- #1
thisguy12
- 3
- 0
If you were to pick two random numbers on the interval [0,1], what is the probability that the sum of their squares is less than 1? That is, if you let [itex]Y_1[/itex] ~ [itex]U(0,1)[/itex] and [itex]Y_2[/itex] ~ [itex]U(0,1)[/itex], find [itex]P(Y_1^2 + Y^2_2 \leq 1)[/itex]. There is also a hint: the substitution [itex]u = 1 - y_1[/itex] may be helpful - look for a beta distribution.
Here's what I've done so far:
I know that the density function for [itex]Y_1[/itex] and [itex]Y_2[/itex] is the same, [itex]f(y_1) = f (y_2) = 1[/itex] on the interval [0,1].
[itex]P(Y_1^2 + Y_2^2 \leq 1) = P(Y_2^2 \leq 1 - Y_1^2) = P(-\sqrt{1 - Y_1^2} \leq Y_2 \leq \sqrt{1 - Y_1^2}) = \int^{\sqrt{1 - Y_1^2}}_{-\sqrt{1 - Y_1^2}} dy_2 = 2\sqrt{1 - Y_1^2}[/itex]
And that's where I get stuck. I thought that maybe be a beta distrbution with [itex]\alpha = 3/2[/itex], [itex]\beta = 1[/itex], but the beta function, [itex]\beta(3/2, 1) = 2/3 \neq 1/2[/itex].
Here's what I've done so far:
I know that the density function for [itex]Y_1[/itex] and [itex]Y_2[/itex] is the same, [itex]f(y_1) = f (y_2) = 1[/itex] on the interval [0,1].
[itex]P(Y_1^2 + Y_2^2 \leq 1) = P(Y_2^2 \leq 1 - Y_1^2) = P(-\sqrt{1 - Y_1^2} \leq Y_2 \leq \sqrt{1 - Y_1^2}) = \int^{\sqrt{1 - Y_1^2}}_{-\sqrt{1 - Y_1^2}} dy_2 = 2\sqrt{1 - Y_1^2}[/itex]
And that's where I get stuck. I thought that maybe be a beta distrbution with [itex]\alpha = 3/2[/itex], [itex]\beta = 1[/itex], but the beta function, [itex]\beta(3/2, 1) = 2/3 \neq 1/2[/itex].