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Chapter 1: Problem 21

If the pdf of \(X\) is \(f(x)=2 x e^{-x^{2}}, 0

Short Answer

Expert verified
The pdf of \( Y \) is \( g(y) = e^{-y} \) for \( y > 0 \), and 0 for \( y \leq 0 \).

Step by step solution

01

Compute the Cumulative Distribution Function (CDF) of \( X \)

The CDF of \( X \) can be calculated by integrating the pdf of \( X \) from \( -\infty \) to \( x \). Here, the pdf \( f(x) \) is only defined for \( x > 0 \), so the CDF \( F(x) \) will be: \[ F(x) = \int_0^x f(t) dt = \int_0^x 2 t e^{-t^2} dt \] Using a subsitution of \( u = t^2 \), \( du = 2t dt \), we find: \[ F(x) = \int_0^{x^2} e^{-u} du = 1 - e^{-x^2} \] for \( x > 0 \), and 0 for \( x \leq 0 \).
02

Find the Cumulative Distribution Function (CDF) of \( Y \)

Now, using the transformation \( Y = X^2 \), we can find the CDF of \( Y \) by substituting \( Y = X^2 \) into \( F(x) \). This gives: \[ G(y) = P(Y \leq y) = P(X^2 \leq y) = P(-\sqrt{y} \leq X \leq \sqrt{y}) \] Since the pdf is only defined for \( x > 0 \), this can be written as: \[ G(y) = P(0 \leq X \leq \sqrt{y}) = F(\sqrt{y}) - F(0) = 1 - e^{-y} - 0 = 1 - e^{-y} \] for \( y > 0 \), and 0 for \( y \leq 0 \).
03

Compute the Probability Density Function (pdf) of \( Y \)

Finally, the pdf of \( Y \) can be determined by differentiating the CDF of \( Y \). This gives: \[ g(y) = \frac{d}{dy} G(y) = e^{-y} \] for \( y > 0 \), and 0 for \( y \leq 0 \).

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Most popular questions from this chapter

A hand of 13 cards is to be dealt at random and without replacement from an ordinary deck of playing cards. Find the conditional probability that there are at least three kings in the hand given that the hand contains at least two kings.

Consider \(k\) continuous-type distributions with the following characteristics: pdf \(f_{i}(x)\), mean \(\mu_{i}\), and variance \(\sigma_{i}^{2}, i=1,2, \ldots, k .\) If \(c_{i} \geq 0, i=1,2, \ldots, k\), and \(c_{1}+c_{2}+\cdots+c_{k}=1\), show that the mean and the variance of the distribution having pdf \(c_{1} f_{1}(x)+\cdots+c_{k} f_{k}(x)\) are \(\mu=\sum_{i=1}^{k} c_{i} \mu_{i}\) and \(\sigma^{2}=\sum_{i=1}^{k} c_{i}\left[\sigma_{i}^{2}+\left(\mu_{i}-\mu\right)^{2}\right]\) respectively.

Cast a die a number of independent times until a six appears on the up side of the die. (a) Find the pmf \(p(x)\) of \(X\), the number of casts needed to obtain that first six. (b) Show that \(\sum_{x=1}^{\infty} p(x)=1\). (c) Determine \(P(X=1,3,5,7, \ldots)\). (d) Find the cdf \(F(x)=P(X \leq x)\).

Let \(X\) have the pdf \(f(x)=(x+2) / 18,-2

Let \(X\) have the pdf \(f(x)=2 x, 0
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