Let $X_1 \\sim \\exp(\\lambda)$ and also $X_2 \\sim \\exp(\\lambda)$ be two independent exponentially distributed random variables. Discover the mean and also variance of random variable $Y=X_1 + X_2$.

$x=x_1 + x_2$

$$f_x(X)= \\int_-\\infty^+\\infty f_x_1(x_1) - f_x_2(x-x_1)dx = \\dots$$

$$\\dfracλ^2Γ(2) x^2-1e^-λx , x>0$$

$E(x)=\\dfrac2λ$ $V(x)=\\dfrac2λ^2$

I to be trying to find if this can be resolved easier...

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Parhs ns think the what Henry is do the efforts to acquire at is the your difficulty has $\\lambda_1$ and $\\lambda_2$ yet no $\\lambda$. Her solution has actually $\\lambda$, however no $\\lambda_1$ or $\\lambda_2$. Over there is no connection in between your problem and also your solution, i m sorry is an extremely confusing. You have actually not expressed yourself an extremely clearly. $\\endgroup$
If $X$ and also $Y$ room any random variables, and $a$ and also $b$ space constants, then$$E(aX+bY)=aE(X)+bE(Y).$$

If $X$ and also $Y$ room independent arbitrarily variables, and also $a$ and $b$ space constants, then$$\\textVar(aX+bY)=a^2\\textVar(X)+b^2\\textVar(Y).$$

In our case we have actually $a=b=1$, but the general expressions may be valuable to friend later.

So you don\"t need to uncover the distribution of the random variable $X_1+X_2$, every you require is formulas for the mean and also variance of one exponential arbitrarily variable $T$ through parameter $\\lambda$. These are respectively $\\dfrac1\\lambda$ and $\\dfrac1\\lambda^2$. Thus your mean and also variance are respectively$$\\frac1\\lambda_1+\\frac1\\lambda_2\\quad\\textand\\quad \\frac1\\lambda_1^2+\\frac1\\lambda_2^2.$$

Edit: The inquiry was changed. It turns out the $\\lambda_1=\\lambda_2=\\lambda$. The is simply a special instance of the above formulas.

It is not clear whether you were questioning also about how come compute the individual means and variances, or whether you currently know these.We need $E(T)$, and $E(T^2)$, since $\\textVar(T)=E(T^2)-(E(T))^2$.

The expectations have the right to be discovered as usual by integration. There are some shortcuts, but they have tendency to involve more advanced notions. In instance you space interested, let me cite the term moment generating function.

Added: we compute the minute generating duty $m(s)$ the our arbitrarily variable $T$ (sorry around $s$, the traditional $t$ is currently taken). For this reason we want $$E(e^Ts)=\\int_0^\\infty e^ts\\lambda e^-\\lambda t\\,dt=\\int_0^\\infty \\lambda e^-(\\lambda-s)t\\,dt.$$ integrate by substitution, easy. We gain $$m(s)=\\frac\\lambda\\lambda-s=\\frac11-\\fracs\\lambda.$$So $m(s)$ has actually a very nice Taylor development $m(s)=1+\\frac1\\lambdas+\\frac1\\lambda^2s^2+\\cdots$. Native this we deserve to pick increase $E(X^k)$ for any kind of $k$. In particular, $E(X)=\\frac1\\lambda$, and $E(X^2)=\\frac2!\\lambda^2$.

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We can also get the moment-generating function of $X_1+X_2$, since the mgf of an independent sum is the product that the mgf\"s..