# Convergence of $(X_k+1)^2$ using characteristic functions

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?

• $e^{it}$, $e^{2itX_k}$ and $e^{itX_k^{2}}$ are not indepedent. How can you use $EXYZ=EXEYEZ$ without independece? Aug 30, 2020 at 12:04
• The hint sounds strange to me. Aug 30, 2020 at 12:08

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 $$\mathbb P(Y_k \le 2) \to 1$$.
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.