Search Results

This theorem is named "integral transformation theorem " As you can see after few passages you get $$F_Y(y)=F_X[F_X^{-1}(y)]$$ now it is evident that applying both $F$ and $F^{-1}$ to any $y$ they can …

The probability to mark the candies exactly of 5 older kids is the probability to chose 5 older kids among the 35 total, say $$\frac{\binom{15}{5}\binom{20}{0}}{\binom{35}{5}}$$ (an hypergeometric...) …

Note that $$P(S|S>0)=\frac{1}{1-e^{-0.3}}\frac{e^{-0.3}0.3^s}{s!}$$ Thus $$P(S=1|S>0)=\frac{0.3}{e^{0.3}-1}\approx 85.75\%$$

No, it does not. I do not know why you think so. As you know $$Y\sim \chi_{(m)}^2=Gamma\Bigg(\frac{m}{2};\frac{1}{2}\Bigg)$$ Thus $$\frac{1}{m}Y\sim Gamma\Bigg(\frac{m}{2};\frac{m}{2}\Bigg)$$ The proo …

I do not know if your really have to calculate this expectation....but reading the title, if you are only interested in detecting if your MLE is unbiased, the answer is the following By Jensen's inequ …

so I need to score 15% or higher on the exam? Or am I missing something? Your answer is not "totally" to waste. first you did not take 25% in the project score but you took $0.3\times0.85=25.5\%$ …

-$n$ random sample you get $$V(T)\geq \frac{k^2\theta^{2k}}{n}$$ b) Observe that $T=\overline{X}_n$ is unbiased estimator for $\theta$ and it is a function of $S=\Sigma_iX_i$, complete and sufficient statistics … can be useful to calculate the expectation of $$T_k=[\Sigma_iX_i]^k$$ This is easy because $\Sigma_iX_i$ has a known distribution (it's a gamma) and $T_k$ is function of $S$, complete and sufficient statistics …

There are several ways to show that $T=\overline{X}_n$ is the UMVUE for $p$ This is one; I chose this because it allow us to do further considerations about that important estimator: $$\mathbb{E}[\ove …

You can manipulate your expression finding the following critical region $$\Sigma_i \left(X_i+\frac{1}{3}\right)^2 \geq k^*$$ Now, under $H_0$, the distribution is known: it's a noncentral chi-squared …

Yes you did. the lower bound for unbiased estimators of $\lambda$ is $V(T)\geq\frac{\lambda^2}{n}$ Using Lehmann-Scheffé Lemma you can find the UMVUE estimator of $\lambda$ $\hat{\lambda}=\frac{n-1 …

Exponential distribution does not belong to a location - scale family. Actually it belongs to a "scale family" Definition A scale family of distributions has densities of the form $$ \bbox[5px,border …

the likelihood is the following $$p(\mathbf{x}|\theta)\propto \theta^3(1-\theta)^8$$ which is a Binomial model What does it suggest to you?

You know that $$P(X\leq 4)=0.8413$$ Knowing also that $E(X^2)-\mu^2=\sigma^2$ you get $$P\left(Z \leq\frac{4-\mu}{\sqrt{10-\mu^2} } \right)=0.8413$$ Using the tables you get $$\frac{4-\mu}{\sqrt{10-\m …

The gamma density is the following, for $x>0$ $$f_X(x,a,b)=\frac{b^a}{\Gamma(a)}x^{a-1}e^{-bx}$$ it is easy to prove that its integral cannot converge if $a,b$ are not both positive

Any estimator is a function of (only of) the data. thus tipically $$\hat{\theta}=t(\mathbf{x})$$ and thus, by definition, $$\mathbb{E}[\hat{\theta}]=\int_T tf(t)dt$$ Example... given a simple random …