{N}(0,\sigma^2)$ variables then I wish to show that $Z=\frac{2XY}{\sqrt{X^2+Y^2}}\sim\mathcal{N}(0,\sigma^2)$ without using a change of variables. If somehow I can argue that $Z\stackrel{d}{=}X$, then I'm done. So two questions here, the general case and then the particular case. Related posts on Math.S.E: . . Edit. This problem is in fact due to L. Shepp as I found out in the exercises of An Introduction to Probability Theory and Its Applications (Vol. II) by Feller, alongwith a possible hint:  Surely, $U=\frac{XY}{\sqrt{X^2+Y^2}}=\frac{1}{\sqrt{\frac{1}{X^2}+\frac{1}{Y^2}}}$ and I have the density of $\frac{1}{X^2}$ at hand. Let's see what I could do now. Apart from this, a little help with the integral above is also welcome.
106975
Show that $$\lim_{(x,y) \to (0, 0)}\frac{xy^2}{x^2+y^2} = 0$$  I have tried switching to polar coordinates but I'm not getting a single term.  This is what I did.   Putting $$x=r\sin θ,\quad  y=r\cos θ$$ we obtain   $$\left|\frac{xy^2}{x^2+y^2}\right|=|r\cos^2 θ \sin θ|                        =|r\sin θ(1-\sin^2 θ)|                        = |r\sin θ-r\sin^3 θ|$$
23869
If $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ be three points on the parabola  $y^2 = 4ax$ and the normals at these points meet in a point then how will we  prove that $$   \frac{x_1 -x_2}{y_3} + \frac{x_2-x_3}{y_1} + \frac{x_3-x_1}{y_2} = 0? $$  I tried as follows.  Let the normals meet at $(h,k)$. Then, $$am^3 + (2a-h)m + k = 0.$$  After Solving the equation, $$ x_1 y_1(y_2-y_3) + x_2 y_2(y_3-y_1) + x_3 y_3 (y1 - y2) = 0 $$  Substituting $x_k = y_k^2/4a$ for $k \in \{1,2,3\}$  But I am not getting the result.
33229
Find the value of $$\cos ^2\theta+\cos^2 (\theta+1^{\circ})+\cos^2(\theta+2^{\circ})+......+\cos^2(\theta+179^{\circ})$$  Can anyone teach me where to start with? I've no idea.