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I have following function $$f(x_0,x_1)=-x_1Ei(-a-\frac{bx_1}{x_0})\exp(a+\frac{bx_1}{x_0})$$. Where $a,b$ are greater than zero, $x_0,x_1$ can have positive values and $Ei(x)$ is the exponential integral. Now its perspective function is $$f_1(x_0)=-Ei(-a-\frac{b}{x_0})\exp(a+\frac{b}{x_0})$$ If I do following transformation $\frac{1}{t}=a+\frac{b}{x_0}$ then I have $$f_2(t)=-Ei(-\frac{1}{t})\exp(\frac{1}{t})$$ Now $f_2(t)$ is concave with respect to $t$. Based on this I assume that $f_1(x_0)$ is concave with respect to $x_0$. Since $f_1(x_0)$ is the perspective function of $f(x_0,x_1)$ therefore I conclude that $f(x_0,x_1)$ is jointly concave with respect to $(x_0,x_1)$. But when I fix $x_0$ and try to prove the concavity of $f(x_0,x_1)$ with respect to $x_1$ through double derivative test. Then I am unable to prove this. Is the function $f(x_0,x_1)$ not jointly concave because if it were then it should be concave with respect to $x_1$ also. Any help in this regard will be much appreciated. Thanks in advance.

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\begin{align}f_1(x_0)&=-Ei(-a-\frac{b}{x_0})\exp(a+\frac{b}{x_0})\\ &=-Ei\left(\color{red}- \frac1t\right) \exp\left(\frac1t\right)\end{align}

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  • $\begingroup$ I am extremely sorry. It was my typing mistake (I have corrected it, Do you agree that the transformation used in my post is right). It is actually, as you mentioned, a negative sign. So $-Ei(-\frac{1}{t})\exp(\frac{1}{t})$ is a concave function. Since $f_1(x_0)=f_2(t)$ so $f_1(x_0)$ is also concave and therefore $f(x_0,x_1)$ should be concave with respect to $x_1$ also. But I am unable to prove it through double derivative test. $\endgroup$ Commented Feb 5, 2018 at 1:50
  • $\begingroup$ I am curious about my transformation $\frac{1}{t}=a+\frac{b}{x_0}$. Do you think it is right transformation? $\endgroup$ Commented Feb 5, 2018 at 1:51
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    $\begingroup$ I haven't verified, but do you have a proof that if $g(t)$ is concave in $t$, then by a transformation of $\frac1t=a+\frac{b}{x_0}$, the function is concave in $x_0$? $\endgroup$ Commented Feb 5, 2018 at 1:58
  • $\begingroup$ No I do not have a proof. That's why I asked this question. But I describe my understanding below. With $\frac{1}{t}=a+\frac{b}{x_0}$ we have $t=\frac{x_0}{ax_0+b}$ therefore for the range of $x_0$ (from $0\to \infty$) the range of $t$ is $0 \to \frac{1}{a}$ and each individual value of $x_0$ produces a unique value of $t$. Since the function $f_2(t)$ is concave overall values of $t$ (which I can prove) therefore I think it is also concave over $0<t<\frac{1}{a}$ (Please correct me if I am wrong in this). Thanks in advance. $\endgroup$ Commented Feb 5, 2018 at 2:09
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    $\begingroup$ Suppose $f_2(t)=-t$, it is a concave function, but $h(x) = f_2 \left( \frac{x}{x+1} \right) = -\frac{x}{x+1}$ is not a concave function in $x$. $\endgroup$ Commented Feb 5, 2018 at 2:17

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