By the argument in the last video, if the X_ i are i.i.d. with mean \mu and variance \sigma ^2, and if M_ n=(X_1+\cdots +X_ n)/n, then we have an inequality of the form
\mathbf{P}\big ( |M_ n-\mu | \geq \epsilon \big ) \leq \frac{a\sigma ^2}{n},
for a suitable value of a.
a) If \epsilon =0.1, then the value of a is:
To determine the value of a, we need to refer to the argument in the last video. Since we do not have access to the video, we are unable to provide the specific value of a for the given inequality.
However, in general, the value of a depends on the specific bound on the tail probability and the properties of the random variables. It is often determined using techniques such as Markov's inequality, Chebyshev's inequality, or similar methods.
If you have access to the last video or any specific information about the properties of the random variables, we can try to help you further.
To find the value of "a" when ε = 0.1, we need to use Chebyshev's inequality. Chebyshev's inequality states that for any random variable with mean μ and variance σ², the probability that the random variable deviates from its mean by more than a certain number of standard deviations is at most 1/a², where "a" is a positive constant.
In our case, M_n is the sample mean of n i.i.d. random variables X_i with mean μ and variance σ². We want to find the value of a such that the probability that |M_n - μ| exceeds ε is at most aσ²/n.
Using Chebyshev's inequality, we have:
P(|M_n - μ| ≥ ε) ≤ σ² / (a²n)
We are given ε = 0.1, so:
P(|M_n - μ| ≥ 0.1) ≤ σ² / (a²n)
To match the given inequality, we need to have:
σ² / (a²n) = aσ² / n
Canceling out σ² and n, we have:
1 / a² = a
Taking the square root of both sides, we get:
1 / a = √a
Multiplying both sides by a, we have:
1 = a√a
Squaring both sides, we obtain:
1 = a³
Therefore, the value of a when ε = 0.1 is the cube root of 1, which is 1.
So, a = 1 when ε = 0.1.