A random variable $X$ is lognormal if ln $X$, the natural logarithm of $X$, is normally distributed. Find the mean and variance of a lognormal random variable with $ln X ∼ N(µ, σ^2)$  How should I solve this using the moment generating function of normal distribution, without differentiation?
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The computed moments of log normal distribution can be found . How to compute them?
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Let $X_1$ and $X_2$ be independent Normal random variables with mean $\mu_1$ and $\mu_2$, and variances $\sigma_1$ and $\sigma_2$. Let $Y = X_2-X_1 + c$, where $c$ is a constant. For notational simplicity, let $Y = X' + c$, where $X' = X_2-X_1$ The moment generating function of a Normal random variable $X_n\sim N(μ_n,σ_n^2)$ is $$M_{X_n } (t)=\text{exp}\Big(μ_n t+\frac{σ_n^2 t^2}{2}\Big)$$ Then, $$M_{X'} (t)=\text{exp}\Big[t(\mu_1+\mu_2)+\frac{t^2}{2}(\sigma^2_1+\sigma^2_2)\Big]$$ Thus, by the uniqueness property of moment generating function, $X'$ is said to follow a Normal distribution with mean $(\mu_1+\mu_2)$ and variance $(\sigma^2_1+\sigma^2_2)$. How do I proceed from here to derive the following:  First, I want to establish that $Y$ is also a Normal random variable using the moment generating function. The constant $c$ bugs me, but I cannot ignore it. As per my understanding, $Y\sim N\big(c+\mu_1+\mu_2,\sigma^2_1+\sigma^2_2\big)$ (and I could be completely wrong). But I don't understand how to apply the moment generating function to prove that.  Secondly, I want to derive $P(Y \leq a)$, where $a$ is a constant. If I am not mistaken, $P(Y \leq a)$ can be written as $$P(Y \leq a)=\Phi\Bigg(\frac{a-X_2-X_1 + c}{\sqrt{2(\sigma^2_1+\sigma^2_2)}}\Bigg).$$ But how do I arrive at that.   Any help is high solicited and appreciated.
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It has been a long time since I have used calculus, and I am trying to understand how the mean and variance of the Laplace distribution with pdf $$f(x|\mu,\sigma) = \dfrac{1}{2 \sigma}\,e^{-\Large\frac{|x-\mu |}\sigma}$$ are derived.  I know that E$[X] = \mu$ and Var$[X] = 2\sigma^2$, but I don't understand how that happens.   Many thanks.
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The question states:      Consider a set of random variables $X_i$, where $i=1,...n$. Each $X_i$ is   normally distributed with mean $0$ and variance $1$, i.e. $X_i$ are $\mathcal N(0,1)$.   What is the mean and the variance of the random variable $Y$, where   $Y=X_1+...+X_n$.   How do I do this?
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To understand what I mean, let's use two well-known distributions: the normal and lognormal ones.  From the dataset point of view, if you take normally-distributed data and take their exponential, you get lognormally-distributed data (norm -exp-> lognorm).  So the "idea" was: why not using the PDF directy and transform the normal PDF to the lognormal one by taking the exponential of its x-axis. The reasoning was that this x-axis represents my data: taking the exponential of these data or the exponential of the x-axis of the corresponding PDF should do the same.  But I did not even bother to do it: mathematically, this is not how one can transform the normal PDF to the lognormal one!  What I don't understand is why just taking the exponential of the x-axis of the PDF does not work. I mean, not mathematically (I get that there is a $1/x$ in the lognormal PDF!), but what is conceptually wrong with "just stretching" the axis?
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... and why ?  Assuming $X_1$,$X_2$ are independent random-variables with mean $\mu_1,\mu_2$ and variance $\sigma^2_1,\sigma^2_2$ respectively. My basic statistics book tells me that the distribution of the $X_1-X_2$ has the following properties:    $E(X_1-X_2)=\mu_1-\mu_2$ $Var(X_1-X_2)=\sigma^2_1 +\sigma^2_2$   Now let's say $X_1$, $X_2$  are t-distributions with $n_1-1$, $n_2-2$ degrees of freedom. What is the distribution of $X_1-X_2$ ?   This question has been edited: The original question was "What are the degrees of freedom of the difference of two t-distributions ?". mpiktas has already pointed out that this makes no sense since $X_1-X_2$ is not t-distributed, no matter how approximately normal $X_1,X_2$ (i.e. high df) may be.
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The random variable $X$ is exponentially distributed with parameter $\lambda = \dfrac{1}{2}$. Also, define $Y = -2X + 1$. Calculate $E(Y)$ and $Var(Y)$.  So we have   $$f_X(x) = \dfrac{1}{2} e^{-\dfrac{1}{2}x} $$  if $x\ge0$ and $0$ otherwise.  But how to calculate $E(Y)$ and $Var(Y)$ ?
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Show that $(X_1,X_2)$ has a bivariate normal distribution with means $\mu_1, \mu_2$, variances $\sigma _1^2 $ and $\sigma _2^2$, and correlation coefficient $\rho $ if and only if every linear combination $t_1X_1+t_2X_2$ has a univariate normal distribution with mean $t_1\mu_1+t_2\mu_2$, and variances  $t_1^2\sigma _1^2+t_2^2\sigma _2^2+2t_1t_2\rho \sigma_{12}$ where $t_1, t_2$ are non zero constants  My work:  Joint mgf of $X_1 $and $X_2=e^{[t_1\mu_1+t_2\mu_2+\frac{1}{2}(t_1^2\sigma _1^2+t_2^2\sigma _2^2+2\rho \sigma_1 \sigma_1t_1t_2 )]} $  I am confused here to find the proper grouping to get the forward result.  For the backward result, I know I have to find the m.g.f. of $t_1X + t_2Y$, for any $t_1 $and $t_2$ real,and plug in the $E(t_1X + t_2Y)$ and $Var(t_1X + t_2Y)$.  I am struggling to get the correct backward result.  Second part:   Let $Y_i=X_i/\sigma_i, i=1,2$ Show that $Var(Y_1-Y_2)=2(1-\rho)$  I have found, $Var(Y_1)=\sigma_1, Var(Y_2)=\sigma_2$ since $X_1/\sigma_1=\sigma _1$  How can I prove that $Var(Y_1-Y_2)=2(1-\rho)$? Thank you so much for your help!
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Just wondering if it is possible to find the Expected value of x if it is normally distributed, given that  is below a certain value (for example, below the mean value).
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Find the distribution function of this multivariate normal variable where $X \sim N(0, 1)$ $$ \begin{pmatrix} X \\ (-1)^nX \\ \end{pmatrix} $$  I need this for a proof that marginal convergence in distribution does not imply joint convergence in distribution, but I am really rusty.   Any help is much appreciated.   Edit:  Explaining notation.  This is a sequence of two-dimensional random variable vectors, where $n$ takes values in the natural numbers. The distribution function should differ for odd and even values of $n$.
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The Multivariate Gaussian pdf is given by  $$(2\pi)^{-\frac{K}{2}} \det(\Sigma)^{-\frac{1}{2}} \exp({-\frac{1}{2}}(X-\mu)' \Sigma^{-1} (X-\mu)) $$  The wikipedia for multivariate Gaussians is    However I could not find a pdf for the multivariate lognormal distribution. Does it exist? If so, what is it?
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Suppose $X_n,X$ are random variables with $X_n \overset{L^2}{\rightarrow} X$. Does this imply $X_n^2 \overset{L^1}{\rightarrow} X^2$?  $X_n \overset{L^2}{\rightarrow} X$ means $E(|X_n-X|^2) \rightarrow 0$ and what we have to show is that $E(|X_n^2 - X^2|) \rightarrow 0$.
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Let $X\sim N(\mu,\sigma^2)$. Show that $Z=\frac{X-\mu}{\sigma}\sim N(0,1)$ using moment generating functions. \begin{align*} M_Z(t)&amp;=M_{\frac{X-\mu}{\sigma}}(t)\\ &amp;=M_{X-\mu}\left(\frac t\sigma\right)\\ &amp;=e^{-\mu t}M_X\left(\frac t\sigma\right)\\ &amp;=e^{-\mu t}\cdot e^{\frac{t\mu}{\sigma}+\frac{t^2}{2}}\\ &amp;=e^{-\mu t+\frac{t\mu}{\sigma}+\frac{t^2}{2}} \end{align*} I'm not sure how to conclude that $Z\sim N(0,1)$.
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We have the following Random variable:  $Y=\frac{\log Z}{\log Z+\log (1-W)}$ where both $W$ and $Z$ are $U(0,1)$. The question asks to calculate the variance of the random variable $Y$.   I think the random variable $Y$ will follow the exponential distribution with parameter $2$. How I obtained this result is as follows:  First, after a little simplification, the random variable $Y$ turned out to be $\log ZW$ which can further be written as $\log Z+ \log W$. Each of which follows exponential distribution. Hence, the variance of the random variable$Y$ can be obtained as $\frac{1}{4}$. What is wrong in this approach? Because at the back, the answer is $\frac{1}{12}$ as they are saying that $Y$ will follow uniform distribution.
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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.
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Consider a random variable $X$ whose probability density function is given by $$f_{X}(x) = \frac{\lambda}{2}\exp(-\lambda|x-\mu|)\quad\text{for}\quad x\in\textbf{R}, \lambda &gt; 0,\,\,\text{and}\,\,\mu\in\textbf{R}$$  Determine its moment generating function, $\textbf{E}(X)$ and $\textbf{Var}(X)$.  MY SOLUTION  According to the definition of moment generating function, we obtain the following result \begin{align*} M_{X}(t) &amp; = \frac{\lambda}{2}\int_{-\infty}^{+\infty}e^{tx}\exp(-\lambda|x-\mu|)\mathrm{d}x = \frac{\lambda}{2}\int_{-\infty}^{+\infty}\exp(tx-\lambda|x-\mu|)\mathrm{d}x\\\\ &amp; = \frac{\lambda}{2}\int_{-\infty}^{\mu}\exp(tx-\lambda(\mu -x))\mathrm{d}x + \frac{\lambda}{2}\int_{\mu}^{+\infty}\exp(tx-\lambda(x-\mu))\mathrm{d}x\\\\ &amp; = \frac{\lambda}{2}\int_{-\infty}^{\mu}\exp((t+\lambda)x-\lambda\mu))\mathrm{d}x + \frac{\lambda}{2}\int_{\mu}^{+\infty}\exp((t-\lambda)x+\lambda\mu)\mathrm{d}x\\\\ \end{align*}  Then I get stuck. Could someone shed some light on the problem?
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Let $X_1, X_2, \ldots X_n$ be random variables with mean $\mu$ and variance $\sigma^2$. We define the sample mean $\bar{X} = \sum_{i = 1}^n X_i$.  I am having trouble understanding why the variance of $\bar{X}$ is $\sigma^2 / n$.  Here is what I worked out thus far:  $$\begin{gather} \sigma_\bar{X}^2 = E((\bar{X} - \mu)^2) = E(\bar{X}^2 - 2\bar{X}\mu + \mu^2) = E(\bar{X}^2) - 2\mu E(\bar{X}) + E(\mu^2) = E(\bar{X}^2) - \mu. \end{gather}$$  However, I'm not sure what to do with the $E(\bar{X}^2)$.