Let $X_k$ be exponentially distributed with $\lambda=\sqrt k$, that is with the density $f_k(x)=\lambda _k e^{-\lambda _k}$ where $x>0$. Find to which distribution converges $(X_k+1)^2$ and calculate the limit : $\lim_{n\to\infty} P((X_n+1)^2 \leq2)$.
Hint: use characteristic functions.
I know that the characteristic function of exponential distribution is $\varphi _x (t)=\frac{1}{1-\frac{it}{\lambda}}$ and using that I tried to find a characteristic function of $(X_k+1)^2$
$\varphi _{(X_k+1)^2}(t)=E[e^{it(X_k+1)^2}]=E[e^{itX_k^2}*e^{2itX_k}*e^{it}]=E[e^{itX_k^2}]*E[e^{2itX_k}]*E[e^{it}]=
\varphi _{X_k^2}(t)*\varphi _{X_k} (2t)*e^{it} $
Now $\varphi _{X_k} (2t)=\frac{1}{1-\frac{2it}{\sqrt k}}$ and here i got stucked because i do not know what to to with $\varphi _{X_k^2}(t)$
Does this approach makes sense and if yes how can i proceed from here?
Best Answer
Solution using only characteristic functions:
Let $X_k \sim \mathcal Exp(\sqrt k)$ and let $Y_k=(X_k+1)^2$. We'll check the convergence of $\varphi_{Y_k}$.
$$ \varphi_{Y_k}(t) = \mathbb E[\exp(itY_k)] = \int_{0}^\infty \exp(it(x+1)^2) \sqrt k \exp(-\sqrt k x) dx = \int_0^\infty \exp(it(\frac{u}{\sqrt{k}}+1)^2)\exp(-u)du$$ where we substituted $u=\sqrt{k}x$. Now note that function under integral is bounded (w.r.t norm) above by $\exp(-u)$ which is integrable on $(0,\infty)$. Moreover, we have poinwise convergence to $\exp(it)\exp(-u)$ hence by dominated convergence theorem $\varphi_{Y_k}(t) \to \exp(it)\int_0^\infty \exp(-u)du = \exp(it) = \varphi_Y(t)$, where $Y$ is random variable with $\delta_1$ distribution (distribution focused at point $1$, that is $\mathbb P(Y=1)=1$). Indeed $\mathbb E[\exp(itY)] = \exp(it)$. Hence by Levy theorem, $Y_k \to Y$ in distribution.
Now $\mathbb P(Y_k \le 2) \to \mathbb P(Y \le 2)$ since $2$ is a continuity point of $\delta_1$ distribution, so $\mathbb P(Y_k \le 2) \to 1$.
Edit:
Easier way would be to use characteristic function of $X_k$ that is $\varphi_{X_k}(t) = \frac{\sqrt{k}}{\sqrt{k}-it} \to 1$ as $k \to \infty$, so that $X_k \to 0$ in distribution ($0$ in the sense of random variable with $\delta_0$ distribution). Now, since convergence in distribution behaves well with continuous functions, we get $Y_k \to 1$ in distribution. The rest as above.