MathNugget
Junior Member
- Joined
- Feb 1, 2024
- Messages
- 195
Hello.
This is [imath]f(x, y) = y x^{y-1}e^{-y}\mathbb{1}_{(0, \infty)}(y)\mathbb{1}_{(0, 1)}(x)[/imath]
Trying to find the marginal probability density of Y.
so I am supposed to solve:
[math]f_Y(x, y)= \int_{-\infty}^{\infty}y x^{y-1}e^{-y}\mathbb{1}_{(0, \infty)}(y)\mathbb{1}_{(0, 1)}(x)dx = ye^{-y}\mathbb{1}_{(0, \infty)}(y) \int_{-\infty}^{\infty}x^{y-1} \mathbb{1}_{(0, 1)}(x) dx = ye^{-y}\mathbb{1}_{(0, \infty)}(y) \int_{0}^{1}x^{y-1} dx = ...[/math]now I suppose [imath]\mathbb{1}_{(0, \infty)}(y)[/imath] tells me y > 0, so
[math]\int x^{y-1} dx = \frac{x^y}{y}[/math][math]\int_{0}^{1}x^{y-1} dx = \frac{1}{y}?[/math]
Is it [math]f_Y(x, y)=ye^{-y}\mathbb{1}_{(0, \infty)}(y) \frac{1}{y}= e^{-y}\mathbb{1}_{(0, \infty)}(y)?[/math]
This is [imath]f(x, y) = y x^{y-1}e^{-y}\mathbb{1}_{(0, \infty)}(y)\mathbb{1}_{(0, 1)}(x)[/imath]
Trying to find the marginal probability density of Y.
so I am supposed to solve:
[math]f_Y(x, y)= \int_{-\infty}^{\infty}y x^{y-1}e^{-y}\mathbb{1}_{(0, \infty)}(y)\mathbb{1}_{(0, 1)}(x)dx = ye^{-y}\mathbb{1}_{(0, \infty)}(y) \int_{-\infty}^{\infty}x^{y-1} \mathbb{1}_{(0, 1)}(x) dx = ye^{-y}\mathbb{1}_{(0, \infty)}(y) \int_{0}^{1}x^{y-1} dx = ...[/math]now I suppose [imath]\mathbb{1}_{(0, \infty)}(y)[/imath] tells me y > 0, so
[math]\int x^{y-1} dx = \frac{x^y}{y}[/math][math]\int_{0}^{1}x^{y-1} dx = \frac{1}{y}?[/math]
Is it [math]f_Y(x, y)=ye^{-y}\mathbb{1}_{(0, \infty)}(y) \frac{1}{y}= e^{-y}\mathbb{1}_{(0, \infty)}(y)?[/math]
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