Question

One of the following two functions is the p.d.f. of a continuous random variable X. For the one which is not, give a reason why. For the one which is, compute the expected value p = E(X) (as exact fraction), and compute P(X < } rounded to nearest percent. 2 - 2 f(x) = { if x € [0,1] (2x-1 if x € [0,1] 0 g(x) = else else English (Canada) Reflect in ePortfolio Download Open with docReader OlFocus A Alternative formats

Answer 1

For a continuous random variable X, f(X) is a pdf if--

$(i)f(x)\geq 0 \:\forall\: x\:\epsilon \:\mathbf{x};(ii)\int_{\mathbf{x}}f(x).dx=1$

So, as for the first function f(x)

$f(x)=2-2x\geq 0 \:\forall\: x\:\epsilon [0,1]$

and

$\int_{\mathbf{x}}f(x).dx=\int_{0}^{1}(2-2x).dx=[2x-x^2]^1_0=2-1-0=1$

For the second function g(x)

$g(x)=2x-1\geq 0\:\forall \:x\:\epsilon\:[0,1]$

and

$\int_{\mathbf{x}}g(x).dx=\int_{0}^{1}(2x-1).dx=[x^2-x]^1_0=1-1-0=0$

So, for the first function f(x) --

$\mu =E(X)=\int_{0}^{1}x*f(x).dx=\int_{0}^{1}x*(2-2x).dx$$=[x^2-\frac{2}{3}x^3]^1_0=1-\frac{2}{3}-0=\frac{1}{3}$

$P(X\leq \frac{1}{3})=\int_{0}^{\frac{1}{3}}f(x).dx$

$=\int_{0}^{\frac{1}{3}}(2-2x).dx=[2x-x^2]_0^{\frac{1}{3}}=\frac{2}{3}-\frac{1}{9}-0=\frac{5}{9}\approx 0.556$ i.e. 56% approximately