# Tag Info

3

Looks right to me. You can calculate that integral by doing integration by parts three times. Or if you already know $E(X^2),$ once is enough, because integration by parts gives you an expression in terms of $E(X^2).$ That’s how I would expect you to do the question (at least, if you don’t know the integral already.) A more advanced trick lets you calculate ...

3

If $X$ is a random variable with continuous law, then for any real number $y$, $P(X = y) = 0$. If $X$ and $Y$ are independent random variables with continuous laws, then $P(X = Y) = 0$. To prove this, you can use the joint distribution and Fubini-Tonelli: \begin{align*} P(X=Y) &= \iint_{\mathbb R^2}1_{\{x = y\}}\mathcal L_X(dx)\mathcal L_Y(dy)\\ &= \...

3

Another way is to "re-cycle" your result which, btw, should have been: $$f_Z = \alpha_1\alpha_2\frac{e^{-\alpha_1 x}-e^{-\alpha_2 x}}{\alpha_2-\alpha_1}$$ (otherwise, you have a negative valued density). Here is how: setting $\alpha_2=\alpha_1+\epsilon$ and making $\epsilon$ tend to $0$, for a fixed $x$, we get: $$\lim_{\epsilon \to 0} f_Z(x) = \... 3 You have done your algebra incorrectly. In your answer you are dividing by zero (which I might add doesn't mean the convolution vanishes as you say). Also, you didn't mention it, but I'm also going to assume X and Y are independent (otherwise you can't just take the convolution of the respective pdfs). The exponential distribution has pdf$$f(x) = \begin{...

3

$$X(X-1) = X^2-X = X^2 - 2\cdot X\cdot \frac{1}{2} + \left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2 = \left(X-\frac{1}{2}\right)^2-\frac{1}{4} \Rightarrow$$ \begin{aligned} F_Y(t) &= \Pr\left(Y \leq t\right) = \Pr\left(X(X-1) \leq t\right) = \Pr\left(\left(X-\frac{1}{2}\right)^2-\frac{1}{4} \leq t\right) = \Pr\left(\left(X-\frac{1}{2}\right)^... 2 If you are uncomfortable with a manual change of variables on the density, then use the cumulative distribution functions \mathbb P(X \le x) = 1-\exp \left( { - \frac{x}{\lambda }} \right) if x \ge 0, and =0 if x \le 0 \mathbb P(Y \le e^{ax}) = 1-\exp \left( { - \frac{x}{\lambda }} \right) if x \ge 0, and =0 if x \lt 0 \mathbb P(Y \le y) =... 2 The likelihood isL(\theta) = \theta^{-n} \exp \left(- \frac{1}{\theta} \sum x_i \right). \tag{1}$$Your calculation of the derivative of the log-likelihood is incorrect; it should be$$\frac{d}{d\theta}[\log L(\theta)] = -n\theta + \frac{\sum x_i}{\theta^2}.$$This yields the unrestricted MLE$$\hat \theta = \frac{\sum x_i}{n},$$and if \theta were ... 1 Correct. and 3. Incorrect. Density function for 2. (using convolution) is f_V(v)=\alpha\beta\int\limits_0^ve^{-\beta x}e^{-\alpha(v-x)}dx=\frac{\alpha\beta}{\alpha-\beta}(e^{-\beta v}-e^{-\alpha v}). Density function for 3. F_W(w)=P(W\le w)=P(X^2\le w)=P(X\le\sqrt{w})$$=1-e^{-\alpha \sqrt{w}}$, giving$f_W(w)=\frac{\alpha}{2\sqrt{w}}e^{-\alpha \sqrt{w}...

1

This is the maximum of a set of i.i.d. exponentially distributed r.v.s $X_1, \dots, X_m$, each with cumulative distribution function $F(x) = 1 - e^{-\lambda x}$ for $x > 0$. Therefore the c.d.f. of $T_m$ is $$P(T_m \le x) = P(\forall i \, X_i \le x) = \prod_i P(X_i \le x) = F(x)^m = \left(1 - e^{-\lambda x} \right)^m$$ and the probability density ...

1

We assume the random sample to be $\{2,4,5,5,x\}$ Since, we are given the sample variance which is defined to be: $$s^2 = \frac{1}{n-1}\sum\limits_{i=1}^n{(x-\bar x)^2}$$ $$\bar x=\frac{16+x}{5}$$ Therefore, we get, $$6 = \left(\frac{6+x}{5}\right)^2 + \left(\frac{4-x}{5}\right) + 2\left(\frac{9-x}{5}\right)^2 + 16\left(\frac{x - 4}{5}\right)^2$$ which ...

1

Because $X$ is exponential, we know $P(X \geq 0) = 1$, which means $P(Y \geq 1) = 1$, so the PDF of $Y$ will be zero outside $[1, \infty)$. We know that the CDF of $X$ is $$P(X \leq z) = \int_{0}^z \lambda^{-1} e^{-\lambda^{-1} x} \, dx = 1 - e^{-\lambda^{-1} z}$$ Then write down the CDF of $Y$ and use what you know: for $y \geq 1$, \begin{align*} P(Y \leq ...

1

As an alternative, if you want to directly derive $f_Y(y)$, consider that $$y=e^{ax}$$ $$x=\frac{1}{a}\log y$$ $$|x'|=\frac{1}{ay}$$ thus $$f_Y(y)=\frac{1}{ay\lambda}\left[e^{\log(1/y)}\right]^{1/(a\lambda)}=\frac{1}{a\lambda y^{1/(a\lambda)+1}}$$ This is your density setting $a=3$ and $\lambda=1$

1

Yes. Distribution $\mathsf{Exp}(\lambda=3)$ with mean $1/3$ tends to give smaller values than $\mathsf{Exp}(\lambda=1)$ with mean $1.$ So you are correct to reject $H_0: \lambda = 3$ against $H_0: \lambda=1$ for large observed values $X.$ In particular, the critical value for a test at level 5% is $c = 0.999.$ That is, you would reject $H_0$ for observed $X \... 1 You can use the Gamma r.v. density function, i.e., if$X\sim Gamma(n, \lambda)$, then $$\int_{\mathbb{R}_+} \frac{\lambda ^ n}{ \Gamma(n)}e^{-\lambda x} x^{n-1} dx = 1$$ In your case $$\int x ^ 3 \cdot 2e ^{-2x} dx = 2\cdot \frac{3!}{2^4}\int\frac{2^4}{3!}x ^ 3 e ^{-2x}dx = \frac{3!}{2^3}$$ where$\Gamma(n) = (n-1)!$for$n\in \mathbb{N}$. 1 Part I: Intuitive answer. This was already describe by @BruceET. I'll just say it again with some different words... The answer is$1/3$. The exponential distribution is "memoryless". From the time you begin being serviced, think of the two others exponential clocks as being "reset": the additional time required for service by the each ... 1 By the no memory property, it makes no difference how long each of the customers ahead of you have been 'in service' with their tellers. As soon as you start with one of the tellers, by the no-memory property, all three waiting times to finish have the same distribution. So you have one chance in three of being last to leave. The simulation below is not a ... 1 I'll write$\widetilde{\mu}$to denote the random variable and$\mu\$ to denote particular values. We're given the following probability density functions: \begin{align} f_{X\mid \widetilde{\mu}}(x\mid \mu)&=e^{-\mu}{\mu^x\over x!}{\bf 1}_{x\in\{0,1,2,...\}}\\[2ex] f_{\widetilde{\mu}}(\mu)&=e^{-\mu}{\bf 1}_{\mu>0} \end{align} Therefore the ...

Only top voted, non community-wiki answers of a minimum length are eligible