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I have been given a joint probability distribution function and asked to find the marginal probability distributions. $$f(x,y)=\begin{cases} λ^2e^{-λy}&0\le x\le y\\ 0&\text{otherwise} \end{cases}$$ how do i compute the marginal distribution of x if I am not given the Y.

I know that to compute the marginal distribution of x i have to integrate over y. But how i do it if it is not given to me?

Parcly Taxel
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Aayush Aayron Deo
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  • The joint describes both marginals as you can compute the integral over the variable you are not interested in and thus arrive at the marginal for the remaining variable $$ f_X(x) = \int_0^{\infty} f(x, y) dy $$ – Mikal Apr 24 '20 at 08:07

2 Answers2

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When we want a marginal distribution over some variable, we have to integrate over it to remove it. Thus we have the formula $$f_X(x)=\int f_{X,Y}(x,y)\,dy$$ In this case, the marginal distribution of $X$ is $$f_X(x)=\int_x^\infty\lambda^2e^{-\lambda y}\,dy=-\lambda[e^{-\lambda y}]_x^\infty=\lambda e^{-\lambda x}$$ (We can start from $x$ because $f(x,y)$ for $x>y$ is zero.) Similarly, the marginal distribution of $Y$ is $$f_Y(y)=\int_0^y\lambda^2e^{-\lambda y}\,dx=\lambda^2ye^{-\lambda y}$$

Parcly Taxel
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First, it is always a good idea to check that the density integrates to 1: \begin{eqnarray*} \iint_{\mathbb{R}^2} f_{XY}(x,y)dxdy &=& \int_{y=0}^{\infty}\int_{x=0}^{\infty}{\bf 1}_{\{x\leq y\}}(x,y) \lambda^2e^{-\lambda y}dxdy\\ &=& \int_{y=0}^{\infty}\int_{x=0}^y \lambda^2e^{-\lambda y}dxdy\\ &=& \int_{y=0}^{\infty}y\lambda^2e^{-\lambda y}dy\\ &=& \lambda^2y\frac{e^{-\lambda y}}{-\lambda}|_{y=0}^{\infty}-\lambda^2\int_{y=0}^{\infty}\frac{e^{-\lambda y}}{-\lambda}dy\\ &=&0-\lambda^2\frac{e^{-\lambda y}}{\lambda^2}|_{y=0}^{\infty}\\ &=&1 \end{eqnarray*} Now, the marginal density $f_X(x)$ is given by \begin{eqnarray*} f_X(x) &=& \int_{y=0}^{\infty}{\bf 1}_{\{x\leq y\}}(x,y)\lambda^2 e^{-\lambda y}dy\\ &=& \lambda^2\int_{y=x}^{\infty}e^{-\lambda y}dy\\ &=& \lambda^2\frac{e^{-\lambda y}}{-\lambda}|_{y=x}^{\infty}\\ &=& \lambda e^{-\lambda x} \end{eqnarray*} You can verify this integrates to $1$ over its range $[0,\infty)$. And, for fun and excitement, we compute the marginal $f_Y(y)$ for $Y$: \begin{eqnarray*} f_Y(y) &=& \int_{x=0}^{\infty}{\bf 1}_{\{x\leq y\}}(x,y)\lambda^2 e^{-\lambda y}dx\\ &=& \int_{x=0}^y\lambda^2 e^{-\lambda y}dx\\ &=& y\lambda^2 e^{-\lambda y} \end{eqnarray*} which also integrates to $1$ over $[0,\infty)$. ${\bf 1}_{\{x\leq y\}}(x,y)$ is the indicator function which is defined to be $1$ when $x\leq y$ and $0$ otherwise.

SpiritLevel
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