A company buys 100 lightbulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the first of these bulbs to burn out? (See Exercise 10.)
Exercise 10 told us that the minimum value of independent random variables is exponential with the mean \(u/n\). Here, we have 100 lightbulbs with a mean lifetime of 1000 hours.
We have: \(u = 1000\) \(m = 1/u = 1/1000\)
\(n*m = 100(1/1000) = 1/10\)
Expected minimum: \(1/(1/10) = 10\)
The expected minimum, or expected time for the first light bulb to go out, is 10 hours.
Assume that X1 and X2 are independent random variables, each having an exponential density with parameter \(λ\). Show that \(Z = X1 − X2\) has density
\(f_Z(z) = (1/2)λe^{−λ|z|}\)
Let’s call W as \(W = X_1 + X_2\)
To make following along with the MIT video simpler, we’re gonna change X1 to X and X2 to Y temporarily.
Recall the convolution function:
\[f_w(W):=\int_{-\infty}^{\infty} f_x(X) f_y(W-X) dX\]
Rewrite Z as \(Z = X + (-Y)\), which leads to:
\[f_z(Z):=\int_{-\infty}^{\infty} f_x(X) f_{-y}(Z-X) dX\]
\(f_{-y}(z-x) = f_y(x-z)\)
So:
\[f_z(Z):=\int_{-\infty}^{\infty} f_x(X) f_y(X-Z) dX\]
\(f_x(X) = 0, x < 0\)
\(f_x(X) = λe^{-λx}, x \geq 0\)
\[f_z(Z):=\int_0^{\infty} λe^{-λx} λe^{-λ(x-z)} dx\]
\[= λe^{λz} \int_0^{\infty} λe^{-2λx} dx\]
\[= λe^{λz} (-1/2e^{-2xz}\vert_{0}^{\infty}) \]
\[ λ/2 * λe^{λz} \]
\(Z = X - Y\) \(-Z = Y - X\)
Z and -Z have same distribution - must be symmetric around 0
\[f_z(Z):= f_z(-Z)\]
\[f_z(Z):= /2 * λe^{λz}, z<0\]
\[f_z(Z):= λ/2 * λe^{λz}, z \geq 0\] Thus: \[f_z(Z):= λ/2 * λe^{λ|z|}\]
Let X be a continuous random variable with mean \(μ = 10\) and variance \(σ2 = 100/3\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities:
Chebyshev’s Inequality.
\(\Pr\left[|X-\mu|\geq k\sigma \right] \leq \frac{1}{k^2}\)
Used to determine fraction of values within certain distance fo the mean for a wider variety of distributions than the rules of thumb used for normal distributions.
u = 10
variance = 100/3
sd = sqrt(variance)Applied: \(P(|X − 10| ≥ Y) \leq \frac{1}{k^2}\)
So: \(k = Y/\sqrt{100/3}\) \(P = 1/k^2\)
k = 2/sd
1/k^2## [1] 8.333333Or 1.
k = 5/sd
1/k^2## [1] 1.333333Or 1.
k = 9/sd
1/k^2## [1] 0.4115226k = 20/sd
1/k^2## [1] 0.08333333