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Suppose that\({{\bf{X}}_{\bf{1}}}\)and\({{\bf{X}}_{\bf{2}}}\)are i.i.d. random variables, and that each has the uniform distribution on the interval[0,1]. Evaluate\({\bf{P}}\left( {{{\bf{X}}_{\bf{1}}}^{\bf{2}}{\bf{ + }}{{\bf{X}}_{\bf{2}}}^{\bf{2}} \le {\bf{1}}} \right)\)

Short Answer

Expert verified

\(P\left( {{X_1}^2 + {X_2}^2 \le 1} \right) = \frac{\pi }{4}\)

Step by step solution

01

Given information

\({X_1}\)and\({X_2}\)are i.i.d. random variables. Both the variables have a uniform distribution on the interval [0,1].

02

Calculate the probability

\({X_1}\)and\({X_2}\)have the uniform distribution over a square, which has area 1.

The area of the quarter circle is\(\frac{\pi }{4}\)

Hence,\(P\left( {{X_1}^2 + {X_2}^2 \le 1} \right) = \frac{\pi }{4}\).

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Most popular questions from this chapter

Prove Theorem 3.8.2. (Hint: Either apply Theorem3.8.4 or first compute the cdf. separately for a > 0 and a < 0.)

Suppose that thenrandom variablesX1. . . , Xnform arandom sample from a discrete distribution for which thep.f. is f. Determine the value of Pr(X1 = X2 = . . .= Xn).

Question:Suppose that the joint p.d.f. ofXandYis as follows:

\(\)\({\bf{f}}\left( {{\bf{x,y}}} \right){\bf{ = }}\left\{ \begin{array}{l}\frac{{{\bf{15}}}}{{\bf{4}}}{{\bf{x}}^{\bf{2}}}\;{\bf{for}}\;{\bf{0}} \le {\bf{y}} \le {\bf{1 - }}{{\bf{x}}^{\bf{2}}}\\{\bf{0}}\;{\bf{otherwise}}\end{array} \right.\)

a. Determine the marginal p.d.f.’s ofXandY.

b. AreXandYindependent?

Let the initial probability vector in Example 3.10.6 be\(v = \left( {\frac{1}{{16}},\frac{1}{4},\frac{1}{8},\frac{1}{4},\frac{1}{4},\frac{1}{{16}}} \right)\)Find the probabilities of the six states after one generation.

The definition of the conditional p.d.f. of X given\({\bf{Y = y}}\)is arbitrary if\({{\bf{f}}_{\bf{2}}}\left( {\bf{y}} \right){\bf{ = 0}}\). The reason that this causes no serious problem is that it is highly unlikely that we will observe Y close to a value\({{\bf{y}}_{\bf{0}}}\)such that\({{\bf{f}}_{\bf{2}}}\left( {{{\bf{y}}_{\bf{0}}}} \right){\bf{ = 0}}\). To be more precise, let\({{\bf{f}}_{\bf{2}}}\left( {{{\bf{y}}_{\bf{0}}}} \right){\bf{ = 0}}\), and let\({{\bf{A}}_{\bf{0}}}{\bf{ = }}\left( {{{\bf{y}}_{\bf{0}}}{\bf{ - }} \in {\bf{,}}{{\bf{y}}_{\bf{0}}}{\bf{ + }} \in } \right)\). Also, let\({{\bf{y}}_{\bf{1}}}\)be such that\({{\bf{f}}_{\bf{2}}}\left( {{{\bf{y}}_{\bf{1}}}} \right){\bf{ > 0}}\), and let\({{\bf{A}}_{\bf{1}}}{\bf{ = }}\left( {{{\bf{y}}_{\bf{1}}}{\bf{ - }} \in {\bf{,}}{{\bf{y}}_{\bf{1}}}{\bf{ + }} \in } \right)\). Assume that\({{\bf{f}}_{\bf{2}}}\)is continuous at both\({{\bf{y}}_{\bf{0}}}\)and\({{\bf{y}}_{\bf{1}}}\).

Show that

\(\mathop {{\bf{lim}}}\limits_{ \in \to {\bf{0}}} \,\frac{{{\bf{Pr}}\left( {{\bf{Y}} \in {{\bf{A}}_{\bf{0}}}} \right)}}{{{\bf{Pr}}\left( {{\bf{Y}} \in {{\bf{A}}_{\bf{1}}}} \right)}}{\bf{ = 0}}{\bf{.}}\)

That is, the probability that Y is close to\({{\bf{y}}_{\bf{0}}}\)is much smaller than the probability that Y is close to\({{\bf{y}}_{\bf{1}}}\).

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