Central Limit Theorems
In this post, we will explore the Central Limit Theorem (CLT), a fundamental result in probability theory that describes the behavior of the mean of a large number of random variables. The CLT states that, under certain conditions, the distribution of the sample mean approaches a normal distribution as the sample size increases. There are several versions of the CLT, each with its own set of assumptions and conclusions. We will discuss the Lindeberg–Lévy CLT (Classical CLT), the Lyapunov CLT, and the Lindeberg-Feller CLT.
Lindeberg–Lévy CLT (Classical CLT)
Theorem (Lindeberg–Lévy CLT). Let \(X_1, X_2, \ldots\) be i.i.d. random variables with mean \(\mu\) and variance \(\sigma^2 < \infty\). Define the sample mean as \(\bar{X}_n = \dfrac{1}{n} \sum_{i=1}^{n} X_i\). Then, as \(n \to \infty\),
\[\dfrac{\sqrt{n}(\bar{X}_n - \mu)}{\sigma} \xrightarrow{d} \mathcal{N}(0, 1),\]where \(\xrightarrow{d}\) denotes convergence in distribution.
Proof. To prove the Lindeberg–Lévy CLT, we can use the method of characteristic functions. The characteristic function of a random variable \(X\) is defined as \(\varphi_X(t) = \mathbb{E}[e^{itX}]\). The following theorem connects convergence in distribution of the sequence of random variables with pointwise convergence of their characteristic functions.
Lévy’s Continuity Theorem. Let \(X_n\) be a sequence of random variables and \(X\) be a random variable. Then \(X_n \xrightarrow{d} X\) if and only if \(\varphi_{X_n}(t) \to \varphi_X(t)\) for all \(t \in \mathbb{R}\) where \(\varphi_X(t)\) is continuous.
Define the standardized random variables \(Z_i = \dfrac{X_i - \mu}{\sigma}\). It follows that \(Z_i\) are i.i.d. with \(\mathbb{E}[Z_i] = 0\) and \(\mathrm{Var}(Z_i) = 1\). Then, we can express \(\dfrac{\sqrt{n}(\bar{X}_n - \mu)}{\sigma}\) as the sum of the standardized random variables, i.e.,
\[\dfrac{\sqrt{n}(\bar{X}_n - \mu)}{\sigma} = \frac{1}{\sqrt{n}}\sum_{i=1}^{n} Z_i := S_n.\]Now we consider the characteristic function of \(S_n\). Since \(Z_i\) are i.i.d. with mean 0 and variance 1, the characteristic function of \(S_n\) is given by
\[\varphi_{S_n}(t) = \left( \varphi_{Z_1}(t/\sqrt{n}) \right)^n.\]For any fixed \(t\), by the Taylor expansion of the characteristic function \(\varphi_{Z_1}(t)\) around 0, we have
\[\begin{align*} \varphi_{Z_1}(t/\sqrt{n}) &= \mathbb{E}[e^{\frac{itZ_1}{\sqrt{n}}}] \\ &= \mathbb{E}[1 + \frac{itZ_1}{\sqrt{n}} + \frac{1}{2}\left(\frac{itZ_1}{\sqrt{n}}\right)^2] + o(\frac{1}{n}) \\ &= 1 - \frac{t^2}{2n} + o(\frac{1}{n}). \end{align*}\]Therefore, the limiting characteristic function of the sum is
\[\lim_{n \to \infty} \varphi_{S_n}(t) = \lim_{n \to \infty} \left( 1 - \frac{t^2}{2n} + o(\frac{1}{n}) \right)^n = e^{-\frac{t^2}{2}}.\]Since \(e^{-\frac{t^2}{2}}\) is the characteristic function of the standard normal distribution \(\mathcal{N}(0, 1)\), by Lévy’s Continuity Theorem, we conclude that \(\dfrac{\sqrt{n}(\bar{X}_n - \mu)}{\sigma} \xrightarrow{d} \mathcal{N}(0, 1)\) as \(n \to \infty\). This completes the proof.
Corollay (Lindeberg–Lévy CLT with unknown variance). Let \(X_1, X_2, \ldots\) be i.i.d. random variables with mean \(\mu\) and unknown variance \(\sigma^2 < \infty\). Define the sample mean as \(\bar{X}_n = \dfrac{1}{n} \sum_{i=1}^{n} X_i\) and the sample variance as \(\hat{\sigma}_{n}^{2} = \dfrac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X}_n)^2\). Then, as \(n \to \infty\),
\[\dfrac{\sqrt{n}(\bar{X}_n - \mu)}{\hat{\sigma}_{n}} \xrightarrow{d} \mathcal{N}(0, 1).\]Proof. To prove this corollary, we can use Slutsky’s Theorem and Continuous Mapping Theorem.
Slutsky’s Theorem. Let \(X_n\) and \(Y_n\) be sequences of random variables, and let \(X\) and \(Y\) be random variables. If \(X_n \xrightarrow{d} X\) and \(Y_n \xrightarrow{p} Y\), then \(X_n + Y_n \xrightarrow{d} X + Y\), \(X_n Y_n \xrightarrow{d} XY\), and if \(Y \neq 0\), then \(\frac{X_n}{Y_n} \xrightarrow{d} \frac{X}{Y}\).
Continuous Mapping Theorem. Let \(X_n\) be a sequence of random variables and \(X\) be a random variable. If \(X_n \xrightarrow{d} X\) and \(g\) is a function that is continuous at all points in the support of \(X\), then \(g(X_n) \xrightarrow{d} g(X)\). This also holds for convergence in probability and almost sure convergence.
By the Lindeberg–Lévy central limit theorem,
\[\frac{\sqrt{n}(\bar{X}_n-\mu)}{\sigma} \xrightarrow{d} \mathcal{N}(0,1).\]Thus it remains to show that
\[\hat{\sigma}_n \xrightarrow{p} \sigma.\]Once this is established, Slutsky’s theorem will imply that
\[\frac{\sqrt{n}(\bar{X}_n-\mu)}{\hat{\sigma}_n} = \frac{\sqrt{n}(\bar{X}_n-\mu)}{\sigma} \cdot \frac{\sigma}{\hat{\sigma}_n} \xrightarrow{d} \mathcal{N}(0,1).\]We now prove the consistency of \(\hat{\sigma}_n^2\).
First, use the identity
\[\sum_{i=1}^n (X_i-\bar{X}_n)^2 = \sum_{i=1}^n (X_i-\mu)^2 - n(\bar{X}_n-\mu)^2.\]Therefore,
\[\hat{\sigma}_n^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i-\mu)^2 - \frac{n}{n-1}(\bar{X}_n-\mu)^2.\]We now analyze the two terms separately. Since the random variables \((X_i-\mu)^2\) are i.i.d. and satisfy \(\mathbb{E}[(X_i-\mu)^2] = \sigma^2 < \infty,\) the weak law of large numbers gives
\[\frac{1}{n}\sum_{i=1}^n (X_i-\mu)^2 \xrightarrow{p} \sigma^2.\]Because \(\frac{n}{n-1} \to 1,\) it follows that
\[\frac{1}{n-1}\sum_{i=1}^n (X_i-\mu)^2 = \frac{n}{n-1}\cdot \frac{1}{n}\sum_{i=1}^n (X_i-\mu)^2 \xrightarrow{p} \sigma^2.\]Next, by the weak law of large numbers, \(\bar{X}_n \xrightarrow{p} \mu.\) Hence, by the continuous mapping theorem,
\[(\bar{X}_n-\mu)^2 \xrightarrow{p} 0.\]Since also \(\frac{n}{n-1} \to 1,\) we obtain
\[\frac{n}{n-1}(\bar{X}_n-\mu)^2 \xrightarrow{p} 0.\]Combining the two convergence results above, we conclude that \(\hat{\sigma}_n^2 \xrightarrow{p} \sigma^2.\) Because \(\sigma^2>0\) and the square-root function is continuous on \((0,\infty)\), the continuous mapping theorem yields \(\hat{\sigma}_n \xrightarrow{p} \sigma.\) Equivalently, \(\frac{\sigma}{\hat{\sigma}_n} \xrightarrow{p} 1.\)
Finally, by Slutsky’s theorem,
\[\frac{\sqrt{n}(\bar{X}_n-\mu)}{\hat{\sigma}_n} = \frac{\sqrt{n}(\bar{X}_n-\mu)}{\sigma} \cdot \frac{\sigma}{\hat{\sigma}_n} \xrightarrow{d} Z \cdot 1 = Z,\]where \(Z \sim \mathcal{N}(0,1)\). Therefore, \(\frac{\sqrt{n}(\bar{X}_n-\mu)}{\hat{\sigma}_n} \xrightarrow{d} \mathcal{N}(0,1).\) This completes the proof.
Now that we have established the Lindeberg–Lévy CLT, we can explore more general versions of the CLT that relax some of the assumptions made in the classical version. The Lyapunov CLT and the Lindeberg-Feller CLT are two such generalizations that allow for non-identically distributed random variables and provide conditions under which the CLT still holds. Both relax the requirement of identical distribution and provide conditions on the moments of the random variables to ensure convergence to a normal distribution.
Lindeberg-Feller CLT
Theorem (Lindeberg-Feller CLT). Consider the following triangular array of random variables: \(\{X_{n,k} : 1 \leq k \leq n, n \geq 1\}.\) Assume that for each fixed \(n\), the random variables \(X_{n,1}, X_{n,2}, \ldots, X_{n,n}\) are independent. Without loss of generality, we assume that \(\mathbb{E}[X_{n,k}] = 0\) for all \(n\) and \(k\). We will denote the variance of \(X_{n,k}\) as \(\sigma_{n,k}^2 = \mathrm{Var}(X_{n,k})\), assuming that \(\sigma_{n,k}^2 < \infty\) for all \(n\) and \(k\). Define the partial sums as \(S_n = \sum_{k=1}^{n} X_{n,k}\). Without loss of generality, we also assume that the total variance of the sum is normalized to 1, i.e., \(\sum_{k=1}^{n} \sigma_{n,k}^2 = 1\) for all \(n\). If the following condition, known as the Lindeberg condition, is satisfied:
\[\forall \varepsilon > 0, \quad \lim_{n \to \infty} \sum_{k=1}^{n} \mathbb{E}[X_{n,k}^2 \cdot \mathbf{1}_{\{|X_{n,k}| > \varepsilon\}}] = 0,\]then as \(n \to \infty\), we have \(S_n \xrightarrow{d} \mathcal{N}(0, 1).\)
Proof. To prove the Lindeberg-Feller CLT, we can still use the method of characteristic functions, similar to the proof of the Lindeberg–Lévy CLT. We will need the following lemmas in our proof.
Lemma 1. Let \(\{a_{j}\}_{j=1}^{n}\) and \(\{b_{j}\}_{j=1}^{n}\) be two sequences of complex numbers such that \(\lvert a_{j} \rvert \leq 1\) and \(\lvert b_{j} \rvert \leq 1\) for all \(j\). Then, \(\lvert \prod_{j=1}^{n} a_{j} - \prod_{j=1}^{n} b_{j} \rvert \leq \sum_{j=1}^{n} \lvert a_{j} - b_{j} \rvert.\)
Proof of Lemma 1. We can prove this lemma by induction on \(n\). For \(n=1\), the inequality holds trivially. Assume that the inequality holds for some \(n \geq 1\). Then, for \(n+1\), we have
\[\begin{align*} \lvert \prod_{j=1}^{n+1} a_{j} - \prod_{j=1}^{n+1} b_{j} \rvert & = \lvert a_{n+1} \prod_{j=1}^{n} a_{j} - b_{n+1} \prod_{j=1}^{n} b_{j} \rvert \\ & = \lvert a_{n+1} \prod_{j=1}^{n} a_{j} - a_{n+1} \prod_{j=1}^{n} b_{j} + a_{n+1} \prod_{j=1}^{n} b_{j} - b_{n+1} \prod_{j=1}^{n} b_{j} \rvert \\ & \leq \lvert a_{n+1} \rvert \cdot \lvert \prod_{j=1}^{n} a_{j} - \prod_{j=1}^{n} b_{j} \rvert + \lvert a_{n+1} - b_{n+1} \rvert \cdot \lvert \prod_{j=1}^{n} b_{j} \rvert \\ & \leq \lvert \prod_{j=1}^{n} a_{j} - \prod_{j=1}^{n} b_{j} \rvert + \lvert a_{n+1} - b_{n+1} \rvert \\ & \leq \sum_{j=1}^{n} \lvert a_{j} - b_{j} \rvert + \lvert a_{n+1} - b_{n+1} \rvert \\ & = \sum_{j=1}^{n+1} \lvert a_{j} - b_{j} \rvert, \end{align*}\]where we used the fact that \(\lvert a_{j} \rvert \leq 1\) and \(\lvert b_{j} \rvert \leq 1\) for all \(j\) from the third to the fourth line, and the induction hypothesis from the fourth to the fifth line. This completes the proof of Lemma 1.
Lemma 2. Let \(X\) be a random variable with mean 0 and variance \(\sigma^2 < \infty\). Denote the characteristic function of \(X\) as \(\varphi_X(t) = \mathbb{E}[e^{itX}]\). Then, for any \(t \in \mathbb{R}\), we have
\[\lvert \varphi_X(t) - (1 + i t \mathbb{E}[X] - \frac{t^2}{2} \mathbb{E}[X^2]) \rvert \leq t^2 \mathcal{E}(t),\]where \(\mathcal{E}(t) = \mathbb{E}\left[\min \{X^2, \dfrac{|X|^3}{6}|t|\} \right]\).
Proof of Lemma 2. We first prove a pointwise inequality. For every \(u \in \mathbb R\), Taylor’s theorem gives
\[e^{iu} = 1 + iu - \frac{u^2}{2} + R(u),\]where
\[|R(u)| \leq \frac{|u|^3}{6}.\]Hence,
\[\left| e^{iu} - 1 - iu + \frac{u^2}{2} \right| \leq \frac{|u|^3}{6}.\]On the other hand, using Taylor’s theorem only up to first order,
\[e^{iu} = 1 + iu + \widetilde R(u),\]where \(|\widetilde R(u)| \leq \frac{u^2}{2}.\)
Therefore, \(\begin{aligned} \left| e^{iu} - 1 - iu + \frac{u^2}{2} \right| &\leq |e^{iu} - 1 - iu| + \frac{u^2}{2} \\ &\leq \frac{u^2}{2} + \frac{u^2}{2} \\ &= u^2. \end{aligned}\)
Combining the two bounds, we obtain \(\left| e^{iu} - 1 - iu + \frac{u^2}{2} \right| \leq \min\left\{ u^2, \frac{|u|^3}{6} \right\}.\)
Now apply this inequality with \(u=tX\). Since \(\mathbb E[X]=0\), we have
\[\begin{aligned} \left| \varphi_X(t) - \left(1 - \frac{t^2}{2}\mathbb E[X^2]\right) \right| &= \left| \mathbb E\left[ e^{itX} - 1 - itX + \frac{t^2X^2}{2} \right] \right| \\ &\leq \mathbb E\left[ \left| e^{itX} - 1 - itX + \frac{t^2X^2}{2} \right| \right] \\ &\leq \mathbb E\left[ \min\left\{ t^2X^2, \frac{|t|^3|X|^3}{6} \right\} \right] \\ &= t^2 \mathbb E\left[ \min\left\{ X^2, \frac{|t||X|^3}{6} \right\} \right]. \end{aligned}\]Lemma 3. (Decaying Variance Condition). Under the Lindeberg condition, we have \(\max_{1 \leq j \leq n} \sigma_{n,j}^2 \to 0\) as \(n \to \infty\).
Proof of Lemma 3. To prove this lemma, we will use the Lindeberg condition to show that the maximum variance among the random variables in each row of the triangular array must converge to 0 as \(n\) goes to infinity.
\[\begin{align*} \sigma_{n,j}^2 & = \mathbb{E}[X_{n,j}^2] \\ & = \mathbb{E}[X_{n,j}^2 \cdot \mathbf{1}_{\{|X_{n,j}| \leq \varepsilon\}}] + \mathbb{E}[X_{n,j}^2 \cdot \mathbf{1}_{\{|X_{n,j}| > \varepsilon\}}] \\ & \leq \varepsilon^2 + \mathbb{E}[X_{n,j}^2 \cdot \mathbf{1}_{\{|X_{n,j}| > \varepsilon\}}]. \end{align*}\] \[\begin{align*} \max_{1 \leq j \leq n} \sigma_{n,j}^2 & \leq \varepsilon^2 + \max_{1 \leq j \leq n} \mathbb{E}[X_{n,j}^2 \cdot \mathbf{1}_{\{|X_{n,j}| > \varepsilon\}}] \\ & \leq \varepsilon^2 + \sum_{j=1}^{n} \mathbb{E}[X_{n,j}^2 \cdot \mathbf{1}_{\{|X_{n,j}| > \varepsilon\}}]. \end{align*}\]Therefore, by the Lindeberg condition, we have
\[\limsup_{n \to \infty} \max_{1 \leq j \leq n} \sigma_{n,j}^2 \leq \varepsilon^2.\]Since \(\varepsilon > 0\) is arbitrary, we conclude that \(\limsup_{n \to \infty} \max_{1 \leq j \leq n} \sigma_{n,j}^2 = 0\), which implies that \(\max_{1 \leq j \leq n} \sigma_{n,j}^2 \to 0\) as \(n \to \infty\). This completes the proof of Lemma 3.
Now we are ready to prove the Lindeberg-Feller CLT. We will show that the characteristic function of \(S_n\) converges pointwise to the characteristic function of \(\mathcal{N}(0, 1)\), which is given by \(\varphi(t) = e^{-\frac{t^2}{2}}\) for all \(t \in \mathbb{R}\). By Lévy’s Continuity Theorem, this will imply that \(S_n \xrightarrow{d} \mathcal{N}(0, 1)\) as \(n \to \infty\). We denote the characteristic function of \(S_n\) as \(\varphi_{S_n}(t) = \mathbb{E}[e^{itS_n}]\). Since \(X_{n,1}, X_{n,2}, \ldots, X_{n,n}\) are independent, we have
\[\varphi_{S_n}(t) = \prod_{j=1}^n \varphi_{X_{n,j}}(t).\]By Lemma 1, we have
\[\begin{align*} &\quad \lvert \varphi_{S_n}(t) - e^{-\frac{t^2}{2}} \rvert \\ & = \lvert \prod_{j=1}^n \varphi_{X_{n,j}}(t) - \prod_{j=1}^n e^{-\frac{t^2 \sigma_{n,j}^2}{2}} \rvert \\ & \leq \sum_{j=1}^n \lvert \varphi_{X_{n,j}}(t) - e^{-\frac{t^2 \sigma_{n,j}^2}{2}} \rvert \\ & = \sum_{j=1}^n \lvert \varphi_{X_{n,j}}(t) - \left(1 - \frac{t^2 \sigma_{n,j}^2}{2}\right) + \left(1 - \frac{t^2 \sigma_{n,j}^2}{2}\right) - e^{-\frac{t^2 \sigma_{n,j}^2}{2}} \rvert \\ & \leq \sum_{j=1}^n \lvert \varphi_{X_{n,j}}(t) - \left(1 - \frac{t^2 \sigma_{n,j}^2}{2}\right) \rvert + \sum_{j=1}^n \lvert \left(1 - \frac{t^2 \sigma_{n,j}^2}{2}\right) - e^{-\frac{t^2 \sigma_{n,j}^2}{2}} \rvert \\ & =: \sum_{j=1}^n A_{n,j} + \sum_{j=1}^n B_{n,j}, \end{align*}\]where we defined \(A_{n,j} = \lvert \varphi_{X_{n,j}}(t) - \left(1 - \frac{t^2 \sigma_{n,j}^2}{2}\right) \rvert\) and \(B_{n,j} = \lvert \left(1 - \frac{t^2 \sigma_{n,j}^2}{2}\right) - e^{-\frac{t^2 \sigma_{n,j}^2}{2}} \rvert\) for all \(j\) and \(n\).
It suffices to show that both \(\sum_{j=1}^n A_{n,j}\) and \(\sum_{j=1}^n B_{n,j}\) converge to 0 as \(n \to \infty\).
First, we analyze the term \(\sum_{j=1}^n A_{n,j}\). By Lemma 2, we have
\[\begin{align*} A_{n,j} & \leq t^2 \mathbb{E}\left[\min \{X_{n,j}^2, \dfrac{|X_{n,j}|^3}{3!}|t|\} \right] \\ & = t^2 \mathbb{E}\left[\min \{X_{n,j}^2, \dfrac{|X_{n,j}|^3}{3!}|t|\} \cdot \mathbf{1}_{\{|X_{n,j}| \leq \varepsilon\}} \right] + t^2 \mathbb{E}\left[\min \{X_{n,j}^2, \dfrac{|X_{n,j}|^3}{3!}|t|\} \cdot \mathbf{1}_{\{|X_{n,j}| > \varepsilon\}} \right] \\ &\leq \lvert t \rvert^3 \epsilon \mathbb{E}\left[\dfrac{|X_{n,j}|^2}{3!} \cdot \mathbf{1}_{\{|X_{n,j}| \leq \varepsilon\}} \right] + t^2 \mathbb{E}\left[X_{n,j}^2 \cdot \mathbf{1}_{\{|X_{n,j}| > \varepsilon\}} \right] \\ & \leq \frac{1}{3!} \lvert t \rvert^3 \epsilon \sigma_{n,j}^2 + t^2 \mathbb{E}\left[X_{n,j}^2 \cdot \mathbf{1}_{\{|X_{n,j}| > \varepsilon\}} \right]. \end{align*}\]Therefore, we have
\[\begin{align*} \sum_{j=1}^n A_{n,j} & \leq \frac{1}{3!} \lvert t \rvert^3 \epsilon \sum_{j=1}^n \sigma_{n,j}^2 + t^2 \sum_{j=1}^n \mathbb{E}\left[X_{n,j}^2 \cdot \mathbf{1}_{\{|X_{n,j}| > \varepsilon\}} \right] \\ & = \frac{1}{3!} \lvert t \rvert^3 \epsilon + t^2 \sum_{j=1}^n \mathbb{E}\left[X_{n,j}^2 \cdot \mathbf{1}_{\{|X_{n,j}| > \varepsilon\}} \right]. \end{align*}\]The second term on the right-hand side converges to 0 as \(n \to \infty\) by the Lindeberg condition. Hence, for any \(\epsilon > 0\), we have
\[\limsup_{n \to \infty} \sum_{j=1}^n A_{n,j} \leq \frac{1}{3!} \lvert t \rvert^3 \epsilon.\]Since \(\epsilon > 0\) is arbitrary, we conclude that \(\limsup_{n \to \infty} \sum_{j=1}^n A_{n,j} = 0\), which implies that \(\sum_{j=1}^n A_{n,j} \to 0\) as \(n \to \infty\).
Next, we analyze the term \(\sum_{j=1}^n B_{n,j}\). By the Taylor expansion of the exponential function, we have
\[\lvert e^{-u} - (1-u) \rvert \leq \frac{u^2}{2} \quad \text{for all } u \geq 0.\]Therefore, we have
\[\begin{align*} \sum_{j=1}^{n} B_{n,j} & = \sum_{j=1}^{n} \lvert \left(1 - \frac{t^2 \sigma_{n,j}^2}{2}\right) - e^{-\frac{t^2 \sigma_{n,j}^2}{2}} \rvert \\ & \leq \sum_{j=1}^{n} \frac{1}{8} t^4 \sigma_{n,j}^4 \\ & \leq \frac{1}{8} t^4 \max_{1 \leq j \leq n} \sigma_{n,j}^2 \cdot \sum_{j=1}^{n} \sigma_{n,j}^2 \\ & = \frac{1}{8} t^4 \max_{1 \leq j \leq n} \sigma_{n,j}^2. \end{align*}\]According to Lemma 3, we have \(\max_{1 \leq j \leq n} \sigma_{n,j}^2 \to 0\) as \(n \to \infty\). Hence, we conclude that \(\sum_{j=1}^n B_{n,j} \to 0\) as \(n \to \infty\).
Combining the results for both terms, we have
\[\lvert \varphi_{S_n}(t) - e^{-\frac{t^2}{2}} \rvert \leq \sum_{j=1}^n A_{n,j} + \sum_{j=1}^n B_{n,j} \to 0 \quad \text{as } n \to \infty.\]Therefore, by Lévy’s Continuity Theorem, we conclude that \(S_n \xrightarrow{d} \mathcal{N}(0, 1)\) as \(n \to \infty\). This completes the proof of the Lindeberg-Feller CLT.
Acknowledgement. The proof of the Lindeberg-Feller CLT presented here is adapted from the proof in Professor Todd Kemp’s Youtube video “Lindberg-Feller CLT” https://youtu.be/CwqQLYr0nIE?si=VNmgibC8sAA7l7ks.