# Probability Theory : convergence in distribution

I would need some help for a pratical exercise of probability about convergence of random variables.

Consider the following distribution function : $F^{X_{n}}(x) = \frac{e^{nx}}{e^{nx}+1} ; n \geq 1$. Proof there is a sequence of random variables $(X_{n}) ; n \geq 1$ , which law is given for all $n$ by $F^{X_{n}}$. Does this sequence converge in distribution ?

I cannot find such a sequence of random variables. Any ideas ? Thanks in advance for helping

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Since the function $F_1:x\mapsto\mathrm e^{x}/(1+\mathrm e^{x})$ is continuous and increasing from its limit $0$ at $-\infty$ to its limit $1$ at $+\infty$, it is a proper distribution function. Choose any random variable $X_1$ with distribution function $F_1$.

For every positive integer $n$, $F_n(x)=F_1(nx)$ for every real number $x$ hence $F_n$ is the distribution function of $X_n=X_1/n$. Since $X_1$ is almost surely finite, $X_n\to0$ almost surely, hence in distribution. (Alternatively, one can note that $F_n(x)\to0$ for every $x\lt0$ and $F_n(x)\to1$ for every $x\gt0$.)

Finally, one can realize $X_1$ as $X_1=\log(U/(1-U))$, where the random variable $U$ is uniformly distributed on $(0,1)$.

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$$F^{X_{n}}(x) = \frac{e^{nx}}{e^{nx}+1}=\int_{-\infty}^xf(t)dt$$ differentiate to get $$f(x)=\frac{ne^{nx}(e^{nx}+1)-ne^{nx}e^{nx}}{(e^{nx}+1)^2}=\frac{ne^{nx}}{(1+e^{nx})^2}$$ we can check that this is a probability density $$\int_{-\infty}^{\infty}f(x)dx=\int_1^{\infty}\frac{1}{u^2}du=1$$

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Thanks for the answer ! –  Hernium Dec 30 '11 at 22:31
Strictly you also need to check $f(x)$ is non-negative, though this is obvious from your second line. Alternatively you could show that $F^{X_{n}}(x)$ weakly increases from a lower limit of 0 to a higher limit of 1. –  Henry Dec 30 '11 at 23:03