Suppose that $P(X_{1} = x) = P(X_{2} = x) = \frac{1}{a} e^{-\frac{x}{a}}$ and $Y = X_{1} + X_{2}$. I am trying to find $P(Y = y)$. What I've done is \begin{align} P(Y = y) &= \int P(X_{1}=x_{1})P(X_{2}=x_{2})P(Y = y| X_{1} = x_{1}, X_{2} = x_{2}) dx_{1}dx_{2} \\ &= \int \frac{1}{a} e^{-\frac{x_{1}}{a}} \frac{1}{a} e^{-\frac{x_{2}}{a}} \delta(y - x_{1} - x_{2})dx_{1}dx_{2} \\ &= \int \frac{1}{a} e^{-\frac{x_{1}}{a}} \frac{1}{a} e^{-\frac{y-x_{1}}{a}}dx_{1} \\ &= \frac{1}{a^{2}} \int e^{-\frac{y}{a}}dx_{1} \\ &=\frac{1}{a^{2}} e^{-\frac{y}{a}} \int dx_{1} \end{align} Which goes to infinity and even if I renormalize it I get $\frac{1}{a}e^{-\frac{y}{a}}$ which I know is not the correct answer. What exactly am I doing wrong?
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Your notations do no make sense. There is no random varaible $X$ such that $P(X=x)=\frac 1 a e^{-x/a}$ for all $x >0$. – Kavi Rama Murthy Jun 27 '20 at 23:29
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$P(X=x)$ implies discrete distribution, but you seem to have a continuous distribution – Alex Jun 27 '20 at 23:50
2 Answers
The choice of notation and evaluation strategy is so strange that it makes me question where you learned these concepts.
It is far easier to observe $$\begin{align} \Pr[Y \le y] &= \Pr[X_1 + X_2 \le y] \\ &= \int_{x_2 = 0}^\infty \Pr[X_1 + x_2 \le y] f_{X_2}(x_2) \, dx_2 \\ &= \int_{x_2 = 0}^y \Pr[X_1 \le y - x_2] \frac{1}{a} e^{-x_2/a} \, dx_2 \\ &= \int_{x_2 = 0}^y F_{X_1}(y - x_2) \frac{1}{a} e^{-x_2/a} \, dx_2 \\ &= \frac{1}{a} \int_{x_2 = 0}^y (1 - e^{-(y - x_2)/a}) e^{-x_2/a} \, dx_2 \\ &= \frac{1}{a} \int_{x_2 = 0}^y e^{-x_2/a} - e^{-y/a} \, dx_2 \\ &= \frac{1}{a} \left[-a e^{-x_2/a} - x_2 e^{-y/a} \right]_{x_2 = 0}^y \\ &= \frac{1}{a} \left( -a e^{-y/a} - y e^{-y/a} + a + 0 \right) \\ &= -e^{-y/a} - \frac{y}{a} e^{-y/a} + 1 \\ &= 1 - \left( 1 + \frac{y}{a} \right) e^{-y/a}. \end{align}$$
Differentiation with respect to $y$ yields the density $$f_Y(y) = \frac{y}{a^2} e^{-y/a}, \quad y > 0.$$
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You need to remember the support of the exponential distribution, as $p_X(x<0)=0$. The last integral is not infinite, which you can see rewriting your expressions, $$\begin{align} p_Y(y)&=\int\limits_{-\infty}^{\infty}\mathrm{d}x_1\int\limits_{-\infty}^{\infty}\mathrm{d}x_2p_{X_1}(x_1)p_{X_2}(x_2)p_Y(y|x_1,x_2)\\ &=\int\limits_{0}^{\infty}\mathrm{d}x_1\int\limits_{0}^{\infty}\mathrm{d}x_2\frac{1}{a}e^{-\frac{x_1}{a}}\frac{1}{a}e^{-\frac{x_2}{a}}\delta(y-x_1-x_2)\\ &=\frac{1}{a^2}\int\limits_{0}^{y}\mathrm{d}x_1e^{-\frac{x_1}{a}}e^{-\frac{y-x_1}{a}}\\ &=\frac{1}{a^2}e^{-\frac{y}{a}}\int\limits_{0}^{y}\mathrm{d}x_1\\ &=\frac{y}{a^2}e^{-\frac{y}{a}}, \end{align} $$ agreeing with @heropup's answer.
